Structural Dynamics by Finite Elements
William Weaver, Jr.
Stanford University Paul R. Johnston
Failure Analysis Associates
l'RI N ncr 11/\11 , INC., I nglc'wood Cliffs, New Jersey 07632
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Preface
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IX
Introduction to Structural Dynamics 1.1 Structural Dynamics Concepts 1. 2 Dynamic Influences 4 · I . 3 Discretization by Finite Elements 1.4 Computer Programs 9 References 9
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Systems with One Degree of Freedom 2.1 2.2 2.3 2.4 2.5 2 ·6 2·7 2.8
Introduction 10 Free, Undamped Vibrations 12 Harmonic Forcing Functions 19 Effects of Damping 25 Periodic Forcing Functions 35 Arbitrary Forcing Functions 38 Step-by-Step Response Calculations Response Spectra 51 Rl'fl'n.lllt'CN 59 Prohl1~n,11 59
10
45
Ill
Contents
Iv
Finite Elements and Vibrational Analysis
73
Contents
6
Normal-Mode Method of Dynamic Analysis 4.1 4 .2 4.3 4.4 4 .5 4 .6 4 .7 4. 8 4 ,l) 4. I()
7
138
Introduction 138 Principal and Normal Coordinates 139 Normal-Mode Response to Initial Conditions 147 Normal-Mode Response to Applied Actions 152 Normal-Mode Response to Support Motions 157 Damping in MDOF Systems 164 Damped Response to Periodic Forcing Functions 168 Damped Response to Arbitrary Forcing Functions 172 Step-by-Step Response Calculations 175 Program NOMO for Normal-Mode Response 177 References 192 Problems 192
Direct Numerical Integration Methods 'i. I
Introduction
195
5.2 Extrapolation with Explicit Formulas 5.J 5.4
5.5 5. Cl 5.7
197
203 Iteration with Implicit Formulas 211 Direct Linear Extrapolation 2 17 Ncwmark's Generalized Acceleration Method 223 Numericul Stahility und Accurucy 225 J>rogrulll DYNA for Dynamk Rl·sponSl' 7'<, Rl'lb l'lll'l'~ l I1 l '111llh•111~
Framed Structures 6. 1 6.2 6.3 6.4 6.5 6.6 6.7 6.8 6.9
73 3. 1 Introduction Stresses and Strains 75 3.2 Equations of Motion for Finite Elements 78 3.3 One-Dimensional Elements 82 3 .4 95 3.5 Transformation and Assemblage of Elements 105 3.6 Vibrational Analysis 112 3.7 Symmetric and Antisymmetric Modes 118 3.8 Program VIB for Vibrational Analysis References 124 Problems 125
4
V
195
Introduction 241 Plane Frames 244 Grids 249 Space Trusses 253 Space Frames 259 Programs for Framed Structures Guyan Reduction 282 Constraints Against Axial Strains Programs DYPFAC and DYSFAC References 303 Problems 303
264 290 299
Two- and Three-Dimensional Continua 310 7 .1 7. 2 7. 3 7.4 7. 5 7.6 7. 7 7 .8 7. 9 7. 10
8
241
Introduction 310 Stresses and Strains in Continua 310 N atura1 Coordinates 318 Numerical Integration 326 Isoparametric Quadrilaterals for Plane Stress and Plane Strain 333 Program DYNAPS for Plane Stress and Plane Strain 340 Isoparametric Hexahedra for General Solids 345 Program DYNASO for General Solids 351 Isoparametric Elements for Axisymmetric Solids 357 Program DYAXSO for Axisymmetric Solids 365 References 369
Plates and Shells 8.1 8.2 8.3 8.4 8.5 8.6
8.7
· 370
Introduction 370 Element for Plates in Bending 371 379 Program DYNAPB for Plates in Bending Element for General Shells 382 390 Program DYNASH for General Shells Element for Axisymmetric Shells 394 406 Program DY AXSH for Axisymmetric Shells 410 Rcf'L-n:nccs
Contents
vii
Contents
111
')
Rigid Bodies within Flexible Structures 9. 1 9. 2 9. 3 9.4 9.5 9.6
to
Introduction 411 Rigid Bodies in Framed Structures 413 Program DYRBPF for Rigid Bodies in Plane Frames Rigid Laminae in Multistory Buildings 425 Rigid Bodies in Finite-Element Networks 434 Program DYRBPB for Rigid Bodies in Plate-Bending Continua 438 References 443
Substructure Methods
444
IO .1 Introduction 444 10.2 Guyan Reduction Methods 445 10.3 Modified Tridiagonal Method for Multistory Buildings 457 10 .4 Programs DYMSPF and DYMSTB 463 10.5 Component-Mode Method 468 471 10.6 Component-Mode Method for Trusses 10.7 Programs COMOPT and COMOST 477 References 482
Notation
483
<~< 11wral References
Appendix A
Appendix 8
491
Systems of Units and Material 495 Properties A. l Systems of Units A. 2 Material Properties
495 497
Eigenvalues and Eigenvectors
498
n.1
n.2
Inverse llcmtion 498 Tra11sfo1111ntion Methods lh•fo1l'lll'l'H ~ 17
505
411
424
Appendix C
Flowchart for Program DYNAPT 519
Answers to Problems Index
579
564
124
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'I . Timoshenk o, S. P., Young, D. II. , nmI W(i:uvc1,. W., Jr ., Vibration Problems in • . . " I Wiley New York I n4. 4 th c, f•11g111('er111g, ·• ' ' E · 2nd ed , . W., Jr ·• Matrix Algebra for ngmeers, ., IO. Gere, J. M., and Wcaver,. Brooks/Cole, Monterey' Calif., 1983. . N Y k 1973 . Alge bra, 3rd ed . ' Macnullan, ew orD., V I. Hohn F E. Elem entary Matrix ' · P.' G., "Symmetry in Structural Meehamcs, · " ASCE ' J . Struct . 1v., o · ll.· Glockner, 99, No . ST! , 1973, pp. 71- 89.
11
Figure 3.18 (cont.)
the truss is aluminum , we give parameters the following numerical values: E = 6.9 L=5m
X
107 kPa
PROBLEMS
p = 2.62 Mg/ m 3
A = 6
X
10- 3 m 2
1.4-1.
. 1 I ment with a linearly distributed load (force per Figure P3.4-1 shows an axia e e - b + (b - b1)x/ L. Find the equivalent i unit length) given by the formula bx - .1 • nodal loads Pb(t) = {pb1, Pb2} due to this mfluence. f"' { f , {, 1= [ i-{
-f.
where SI units are implied (again, see Appendix A). To process this truss with Program VIBPT, we must analyze half the structure twice. In the first analysis, restraints on the plane of symmetry allow only symmetrk· modes of vibration; and the second analysis uses restraint data for only antisymmetrk modes. Figures 3.18(b) and (c) illustrate the mode shapes corresponding to the first and second angular frequencies w 1 = 79.55 s - 1 and wz = 168.9 s - 1. We see that the first mode is symmetric with respect to the plane of symmetry, while the second mode is antisymmetric.
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REFERENCES
Figure P3.4-1
1.4-2.
I. Weaver, W., Jr. , and Johnston, P. R., Finite Elements for Structural Analysis, Prentice-Hall, Englewood Cliffs, N.J., 1984.
~'l
o - I
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-T.. ),
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~
I
b, - L1}-z::]
A parabolically d!stributedd ~oa:. (fo;~\:zer D~~~~::r:~e ~~sui:~e:~:~~~ bx = bz(x/L)2, as illustrate m ig. ·. · loads Pb (t) = {pb1, p ,,2} resulting from this body force.
2. Zienkiewicz, 0. C. , The Finite Element Method, 4th ed., McGraw-Hill, Maidenhead, Berkshire, England, 1987. 3. Cook, R. D. , Concepts and Applications ofFinite Element Analysis, 2nd ed. , Wiley. New York, 1981. 4. T imoshenko, S. P., and Goodier, J. N. , Theory of Elasticity, 3rd ed. , McGraw-Hill. New York, 1970. 5. Weaver, W., Jr. , and Gere , J. M. , Matrix Analysis of Framed Structures, 2nd ed., Van Nostrand Re inhold, New York, 1980. 6. Oden, J. T. , Mechanics of Elastic Structures, McGraw-HiII, New York, 1967. 7. Archer, J. S. , "Consistent Matrix Formulations for Structural Analysis Using FiniteElement Techniques," AIAA J. , Vol. 3, No. 10, 1965, pp. 1910- 19 18 . 8 . Clough, R. W. , "Analysis of Structural Vibrations and Dynamic Response ," Rec. Adv. M at. Methods Strucr. Anal. Des., ed. R. II. Onll ughe r, Y. Yamada, and J. ·1~ Oden, University of Al abama Press, ll1111tsville, Alu ., I C J7 I , pp . 25 45.
I ..I
.t
Figure P3.4-2 . h h odes as shown in Fig. P3.4-3(a). In Assume
.;
Finite Elements and Vibrationa l Analysis
130
1,5-7.
y
Figure P3.5-7 shows a two-element beam for which the parameters E, I, A, and pare constant along the length. Assemble the stiffness matrix Ss and the consistent-mass matrix Ms (for translational inertias) in rearranged and partitioned forms.
,I
I
0.6L
131
Problems
Chap. 3
CD 0----------0 2
y
L
©
0 0
4
/ ,k
/
1·
·I
0.8L
CD
1
X
l
\.
~
0
3
~
L
L
Figure P3.5-4
3.5-5.
2
X
.,
Figure P3.5-7
For the plane truss shown in Fig. P3.5-5, repeat Prob. 3.5-1. Assume that the cross-sectional areas of members I and 2 are 0.8A, those for members 3 and 4 are 0.6A, and those for members S and 6 are equal to A.
I ~-It Repeat Prob. 3.5-7 for the two-element beam shown in Fig. P3.5-8. y
y
CD
1~
,I
,/ ~,
2
~
0
X
~3
L~
Figure P3.5-8
I ~ 11, For the two-element beam shown in Fig. P3.S-9, repeat Prob. 3.5-7. y
,---0.BL-- --.
i..l
Figure P3.5-5
3.5-6.
1 0
Figure P3.5-6
Repeat Prob. 3.5- 1 for the plane truss shown in Pig. P3.5-6. In this case let the cross-sectional urcus for llH.lmhcrs I ll11011gh 4 tw NJUIII tn /\ , while those fo1 nw111lm1s 'i 111HI <, 11n• l'q1111l to V? A,
/
1
G)
4ii
0
r,-3
.. - -L- -\- -L - - 1
-X
Finite Elements and Vibrational Analysis
134
Chap.3
3.6-7. Repeat Prob. 3.6-5 for the two-element continuous beam shown in Fig. P3.6-7. Cross-sectional properties I and A are constant, and the free displacements at joints 1 and 2 are both rotational. y
Chap. 3
.t7-3.
135
Problems
For the symmetric continuous beam shown in Fig. P3.7-3, find the angular frequencies and mode shapes for (a) symmetric and (b) antisymmetric distortions. Use only half the structure, assuming that each of the four elements has the same values of I and A. y
2___®=2---~ _3- - x
, ____ (D=,_ _ _..
,/
4
i-1,/- L - ~ . - - - -- L ~ Figure P3.6-7
3.7-1. The fixed-end beam shown in Fig. P3.7-l consists of two prismatic flexural elements having the same values of I and A. Using only half the beam, find the angular frequencies for the (a) symmetric and (b) antisymmetric modes. y
1.7-4. The symmetric plane truss shown in Fig. P3. 7-4 has cross-sectional areas for members 1 and 2 equal to V2 A , whereas those for members 3, 4 , and 5 are equal to A. Using only half the structure, calculate the angular frequencies and the (a) symmetric and (b) antisymmetric mode shapes.
I
2 1~
z
/ 1·
Figure P3.7-3
CD
ISym.
•
0
X
i-Sym.
I
e
e
3~
L= 2C
'l
Figure P3. 7-1 y
3.7-2. Figure P3.7-2 shows a prismatic continuous beam composed of four flexural elements. Determine the angular frequencies and mode shapes for (a) symmetric and (b) antisymmetric deformations, using only half the beam. y
L
-~-...::.:®~-6XX'.
0-2_0.::::::_4
01)/'.
/
l_,
~ L - 1 4l ~
Figure P3.7-4
For the symmetric plane truss shown in Fig. P3.7-5, determine the angular frequencies 11nd the (n) symmetric and (b) antisymmetric mode shapes. Use only hull' the structure, and assume that the cross-sectional areas of members 1 and J llll' O.M . thut for mc111hcr 3 is 0.8A , und those for members 4 and 5 are equal to A.
496
Systems of Units and Material Properties
Table A. I presents conversion factors for calculating quantities in SI units from those in US units. The factors are given to four significant figures, which usually exceeds the accuracy of the numbers to be converted. Note that stress is defined in SI units as the pascal. That is, 1 Pa TABLE A.I
=
1 N/m 2
Conversion of US Units to SI Units
Quantity
US Units
x Factor
Length Force Moment Stress Mass
inch (in.) kilopound (kip or k) kip-inch (k-in .) kip/inch2 (k/in. 2 or ksi) kip-sec2 /inch (k-s 2/in.)
2.540 4.448 1.130 6.895 1.751
X X X X
10-2 10- 1 103 10 2
System
ST
us
Length
(GN/m2 or GPa), acceleration ii in meters per second squared (m/s2 ), and so on. [Note that the force kilo newton corresponds to the mass megagram (Mg).] Also, in US units we give force Pin kilopounds (kips o r k), length Lin inches (in.), modulus E in kips per square inch (k/in. 2 or ksi), acceleration ii in inches per second squared (in./s2), and so on.
= SI Units
meter (m) kilonewton (kN) kiloncwto n · meter (kN · m) kilopascal (kPa) megagram (Mg)
Consistent Systems of Units
Force
497
Material Properties
A.2 MATERIAL PROPERTIES
For any numerical problem in structural mechanics, we must use a consistent system of units. By this we mean that all structural and load parameters must be expressed in the same units within each system. Some examples of consistent units for force, length, and time appear in Table A.2. For instance, in SI(l) w1.· must express an applied force P in newtons (N), a length L in millimeters (mm), the modulus of elasticity E in newtons per square millimeter (N/mm2 ), an acceleration ii in millimeters per second squared (mm/s2 ), and so on. TABLE A.2
Sec. A.2
Time
(I) (2) (3)
newton kilonewton meganewton
millimeter meter kilometer
second second second
(I) (2) (3)
pound kilopound megapound
foot inch yard
second second second
When programming structural dynamics for a digital computer, it is l'" pecially important that the system of units for input data be consistent. Othl'I wise, units would have to be converted within the logic of the program , thcrl•hy restricting its usage. For example, if in US units the length L were given in h-rt and the modulus E were expressed in pounds per square inch, the program woulll need to convert either L to inches or E to pounds per square foot. For all of the numerical examples and problems in this book wc IISl' 1.•11 hrr S1(2) or US(2) in Table A. I . Thus, in SI units wc take force I' in kilo11l'Wl1111, (kN), 11.!ngth f, in llll'lcrs (111), mocl11l11s /\' in gip1111l'Wl
To analyze solids and structures composed of various materials, we need to know certain physical properties. For structural dynamics, the essential material properties are modulus of elasticity E, Poisson's ratio 11, and mass density p. Table A.3 gives these properties in both US and SI units for some commonly used materials. Note that the shearing modulus G is not listed in the table because it can be derived from E and 11. TABLE A.3
Properties of Materials"
Modulus of Elasticity E GPa
Poisson's Ratio v
69 103 103 97 25 45 207 207 117 379
0.33 0.34 0.34 0.25 0.15 0.35 0.31 0.30 0.33 0.20
Material k/in. 2 Aluminum Brass Bronze Cast iron Concrete Magnesium Nickel Steel Titanium Tungsten
1.0 1.5 1.5 1.4 3.6 6.5 3.0 3.0 1.7 5.5
X
10• 104 10• 10• 103
X
103
X X
10• 104 104
X
104
X X X
X
X
Mass Density p k-s 2/ in. 4 2.45 8. 10 7.80 6.90 2.25 1.71 8 .25 7.35 4.20 1.80
X X X X
X X X X X X
10- 7 10-7 10- 7 10- 7 10- 7 10- 7 10- 7 IQ- 7 10- 7 10 6
Mg/m3 2.62 8 .66 8.34 7 .37 2.40 1.83 8 .82 7.85 4.49 19.2
"Numbers in this table are taken from J. M. Gere and S. P. Timoshenko, Me chanics of Materials, 2nd ed . , Brooks/Cole, Monterey, Calif., 1984.
Preface
Structural Dynamics by Finite Elements represents a culmination of the two topics identified in its title. Structural dynamics continues to grow as an essential subject for structural engineers , and the best method for handling structural dynamics problems is with finite elements. The solids and structures discussed herein are subjected to time-varying influences that cause accelerations and velocities as well as displacements, strains, and stresses. To analyze such a problem, we discretize the structure (or solid) by dividing it into a network of elements having finite sizes. Then equations of motion are written for the nodes (or joints) of the discretized continuum, which include inertial and damping actions. This finite number of differential equations may be solved on a digital computer to find approximate time-varying nodal displacements and stresses within the finite elements. This publication is intended to be used as a textbook for a graduate-level course on structural dynamics in civil, mechanical, or aeronautical and astronautical engineering. We have tried to present the material in a clear, forthright manner for either a university student or a structural analyst in industry. As background for studying this book, the user should have had the mathematics and solid mechanics usually offered in schools of engineering at the undergraduate level. Included among the former courses are differential equations, matrix algebra, and computer programming. Other desirable courses are vibration theory, matrix analysis of framed structures, and finite elements. Although it is not necessary to know the theories of elasticity, plates, and shells, previous exposure to those topics will give the reader greater perspective on the present work . he
"
Preface
( 'hapter 1 introduces structural dynamics by compuring time-varying ~e11p1111Nl'S to dynamic loads against results for static loads. Next, we describe 11111111u>nly encountered dynamic influences, which are initial conditions, a~plil•d nctions, and support motions. Then discretization by finite elements 1s dl l'l t'IISNcd for framed structures, two- and three-dimensional continua, plates, 111111 11hclls . Such analytical models are processed by digital computer programs, 111'1 11ll'lllioned in the last section of this introductory chapter. In preparation for later work, we describe and analyze systems with only mil' degree of freedom in Chapter 2. This material would be a useful review for 11 pl'l'Non who has had a previous course on vibration theory for one-degree 11yNll•111s. It also serves as an introduction to the subject for anyone without such 11 h11ckground. Moreover, this book on structural dynamics becomes more selfl'1111t11i11cd by including material on one-degree systems. We present derivations for finite elements and vibrational analysis in Chaph•1 \, where all of the discretized structures have multiple degrees of freedom. 1~111·1HY consistent stiffness, mass , and load matrices are developed for onedi11u111sional elements for later use in framed structures. From the homogeneous 101111 of the nodal equations of motion for an assembled structure, we can solve 1h1• l'lgcnvalue problem for vibrational frequencies and mode shapes. If planes 111 Hy111111etry exist, appropriate restraints at nodes on those planes allow us to find 11y11111mtric and antisymmetric modes, using only part of a structure. At the end 111 <'lluplor 3, we describe Program VIB for vibrational analysis of finite-element m•tworks, with specialization to Program VIBPT for plane trusses. ( 'huptcr 4 contains the normal-mode method for dynamic analysis of dlN1 1l'li:1.cd structures. By this approach, we transform the equations of motion 111 11111111111 coordinates, where each flexible-body or rigid-body mode has unit 11111NH und responds as if it were a system with only one degree of freedom. We d1 t11·11sN normal-mode responses to initial conditions, applied actions, and sup11111 I 111otions for structures with or without damping. Step-by-step response l 11il'11l111io11s ure explained and coded in Program NOMO for normal-mode 111111lyNis. This program is specialized to become NOMOPT for plane trusses, wll1l'11 i11l'l11des VIBPT from Chapter 3. 111 C'lwpter 5 we cover direct numerical integration methods.for calculat~ng dy1111111i1· r<.~sponscs. These approaches m~y all be. characterized a~ fim~edtlll' ll' lll'C approximations with respect to time. We discuss extrapolation with ,•xplkil loa111ulas, iteration with implicit formulas, and direct linear extr~po!11111111 , with solution for incremental displacements. Newmark's generalized 11n'l'll'l'tllio11 method is extended by Hilber's approach and applied in Program I >YNA lor dynamic responses of structures. Then this program is specialized to t,,•1•111111• J)YNAPT for plane trusses, which includes VIBPT from Chapter 3 and N< >M< WI ' fro111 Chapter 4. Thus, in several stages we construct a program that wlll l11111dll• 1101 only vihrntionul 111111lysis but ulso two types of dynamic response l llll'll lllt iollN <'011n•pl~ d!'V!'lop1•d i11 p11•vil111 Nl h11ptl•1:, loa pl111ll' la llHHl1H nn: l'Xll•11dl•d to
Preface
xi
all other types of framed structures in Chapter 6. The programs described are DYNACB for continuous beams, DYNAPF for plane frames , DYNAGR for grids, DYNAST for spaces trusses, and DYNASF for space frames. We also discuss methods for reducing the number of degrees of freedom for beams, grids, plane frames, and space frames. Guyan reduction may be used to eliminate joint rotations from the equations of motion for these four types of framed structures. Also, constraints against axial strains can be imposed in plane and space frames; and Programs DYPFAC and DYSFAC have been coded for this technique. Chapter 7 describes finite elements to be used for dynamic analyses of twoand three-dimensional continua. We employ isoparametric quadrilaterals for solving two-dimensional problems in plane stress or plane strain and present Program DYNAPS for that purpose. Isoparametric hexahedra are applied in calculations for general solids, and we have coded Program DYNASO for obtaining their dynamic responses. Axisymmetric solids require the use of ring clements having cross sections that again are chosen to be quadrilaterals. Program DYAXSO for dynamic analyses of axisymmetric solids completes the set of programs documented in this chapter. In Chapter 8 the finite elements for analyzing plates in bending and general 11nd axisymmetric shells are based on those for general and axisymmetric solids in Chapter 7. These specializations automatically include the effects of shearing deformations and rotary inertias. The programs described are DYNAPB for plates in bending, DYNASH for general shells, and DYAXSH for axisymmetric Hhclls. Chapter 9 explains the effects of including rigid bodies in the analytical models for framed structures and other discretized continua. For framed structures the convenient approach is a member-oriented technique with rigid offsets nl lhc ends of each member. On the other hand, for finite elements with more lh1111 two nodes, a body-oriented method appears to be mandatory. Sample paograms discussed in this chapter are DYRBPF for rigid bodies in plane frames 111,d DYRBPB for rigid bodies connected to plates in bending. We also describe 1IHid laminae in multistory buildings in preparation for the next chapter. The topic of Chapter 10 consists of . substructure methods for dynamic 111111lysis. We divide the subject into Guyan reduction methods and the 111111ponent-mode technique. Within the first approach, emphasis is placed on the 111otli fled tridiagonal method, which is applied to multistory buildings. Programs I lVMSPP and DYMSTB have been coded to handle multistory plane frames and th I buildings with rectangular framing patterns. On the other hand, the theory 111 thl· component-mode method is explained and implemented for the analysis 111 plnnl' und space trusses having only a few members. Programs COMOPT and I '<>M< >ST represent the codes for such analyses. At tlu: hnck of the book we give a list of notation, general references, 11ppl'11dil'l'H, 11ml 1111swl'l's to prnhlcms . Appendix A describes systems of units (SI 111111 l JS) 1111d physku l prop,~1tlt•1, fol' v111l111111 mntcrinls . In Appendix B we divide
1
Preface
KIi
Nolution of eigenvalue problems into the topics of inVl'l'Sl' ill'rttlion (for few 11uidcs) and transformation methods (for many mod<.!:-.) . 'fhl' lutLer methods include those of Jacobi, Givens, and Householder. Our finul approach consists ol I louseholder transformations of the coefficient matrix to 11idiugonal form, lollowed by iteration with the QR algorithm. Spectral shifting may be used to l111prove the convergence of either inverse or QR iteration . Last, Appendix C contains the detailed flowchart for Program DYNAPT. As mentioned in Chapter 1, all of our computer programs are coded in FORTRAN. These codes and data for examples have been assemb led on a 11111gnetic tape, a copy of which can be obtained from Paul R . Johnston for a nominal fee. His business address is: Failure Analysis Associates, 2225 East llayshore Road, Palo Alto, California, 94303. We wish to thank graduate students and teaching assistants at Stanford who hnve directly or indirectly contributed ideas for this book. Needless to say, our wives Connie and Terry have shown much patience and consideration while we wen: engrossed in its development. As before, Suzanne M. Dutcher did an outstanding job of typing the manuscript, and working with her was a great pleasure. Failure Analysis Associates of Palo Alto allowed us free computer mage, for which we are most grateful. Also, a Ford Foundation grant from the l'rnvost at Stanford provided funds to partially offset our expenses.
Introduction to Structural Dynamics
1. 1 STRUCTURAL DYNAMICS CONCEPTS WILLIAM WEAVER, JR. PAUL
R.
JOHNSTON
i\(.'knowledgement
This book was written in collaboration with C. Lawrence Loh, Staff E11gint!er, Engineering Information Systems, Inc., San Jose, California. Larry rnmposed some of the computer programs, provided computer examples, and dll'cked the Answers to Problems.
Il' a solid or a structure is loaded very gradually, it is said to be in a state of static ,•quilibrium, for which static actions and reactions equilibrate each other. In such 11 situation, time has no significant influence; and static analysis may proceed without considering this variable. On the other hand, if forces are rapidly 11pplied , the solid or structure is said to experience dynamic loads . In th is case we can say that a state of dynamic equilibrium exists, for which time-varying lll'lions and reactions equilibrate each other at every instant. To examine some of the differences between static and dynamic analysis, h•t us consider the prismatic cantilever beam in Fig. 1. l(a). This beam is loaded with a concentrated force P(t) in they direction at its free end (point 2) . If the loud is applied slowly , the static displacement at point 2 given by elementary ht•11111 theory is · (1)
111 which m is lhc flexural rigidity of the cross section. The response curve 1,ilwll•d I in Fig. I. I (c) shows that by gradual application, the load produces an ,1w111plotk value of (1J2).i, In add itio n, the displacement v (x) at any point along IIH' lt•11µth of the beam is a function of r only. Al tht· olhl'I' l'Xlrc111t.·, suppose thut the force in Fig. I. I (a) were applied l11~1111111111t·11111,ly 111 tl11s l.'ll~l' thl' h l'HIII will not only displace but will also 1111 rlr111lt• 111 l'Vl'IY p111111 111011µ 1h ll•11pth , 'l'ht.•1·dort· , the displacement D(x, t)
1
Preface
11111rn of' eigenvalue problems into the topics of invcl'Sl' lll'f"lllion (for few ul1•N) nnd transformation methods (for many modes) . Tlw laller methods l111ll' those of Jacobi, Givens, and Householder. Our finul approach consists 11011,dlOlder transformations of the coefficient matrix lo 11 idiagonal form, lh,wl•d by iteration with the QR algorithm. Spectral shifting may be used to 1prnv1· the convergence of either inverse or QR iteration . Lusl , Appendix C 11111111!-. the detailed flowchart for Program DYNAPT. I\~ mentioned in Chapter 1, all of our computer programs are coded in >l<'IRAN . These codes and data for examples have been assembled on a 111•m•lil' tupc, a copy of which can be obtained from Paul R. Johnston for a 111111wl Ice. His business address is: Failure Analysis Associates, 2225 East 11yi,ltrnc Road , Palo Alto, California, 94303. We wish to thank graduate students and teaching assistants at Stanford who uvr dirl'Ctly or indirectly contributed ideas for this book. Needless to say, our 1v1·'1 C'onnie and Terry have shown much patience and consideration while we 1·11• 1•11grnsscd in its development. As before, Suzanne M. Dutcher did an 11hl11mltng job of typing the manuscript, and working with her was a great h•11,111 l.'. Failure Analysis Associates of Palo Alto allowed us free computer 1•1111-11• , Im which we are most grateful. Also, a Ford Foundation grant from the 111,v11,1 111 Stanford provided funds to partially offset our expenses.
Introduction to Structural Dynamics
1.1 STRUCTURAL DYNAMICS CONCEPTS WILLIAM WEAVER, JR. PAUL
R.
JOHNSTON
At·knowlcdgement
This hook was written in collaboration with C. Lawrence Loh, Staff I t1Hlnl'l't , Engineering Information Systems, Inc., San Jose, California. Larry 111111pw,1•d some of the computer programs, provided computer examples , and 1h1•1 k1•d tlw Answers to Problems.
If a solid or a structure is loaded very gradually, it is said.Jo be in a state of static equilibrium, for which static actions and reactions equilibrate each other. In such a situation, time has no significant influence; and static analysis may proceed with?ut considering this variable. On the other hand, if forces are rapidly apphed, the solid or structure is said to experience dynamic loads . In this case we_ can say that ~ state of dynamic equilibrium exists, for which time-varying actions and reactions equilibrate each other at every instant. To examine some of the differences between static and dynamic analysis, let us consider the prismatic cantilever beam in Fig. 1. l(a). This beam is loaded with~ conc~ntrated force P(t) in they direction at its free end (point 2). If the loud ts applied slowly, the static displacement at point 2 given by elementary hcum theory is ·
(v2\1
=
PL 3 3£/
(])
111 which EI is the flexural rigidity of the cross section. The response curve lulwlcd I in Fig. 1. I(c) shows that by gradual application, the load produces an 11Ny111plotic value of (v2)<1. In add ition, the displacement v(x) at any point along tlt1· l1•ngth of the beam is a function of x only . /\1 lhl' olhl·r l'Xlrl'mc, suppose that the force in Fig. l. l (a) were applied 1111,l1111111111•m1sly 111 1hi1, l'H/'ll' 1h1· lwa111 will not only displace but will also 1111 l'lrtilll' ,,1 rv1•1y p111111 al1111F tis l1•nr11t Thl'tcforc, the displacement v(x, r)
111111111111 tlon to !itru1,111r1I Uvn•mh 11
a
<.hnp. 1
t
,~ 1
/ ~
P(t)
'2
X
L
3
(a)
(2a)
y, v
the inertial force, P(t)
P(t)
m!2t
1~--~~~~~~---
,/
!1tr111 t111nl lly111011h1• <.11111 npt1
lwrn1m•s II fu1K'lto1111f holh spm.'l' (.t) und time (t). Although it is possible to write n pmliul di ffcrcntiul equul ion of motion for dynamic equilibrium of this simple structure [ 11, * that approach will not be pursued in this book. Instead, we will nlwuys discrctize the structure, as discussed later in Sec. 1.3. A crude discretization of the cantilever beam problem is represented by the unalytical model in Fig. l.l(b). Appearing at point 2 is a concentrated mass m, representing some fraction of the distributed mass of the beam. Also shown at point 2 is a hypothetical dashpot damper that generates a dissipative force in proportion to velocity. The three types of forces in the figure opposing the applied load are the elasticity force,
Y, V
v(x,t)
"" 1.1
c~
t ilt
•
PE
L
(2b)
and the dissipative force, X
PD = cv(t)
where c is a damping constant. From D' Alembert's principle we have
PM P0
(b)
(2c)
P(t) - Pe - PM - PD
=0
(3a)
which expresses dynamic equilibrium of the mass m. Substituting Eqs. (2) into Eq. (3a) and rearranging yields (3b)
0 (c)
li'IJ(lll't' 1.1 (u) Beam with distributed mass; (b) lumped-mass approximation; (c) Nllllll 1111d dynumic rcspon~cs.
Solution of this ordinary differential equation of motion by the theory in Chapter 2 produces the responses in Fig. l. l(c) labeled 2 and 3. Curve 2 is the response when the damping constant c is zero, and curve 3 represents a case of nonzero damping. For perspective, let us review what has been shown by this example. First, we replaced a structure having an infinite number of degrees of freedom with an analytical model having only one degree of freedom. That is, in the former case the mass was distributed over an infinite number of points along the length of the beam, whereas in Fig. l. l(b) the single concentrated mass m exists only at the end. This simplification separates the variables of space and time both physically and mathematically. Therefore, Eq. (3b) is characterized as an ordinary differential equation instead of a partial differential equation. Second , we described the response at point 2 due to the applied force for both gradual and instantaneous rise times. In Fig. l. l(c) we can see that the maximum value of curve 2 is twice that of curve l. This figure dramatically illuslruks lhe difference between responses due to static and dynamic loads of ~N11111h1'1~ 111 h111t' k1' I~
l11dl1•11tl'
11•fc1cm·1·1 111
lhl• end of 1hc chapter.
l1111111h1t lh111 111 hl11111111nl llv11n111l1 11
< l111p I
I'
lht· ~11111l' 11111g1111t11k . <'111w .> 11st ill11h· ~ 111,11111 llw displm•1•d p11111111111 (r 1),1 wllh 11 p,·1111d '/' L'quul lo lhc 11a111ral 11,·,·wrl t!/ 1•i/1mtlo11 for lht· 111111ly11c11I 111odd. 'l'hinl, we included the possih1ll1y of <.l ccuying motion L'Ullscd by the pll'Nl'lll'C of u hypothetical damping mechanism . Curve 3 in Fig. I . I (c) shows
fill' cffct:l of a dissipative force is to suppress the response as time passes. Allltough this influence is of interest to us, it is not nearly so important as the 1•ll1•l'I of lhe inertial force due to the presence of the mass. This simple example also demonstrates the steps an engineer takes when 1o11lvi11g a slructural dynamics problem . They are: 111111
I . lktine the problem. 1
<'rnnposc an analytical model. C'uk'u late the response.
ll y lh1• word re.wonse we mean time-varying displacements, stresses, internal 1H 1111111,,
1111d so on, that may be of interest. Such quantities may be expressed as values, or response spectra, depending on the nature
111111• ltl i.. lol'ics, maximum 111 llt1• problem .
(el
1,2 DYNAMIC INFLUENCES
V111 irn1.~ 11nlurnl and man-made influences may cause dynamic responses in Nl11wlt11l'S. The rnost common types of influences are initial conditions, applied ,1, ·t/011.1·, 1111d s11pport motions . Figures 1.2(a) and (b) show what we mean by 111lli11 I 1•1111ditions that result in dynamic responses. In the first case, a crane
' (f )
(g)
P(t)
(11)
lb)
le) (11)
Figure 1.2
Dynamic innuences.
l1'l1111r11 1,2
(('(l11 / .)
6
Introduction to Structural Dynamics
Chap. 1
1111ddcnly drops its load and rebounds from an initial condition of static-load tltsplucements. In the second instance , a truck backs into a loading platform at so11,c initial velocity. Consequently, the frame and body of the truck respond tlynumically. In Figs. l .2(c) and (d) we illustrate two types of applied actions in the lrnms of lateral loads that act on a building. Curve 1 in Fig. 1.2(d) depicts the 1111 cc of an air blast that rises very suddenly, decreases more slowly, and also hl'l'Omcs negative. On the other hand, curve 2 represents the force of a wind gust Ihat rises and falls gradually, but still fast enough to induce significant accelerntions. Some other types of applied actions appear in Figs. 1.2(f)-(h). The machinery force acting at the third level of the frame in Fig. l .2(f) follows a 11i111plc harmonic function, as shown by curve 1 in Fig. l .2(g). On the other hnnd , the moving load on the plate in Fig. l .2(h) may have any time variation. Two kinds of support motions are indicated in Figs. l .2(c) and (e)-(g). 'lhc first type consists of rigid-body ground accelerations due to earthquake, 111,tninst which most modern buildings and other structures are designed. Figure 1,2.(c) shows a typical time history of such an aperiodic ground acceleration. A N<'l'ond type of restraint motion is the independent displacement implied at the 111,t ht hund support in Fig. l .2(f). The time history specified for such an induced d111placcment can be arbitrary. For example, it may vary according to the squarewnvc pattern labeled 2 in Fig. l.2(g).
1.3 DISCRETIZATION BY FINITE ELEMENTS
To Nt•t up an analytical model for dynamic analysis, we will discretize solids and NII lll'htrcs using the method of fmite elements. The applications include framed 11t111l'l11rcs [2J, two- and three-dimensional continua, plates, and shells [3]. The 1111111· t·lernent method enables us to convert a dynamics problem with an infinite 1111111lll•1· of degrees of freedom to one with a finite number in order to simplify 1h1· 1mlution process. Ordinary differential equations of motion can be written for 11,•lt'l'kd points (called nodes) on the analytical model. The primary objectives ol dynamic analysis by finite elements are to calculate approximately the re11po11Nc11 ut such nodes or at other selected points. Framed structures usually are automatically discretized by virtue of the l'o1111111,n dcflnitions for members and joints. That is, a frame member may be rnnsidcrcd as a one-dimensional finite element spanning the distance between two Joints . On the other hand, a joint (or node) is defined as a point where 1111•111lwrs join, a point of support, or a free end of a member. However, we need 11111 ll'Slrkt ourselves to these purticular definitions. Figure 1.3 shows a plane l1111m• l'on11lsling of 1hrcc members (indicutcd hy encircled numbers) and four !111111 ~ II d,•sirt•d , wt• could 1n1hdividt• t•u,·h ancmhcr inlo four purls and view each flql,llll'III IHI (I llt'W llll'llllll'I . Thon lilt' IIIClll' lt'hm·cl IIIIUlylkul lllOdl•l would have I) tlll' llllll'r~ und I I 1111111 ~ Sud, 111t'll11t'111t•111 w1111ld p111vid1• IIIOll' po1111~ whl'lt'
Sec. 1.3
Discretization by Finite Elements
7
y
0..
2.,.__ _ _ 2 _ _ _ _3
---x
/
z
Figure 1.3 Plane frame with subdivided members.
masses ~ould be ~ocate~ for the purpose of writing more equations of motion. Thus, ~1th these hberahzed definitions for members and joints we are able to use ~e finite-element method more effectively. Although framed structures are discussed throughout the book, Chapter 6 is devoted exclusively to this topic. Whereas framed structures fit rather neatly into the theory, finite elements have much more powerful applications in two- and three-dimensional continua. For exa~ple, ~he hfperbolic paraboloidal (or hypar) shell structure in Fig. 1.4 mar be d1scretized_mto qu~drilateral elements that are curved in space. We can denve energy-consistent stiffnesses, masses, and equivalent nodal loads for such clements and assemble them into an analytical model having a large but finite number of degrees of_f~eedom. By so doing, we are able to simulate accurately the response of the ongmal structure. The theory of finite elements appears first z
1,·1111111• 1.4
I l y p111 ~hl'II tli~l 11'1111•11 hy
lilllll' CICllllllllN .
Introduction to Structural Dy namics
8
Chap. 1
in Chapter 3 for one-dimensional elements. Later, Chapters 7 and 8 contain applications to two- and three-dimensional continua, plates, and shells. Occasionally , we may encounter a structure in which one or more parts are very stiff compared to other parts. In such a case, it may be convenient to inc_lu~e rigid bodies within the analytical model. For example, the rectangular sohd m Fig. l . 5 could be divided into three-dimensional finite elements, while the pl_ate is modeled by two-dimensional elements . However, a more clever modeling procedure would treat the block as a rigid body , which is connected to the flex ible plate. The resulting model would have fewer degrees of freedom (?ue to rigid-body constraints) and better numerical conditioning than the first choice . This topic of rigid bodies within flexible continua is covered in Chapter 9. z
/
X
Figure 1.5 Rigid body supported by discretized plate. I( is also possible to analyze a structure by dividing it into substructures 1h11t 111t· handled one at a time. Substructuring conserves core storage in a digital t'1H11putcr nnd also allows several groups of analysts to work on diffe_ren~ parts ol II structure simultaneously . Figure 1.6 shows half of a symmetnc aucraft ~t111t·tmc di vided into five types of substructures that are joined at common (i11tt·1fun:) nodes. Five groups could analyze the substructures individually while 11 sixth group handles the assembly and solution process for the entire structure.
/
M111111• I .<,
/
/
!-:v 111111111 l, 111 1111111
.....
/ I
I
I
il1 vhh d 11 1111 - 1111~1111, 1111,•,
Sec. 1.4
Computer Programs
9
The wing group may choose to model the engines as rigid bodies connected to a flexible substructure. Chapter 10 contains descriptions of various substructure methods that have proven to be useful for structural dynamics.
1.4 COMPUTER PROGRAMS
Structural dynamics by finite elements is a computer-based method, so we include and explain digital computer programs that perform the calculations. Our programming philosophy consists of starting in Chapter 3 with the relatively simple task of vibrational analysis for plane trusses (Program VIBPT). In Chapter 4 we extend the vibrations program to include dynamic responses by the normal-mode method (Program NOMOPT). Then we further extend the program in Chapter 5 to calculate dynamic responses by a direct numerical integration approach (Program DYNAPT). In subsequent chapters this program is revised many times so that it applies to other types of framed structures and continua that are discretized by finite elements. All of our coding is written in FORTRAN, and we present FORTRANoriented flowcharts for the main programs in VIB , NOMO , and DYNA. We also include a detailed flowchart for Program DYNAPT in Appendix C. Various tables for preparation of data appear throughout the book in order to show how a user must interpret problems to the computer. We show some computer output in the form of line prints, but the amount of information for structural dynamics Is so large that we decided to minimize printing. Instead, we put emphasis on rnmputer plots to show time histories of forcing functions, displacements , stresses, and so on. A plot is much more informative than a print, so the more desirable choice of output became obvious . Program notation is included in the h~t of notation near the end of the book.
REFERENCES T i111oshc nko, S. P. , Young, D. H. , and Weaver, W., Jr. , Vibration Problems in l:'11gi11eeri11g, 4th ed ., Wiley , New York, 1974. ' Weaver, W ., Jr. , and Gere, J. M. , Matrix Analysis of Framed Structures, 2nd ed ., Van Nostrand Reinhold, New York, 1980 . Wr11w 1, W., Jr., and Johnston , P. R., Finite Elements for Structural Analysis, l '1l·ntit·c ll ull , Englewood Cliffs, N . J ., 1984.
Sec. 1.1
2
Introduction
11 D(t)
~ j.
D(t)
t
em L
./
t
•
c_{
&
).
L 2
~
2
(a)
(b)
Systems with One Degree of Freedom
D(t)
/1-lom____ /
/ /
.,,
I I
/
/
(c)
2.1 INTRODUCTION /
1•111 preliminary or approximate analyses, simple structures may often be ideal1/t•
Nhows a few cases where mass m or mass moment of inertia Ir is somehow 1111Nod11tcd with a single point or axis. In each case the translation of a point or rnllltion about an axis varies with time. For the beams in Figs. 2. l(a) and (b), tht• 11111sscs may be either attached concentrations or some fractions of the
lht• 111•1• 1•11<1 Ill
(d)
(e )
li'l1turc 2.1
Structures modeled as SDOF systems: (a) cantilever beam; (b) simple
lwum ; (c) plane frame; (d) space frame; (e) fixed plate.
wh11·h. is th~ stati~ translation caused by a unit force applied in the positive dl1n 111:11 of the dis~lace~~nt D ._ In this expression the symbol E represents \ 111111p 1:1 modulus of elast1c1ty, I 1s the moment of inertia* of the cross section 111 llll' ht•n111 , and t, is its length. S11//i11·.1'.1'('.I'
nrc found as static holding actions, corresponding to the discaused by inducing unit values of those displacements. At
plt111 1111• 111s 11how11 ,
I. ' 11'/
' I It,· 111111111•111 ut hw11l11 111 1111• l lllHN Hrt'li1111 IH 11 1111N11omcr for the second moment of area 1,1 1111 1111~~ ~,·, 111111 Wllh 11·~111·1 I lo 1hr 111•1111111 uxiH
12
Systems w ith One Degree of Freedom
Chap. 2
Soc. 2.2
Free, Undamped Vibrations
13
lhl· free end of the cantilever beam in Fig. 2.1 (a), the stiffness k is a static force 111 the direction of D that is required to produce a unit value of D. Thus, we have k = 3EI = L3
!
o
Nole that the stiffness k is the reciprocal of the flexibility 6. Similarly, if we assume that the simple beam in Fig. 2.1 (b) is prismatic, then its flexibility is u(t),ii(t) (a)
which is the translation at the center due to a unit force corresponding to D. Furthermore, the stiffness of this beam has the value k
= 48£/ =! L3 o
which is the holding force at the center required for a unit amount of D. We could also find flexibilities and stiffnesses for the other structures in Ftg. 2. 1. However, the analyses required would be more complicated than those tor lh1.: statically determinate beams. Although other motions of the structures are possi ble, we will restrict our attention in this chapter to systems idealized as l111ving onl y one degree of freedom. Analytical models having coupled multiple d1·11H'l'S of' freedom will be discussed in subsequent chapters. Topics of interest for SDOF systems consist of free and forced harmonic 1110111111s with and without damping, response to arbitrary time-varying loads or ,.11ppo, 1 motions, and response spectra for dynamic loads. These subjects are l'OVl'll' d in the ensuing sections of this chapter and will be used throughout the hook .
--· ku
mii
u(t),ii(t) (b)
Figure 2.2
(a) Mechanical analogue for SDOF system; (b) partial free -body dia-
grum .
1111•1tial ~estoring force mu also arises due to the presence of the mass. From the po1l111l free-body diagram in Fig. 2.2(b), we see that - ku - mu
2.2 FREE, UNDAMPED VIBRATIONS Figure 2. 2(a) shows a mechanical analogue for any of the SDOF systems dt•sl·nbcd in the preceding section. The mass m and the spring constant k are
d1•ll'1111incd from the properties of a given elastic structure , and the symbol u (t) dl·notes the single displacement coordinate. The figure also indicates the accel1•1at ion ii (t), which is the second derivative of u (t) with respect to time (II d 2 u/ dt 2). If the displacement is rotational, then m, k, and u may be H·plt1l'l'd by /,. k,., and 60 (a smnll angle of rotation). We assume that the sys1c111 in Fig. 2.2(a) is initially at rest inside or outside 11 ~rnvi1111 io1111l field I 11. Tht•11 lt•l 11s disturb it in some manner from its position ol .,,111k l'qtdlihr i11111 1>111• lo th,• d1Hpl11n·111cnt II relative to the static position , 1t ll'S1t11111 ~ t1111·t· 1•q1111l 111 fo ill' vrl11p1, 111 tlw s p1111 g, 11s shown in Fig. 2.2(h) . An
=O
(1)
\\ hid, is. an applicatio_n of D' Alembert's principle for dynamic equilibrium. l~111 111111grng Eq . (1) slightly, we have the following equation of motion:
mu+ ku _=
O
(2)
I lr h1 Nt'l'l~nd-order differential equation may be simplified further by introducing 1111 IIO(H(I Oll
(J) 2
k
= -
m
(3)
= 0
(4)
whit h prncluces
u + w 2u
',111 Ii 1111 1•q1111tio11 will be satisfied if we take the general solution 11
<'1 cos
Ml
I C, sin "'t
(5)
Systems with One Degree of Freedom
14
Chap. 2
Sec. 2.2
Free, Undamped Vibrations
where C 1 and C2 are arbitrary constants. Thus, the response consists of the sum of two harmonic functions that repeat themselves with time. This oscillatory motion is called free vibration, for which the angular frequency w (rad/sec) is the constant
15
u
(6)
w=~ as given by Eq. (3). The natural frequency f (cycles/sec) is
f=..!!!.._=_1 {k 21T
(7)
21T "{;;;
and the natural periodJ (sec) becomes T
= = 2: = 21r ~
7
(8)
(a)
The last expression is the time for which the vibration repeats itself. To determine the constants C 1 and C 2 in Eq. (5), we must consider the initial conditions of a given problem. Assume that at the time t = 0 the mass m has an initial displacement u0 from its position of equilibrium and that its initial vl'locity is u0 . Substituting t = 0 into Eq. (5), we obtain
C1
u
= Uo
If we also take the derivative ofEq. (5) with respect to time and substitute t
= 0,
we find that (b)
u
I
A cos (wt - a)
Thus, Eq. (5) becomes
Uo.
u = u0 cos wt + - sm wt
(9)
w
1lcncu, the cosine term depends only on the initial displacement, whereas the sine term depends only upon the initial velocity. E2ch of these harmonic terms c:un be represented graphically, as shown in Figs. 2.3(a) and (b), by plotting displacements against time. The total displacement u of the vibrating mass mat uny time t is equal to the sum of the ordinates of the two curves, yielding the third harmonic curve shown in Fig. 2.3(c). An allernative way to represent vibrations is with rotating vectors. Imagine two orthogonal axes u and u/ w, as shown in Fig. 2.4, which define the phase pl,1111•. Let a vector OP of mngnitudc u0 rotatu with constant angular velocity w 11101111<1 thl' flxt•d point O JI at tillll' t O till' wctrn OP coincides with the 11 nxi~. tht• 1111gk it 11111kt•s with tht· t11111w nxis nt 1111y 1111111 tinw tis l'qunl to (I)(. The p11111·l·ti111111I tl11'i v1·rl111 1111thr111a,w1 i'i l''lll1tl l111111 l 1111 ,,,, 1111d u•p1t•st•11ts till' flr~t
A
T
te
_l_ 0
T
T
4
(c)
l•'l1111n• 2 ..\
RC'sponhl' funl'tionN: (n) cosine; (b) sine; (c) combined.
Systems with One Degree of Freedom
10
Chap. 2
17
Free, Undamped Vibrations
Sec. 2.2
te
u
a
(13)
= -(J)
This lag time is indicated at the upper left in Fig. 2.3(c). R
Example 2.1
/
/ /
/
&"/
/ /'
/
The_ rectangular plane frame in Fig. 2.5(a) has a very stiff girder of mass m and rather flexible columns, each having length Land constant bending rigidity El. Neglecting the muss of the columns and their axial strains, find the values of w, f, and T for this structure, treating it as a SDOF system.
~ sin wt w
---
A
COS
(wt - a)
p/
y
/
~
I
'o Uo w
_
"\ \
\ w
I i , - - - - - - - m_ _ _ _ _--,1 -
u(t)
1
\ 0
Figure 2.4 Phase-plane .representation.
in Eq. (9). If we also take another vector OQ of magnitude u0 / w and 1w1 pl•ndicular to the vector OP, we see that its projection on the u axis gives the Hl'l'1111d h.:nn of Eq. (9) . The total displacement u of the vibrating mass is found hy 11dding the projections OJl the u axis of the two perpendicular vectors OP and 00, rnluting with angular velocity w. 'l'hc sumc result will be obtained if we consider the rotating vector OR (the H11111 ol wctors OP and OQ) and take the projection of this vector on the u axis. llrn111 l•'ig . 2.4 the magnitude A of this resultant vector is
/ -
11•1111
-
-x
/
/
'
(a)
y
(10)
11ml till' angle that it makes with the u axis is wt - a, where a= tan- 1
-
uo
(11)
WUo
2r
'l'ltw,, we cun express Eq . (9) in the equivalent form 11
• /\
cos (wt
a)
(12)
wlll'll' (hl' {llllf)littul<·
11111111 11\ 1,x11111ph·~ (11) 111111Nl111lu1111l ; tlti 1111,1111111111
I
/
ill)
,y111111111
1ft
with 01111 l>nu1nn 111 I 1nnil11111
19
I ln11n1111l1 I 1111 1t1u I 111111111111
Cl111p, 'I
HARMONIC FORCINO FUNCTIONS lly 1runsl111ing the girder u unit tllHluncc (u Ntlll11w1H l'Onstunt us the following holding force: k
I) 111 1h1•
= 2 (12EI) = 24£1 LJ
L3
1
dl11•1 liou , we tlnd lhc '1111 111 lhl• 111011.1 in~port.unt types of forcing functions that we must study is the l111pli• h111111on1c f unc~1on P sin Ot (or P cos Ot), where n. is the angular
(a)
111 q111 m•y ol the funcllon. The first of these functions appears in Fig. 2.6(a) , 11ppl li•d lo Ih~ undarn.ped ~DOF system of the preceding section. From the partial Ii, 1 hody d1ngrum JO Fig. 2.6(b), we can write the differential equation of 11111111111 Ior this case, as follows:
' lfa111 tht• 11n~ul11r frequency w from Eq. (6) is
w= 'J; ik=IL 'J-;i f6Ei 1111!1
(b)
mu + ku = P sin Ot
tlll' 1111turnl frequency f from Eq. (7) becomes
I II i.l111plif'y thi s equation, let us divide both sides by m to obtain
f=.!:!...=-1 {6Ei 21r
1rL
'J -;;;f,
u+
(c)
w2u
= Pm sin Ot
/\IH11 , the notuml period T from Eq. (8) is seen to be
T=7= 1rL~
p
Pm= -
(d)
m
I h 11 lhl· symbol p,,. represents a force per unit of mass. 1r.x11m11h.12.2 S11pp0Hc thnt the rigid disk in Fig. 2.5(b) has a mass moment of inertia lry with respect lo 1h1• y nxis. The disk is supported at its center by a massless flexible rod of length L h11vl111t II drculur cross section with radius equal to r. If the disk can rotate about the y 11xlN without trunslation, find w, f, and T for this problem as a SDOF system. Rolntlon of the disk a unit amount (80 = 1) about they axis yields the stiffness l'IHllltlllll IIS the holding moment
/
(e)
k = GI
'
L
111 whk h <, is the shearing modulus and
1rr4 J=2
(a)
(f)
h1 1h1i p11lt1r moment of inertia of the cross section of the rod. Replacing m and k with l ry 111111
k, in Eqs. (6), (7), and (8), we find that (g)
111111 2
I-t&G
w r J -=-=21T 21T -21,y/,
(h)
---·
ku
mu
- P s i n flt
u(t),iJ(t)
(ll)
1111d '/ '
I
f
271
(1)
{?t,. /,
,.J \/
,,c;
(i)
1<111111·,, 1.t1 (11) ll111111onlt' fo1d 111-1 f'unl·llon npplicd to SDOF system; (b) partial 111 1 liudy dlllHIIIIII,
(2)
(3)
y1111111111 with 01111 1>11u11u1 111111111110111
20
Chnp. 'J
Thu total solution for Llq . (l) rnnsists of lhe su111 111 llll' ~1·m•rul solulion IOI' the homogeneous equation, plus the particular solul1011 lhnt :mtislies the whole equation . The form of such a particular solution is
= C3 sin
u
11111..
Harmonic l·orol1111 I 111wll111111
2.3
/3 3
(4)
flt
Suhstituling this expression and its second derivative with respect to time into l\ll , (2), WC have 2
21
I -
I I I I I I I I I
'l'hncforc,
C3
=
Pm W
2
-
(5)
n2 ,H
I I
'l'lwn the total solution becomes
u
= C 1 cos
wt
+ C2 sin
wt
+ C3
sin flt
(6)
Till' first two terms on the right-hand side of Eq. (6) are called the free part of till' response (with angular frequency w), and the third term is referred to as the /on'1•d part (with angular frequency fl). Let us rewrite the forced part of the solution in Eq. (6) as
u
= w2
'l'hc11, by using Eq. (3) and w 2 =
p,,, -
.
n
fl2 sm :ut
(7)
k/m, we may cast this expression into the form
u = [l-(~/w) 2 ]fsinflt
(8)
'l'hc second factor on the right-hand side of this equation is the displacement of llll' mass in Fig. 2.6(a) if the forcing function were applied statically. The 11hsol11tt.: value of the term in brackets is called the magnification factor /3. Thus, Wl' h11ve
f3
_ _ _ _ _ _ J_ _ _
= \ 1 - (~/w) 2
1
(9)
whkh reprt.:scnts the ratio of the dynamic response to the static response. The p111 I of lhe motion in which this factor appears is called steady-state forced
1'ihl'lltio11. Figure 2.7 shows values of the magnification factor f3 plotted against the t1c.·qucncy ratio fl/ w, which is the ratio of the impressed angular frequency fl ol the disturbing force to the natural angular frequency w of the SDOF system. I 11)1' s,nall values of the frequency ratio , the magnilicut ion factor is approximately 1111i1y ; und tht.: response is about the same ns t'or the slatic application of the h111111011ic.: load . When this mt io 11pproadws 11111ty , however, the magnification
I o'---~~---1.~~~--L~~~...L_~~____J~~~_J_~-
2
fUw
Figure 2.7 Magnification factor.
factor a~d th~ amplitude of forced vibration rapidly increase and theoretically become mfimte for the case of = w. This state is known as the condition of res~na_nce. The infinite response at resonance implies that there is no damping to dis~ipate energy and suppress the response. However, in practical structures there 1s always some damping, as will be discussed in Sec. 2.4. Wh~n th_e frequency of _the d!sturbing force increases beyond the frequency of free vibration, the magnification factor again becomes finite. Its absolute v~lue ~iminishes with the increase of the ratio fl/wand approaches zero when this ratio becomes very large. Thus, when a harmonically varying force of high frequency acts on a SDOF system, the response is very small; and in such cases the mass may be considered as remaining stationary. Considering the sign of the expression I/[ 1 - (fl/ w )2], we see that for ~ < ~ the sig~ i~ positive. Physically, this means that the displacement of the v1bratmg mass JS m the same positive sense as the disturbing for.ce. On the other hand, when _fl > w _the expression is negative, meaning that the displacement of the mass is opposite to that of the force . In the first case the vibration is said to be in phase with the excitation, while in the latter case the response is said to be out of phase. . It is also possible to cause forced vibrations with harmonic ground motions instead of applied forcing functions. Two types of ground motions of interest in ~-tructural .dy~amics. are. ground displacement ug (t) and ground acceleration ~g(t), as mdicated m Fig. 2.6(a). A harmonic version of the first of these influences may be expressed as
n
ux • d sin flt
(10)
y11"1111 with
01111 DM\11 ""
111 I 1m11l11111
C:hnp. ')
wht'll' till' symbol d represents thl' 11111plltuuc or grou11d di.~plun•1m·11t. For this ,·1111t• Wl' write the fo llowing differential equation of motio11. mu
+
k (u - Ug)
=0
( I I)
111 whkh ,, u is the displacement of the mass m relative to the ground. S11h111it11ti11g Eq.\JO) into Eq. (11) and rearranging the latter gives mu
+
ku
=
kd sin flt
(12)
111
~. I
I 11111110111!
11*
u+ w
u
(17)
lt1 th1'Nt' 1t·lutionships the symbol ulf< represents the relative displacement of the 11111 N with rnspect to the ground, and u* is the relative acceleration. Substituting ,, //• 111HI Ii from Eqs. ( 17) into Eq. (11) and rearranging, we obtain mu*
mu*
= pg sin Ot
(13)
u* (14)
+
ku*
(15)
= -ma sin Ot
\\ 111
+ w 1 u*
(19)
=
Pi sin Ot
(20)
11•
Pi
Th INv11 h1c of p11 is an equivalent force per unit of mass, due to ground displace111t•111 . ( 'olllparing Eq. (12) to Eq. (1), we see that kd has replaced P. Therefore, 1t IH l'Vident that occurrence of the harmonic ground displacement d sin flt is ,•q11i v11 1L•1Jl to applying a force kd sin flt directly to the mass m. All previous r111wl11sions regarding the harmonic force may now be applied to the case of 111111111111 k grnund displacement. That is , if we replace Pin Eq. (8) with kd and t 'lllll'l' I k, the result is
(18)
111111 di vision of Eq . (19) by m produces
wht'll' kd Pc= m
+ ku* = -mug
II \VI' 11 IH0 use Eq . (16) in Eq. (18), the result is
'l'lw11 division of both sides by m yields 2
23
I 011 111111 I 11111 lilll in
= -a
(21)
l ltl11 v11 lm! of pf is an equivalent force per unit of mass, due to ground acceler,11 h 111
l(q11111ion (19) is of the same mathematical fonn as Eq. (1), except that u* 1111~ 111pluccd u and - ma has replaced P. Thus, the case of harmonic ground 11 11 It rnl ion a sin flt is equivalent to applying a force -ma sin Ot directly to the 1111111N 111 . Using - ma instead of Pin Eq. (8), we have u*
1 ]ma."' = - [ 1 - (O/w)2 ksmut
(22)
,, hh Ii ii, lhc displacement of the mass relative to the ground. This relative motion flnding the force in the structure,* which is represented by the spring 111 th1• SDOF system. Note that the minus sign before the brackets in Eq. (22) 11111111 tlw n.ilati ve response of m out of phase with the ground acceleration. I IINr t II I for
'l'ht• tl•1111 d si n Ht in this expression is the motion of the mass when the ground dl1,pl111..·c11wnt occurs very slowly (or statically). The premultiplier in brackets is !Ill' 1,111111· f'uctor as that discussed before [see Eq. (9)]. Thus, we need only 1•111111id1•1· th\: displacement of the mass due to the static displacement of the gi 111111d in order to calculate the steady-state forced response of the SDOF Nyt1t1•111. 111 1111111y cases it is more convenient to deal with ground accelerations than I'' 111u1d dl splucemcnts because a measuring device called an accelerometer [2] hns ht•1111 IIH\:d 10 obtain information about the ground motion. For example, 1•111 tliq1111k1• µround motions are usually measured and reported in terms of three 111 lh11prn11tl coinponents of ground acceleration, which are north-south, eastw1•111, 1111d Vl'l'lical. Therefore, we shall reexamine the ground motion problem by Np1•1 ilyi11g II hurmonic ground acceleration , as follows :
a~-
(I
sin
n,
( 16)
wh1111• 1lw symbol
I•,111u11lt• 2.3 '11p1 111N1• th111
the rectangular plane frame in Example 2.1 is subjected to a harmonic force
I' 1111 \lt, 11s indicated in Fig . 2.8. Calculate the steady-state forced vibration of the mass
II O ~w/6. l(1•1·11lllng that the stiffness constant for this frame is k = 24£1/L 3 , we substitute 1111 v111i11•s o l' !land k into Eq. (8) to obtain .
111
1 PL3 u = 1 _ (S/ 6) 2 24£! sin Ot =
3PL3 - - sin Ot 22£/
(a)
N111, th11t lh1• vn lul.! of thl.! mngnifi cution factor is f3 = I36/ LL I, in which the positive sign lhl' vi hrntlon is in ph11sl.l wi th the forcing function.
1111pll1 N 111111
1 w, wlll 111111111 lo thlN 1111111<,1 o l 1•111,,11lutl1111 tl11i 111l111ivc displ uccrncnt of the mass w ith 11 ~1111 I 111 1hr p1111111d 111 llx11111ph• 1 1.1
:u
V•h1t111 with
0110
Chup. 'J
IJ11111nn 111 I 11111!111111
y
I.
'" ~ ,1 I
I
11111111 111 I >1011pil111
26
I) pt rnlitl'l'S
p,,,!l w2 - 112
11(1)
I .-------m--------, -
I hll N,
P sin !lt
C
2
= uo _ p,,,11/w w2
w
(25)
112
-
111111 1111• lolnl solution in Eq. (23) becomes . wt + Pm l'\ (sm . I'\ 11 sm . wt) sm ~ £t - + -iio w w 2 - 3.£2 w 11 111,, 11111ial conditions are u0 = u0 = 0, Eq. (26) reduces to 11
uo cos wt
u = w2 P,,, _ 112 (·('\ sm Ht
1 1 .wt) w sm
-
(26)
(27)
I q1111l 11111 (27) represents the response of the SDOF system to the forcing
Figure 2.8 Frame with harmonic forcing functions.
11t, and we see that it consists of two parts . The first part is 1111 !ilt•ndy state response discussed previously, and the second part is called 1111 11,111si1•11t re.sponse. This name refers to the fact that the term dies out in the 1111 ~t lll'l' or damping, as do all free vibrations. The sum of the two terms is not 1 h111111011 ic motion even though it is composed of two harmonic functions, 1111 1111N 1· lhll components have different frequencies, 11 and w. When the forcing function is P cos 11t instead of P sin 11t, the term cos 11t 11 11l1111•s sin !lt in Eq. (23). In this case the initial conditions result in the l 11ll11w 111g constants of integration: I 11111 11011 I' sin
IC1mm1>lt1 2.4 11 lh1• Nt1111c frumc as that in Example 2.3 has a harmonic ground acceleration ii8
=
a sin
0, (Nl'l' Fig. 2. 8), find the steady-state forced vibration for O = 3w. In this case the relative response, as given by Eq. (22) is:
l mal3 . u* = - I - (3)2 24£/ sm Ot maL3 • = --sm Ot 192£/
(b)
lk1,· w1• Nl'l" tlrnt the value of the magnification factor is /3 = l-1/8 1, so that the vllu11tlm1 ls 11pp11rcntly out of phase with the ground acceleration. However, the two 11111111~ Nig11s l:llllCCI lo yield a positive sign in Eq. (b), meaning that in this situation the vlh1ntio11 ts ul'lunlly in phase with the ground acceleration.
C I = Uo -
W
-
(28)
l'\2 ~L
,11liict1t111ion of' these values into the total solution gives II
111 tht· explanation preceding the examples, only the third term on the il1,1 hl h1111d side of' Eq. (6) was used for studying forced vibrations. Now let us 1•11111111m• the free vibrations associated with the first two terms, which will be dilh·ll' ttl frolll those in Sec. 2.2. Substituting Eq. (5) into Eq. (6), we have
p,,, 2
. wt + - Pm uo cos wt + -Uo sm 2 W
W
II 1111 111illn l conditions are taken to be
u
-
u0 =
AZ
u0 =
( = w 2 p,,, AZ cos ut .I'\
(cos
I'\
ut -
cos wt)
(29)
H
0, this equation becomes
cos wt)
(30)
~£
II
C I cos
WI
. + C 2 s in
wt
+
w2
p,,,
. I'\ 112 Sll1 Hf
(23)
111 1111 11 111slancll the transient part of the response has the same amplitude as the "lllly Hllltt• purl , regardless of the ratio 11/w.
I >illl•n•111inting this equation with respect to time yields
ti
C 1w sin
CJJI
I C1 cJJ cos"'' I
,,llln ,
n l
(1/
~£
cos
{\
(24)
al
flu1111 il1l' i11iti11l rn11dilio1111 11 11 111 litm· I 0 , l \q ( .> \) giws C, iN illl' hllllll' llh lil•l1>1l' I loWl'Vl' I , H1tl11,li1t1ti1111111 fi ti11 Ill tiilll' I
which 0 into Eq . 110,
n
F CTS OF DAMPING
111 pH v11111s tllst·ussions or free und fon:etl vibrations for SDOF systems, we did 11111 1 c111N1dt•1 lht· l'fft·t·ts ol disslpuliVl' i11lh1t·nct!s, such as friction or air rel 1111111• <'011st•q111111tl y, wt• l1rn11d 111111 llw n111pllt11dt· of' frllt! vibration remains
Gy • IMl!lft
with
01111 l>MIJtnM 111
I 111Md11111
Chup, 'J
1·1111Nl11111 wllh lime; but experiem:c shows that the 11111plit11d1• d1111i11l shes with ll1w• 1111d that the vibrations arc graduall y damped out. Si111ll111'1 y, for undamped l111 c1•d vihrutions the theory indicates that the amplitude cun grow without limit 111 11•so1111 11cc. 1lowcver, we know that because of damping there is always some 1111111• 11111plitu<.lc of steady-state response, even at resonance . 'l'o bring our analytical discussion of vibrations into better agreement with 11•1d 11y, do111f)i11g f orces must be included. These forces may arise from several 11111s1•N, such as fri ction between dry or lubricated sliding surfaces , air or fluid 11·NtNl1111c1·, clectric impedance , internal friction due to imperfect elasticity of 1111111,1luls, und so on. Among all of these sources of energy dissipation, the case wh1•H lhL' dumping force is proportional to velocity, called viscous damping, is llw i;it11pkst to deal with mathematically. For this reason resisting forces of a rn111pl k uted nature are usually replaced, for purposes of analysis, by equivalent 1•/,1·,·011s rlo111f)i11g I 1]. This equivalent damping is found by equating the dissip111 lo11 of" energy per cycle of vibration to that for viscous damping . For example, .1·tr 11,·11m ,! da111pi11g due to internal friction can be treated by this approach. Wt• shall now consider the case of a SDOF system that includes viscous tl11111pi11p i11 lhl) form of a dashpot damper , as shown by Fig. 2 .9(a) . Assuming lh111 11 viscous fluid in the dashpot resists motion in proportion to velocity, we
HII ,
:,!.11
I 11111:11 ol l>nt11pl1111
27
w, 111• the hon1oµeneous equ ution for free vibration as ma
+
cit
+
=0
ku
(1)
1111• rnc l"ficient c in this equation denotes the damping constant, which has dl11ll't11,ions of force per unit velocity . Note that in the partial free-body diagram IN1•t• Fiµ. 2.9(b)] the damping force acts in the direction opposite to the velocity. I >lviding Eq. (I ) by m produces ii
+ 2nu +
w 2u = 0
(2)
111 which we define the new constant
1
C
2n = -
(3)
m
1111 rnnvenience in the ensuing solution. To solve Eq. (2), we assume a harmonic function in the following form :
(4)
u = Ce 51
111 which C and s are constants that satisfy Eq. (2). Substituting Eq. (4) into Eq. I
1) ,
we find that s2
+
2ns
+
w
2
=0
;
1111111 which
s
=
-n ± \!n 2
-
w2
(5)
II lhe value of n is less than that of w, then the quantity
wa= w2 IM positi ve,
n2
and we obtain for s two complex roots, as follows:
~u,U,iJ
(a)
wh1•re i = \/=1. Substituting these roots into Eq. (4) , we find two solutions of h 1 (2). The sum or difference of these two solutions multiplied by any constant will ulso be a solution. Thus,
cu -
1\ddinµ these formulas , we obtain the general solution of Eq. (2) for damped free 1•/11mtio11s as ~
11, 11, 1/
(6) (Iii
l'hl· 1111·101 1• "' i11 lhis t'<111111io11 d1•1·u•us1•H wi lh li1IIL', so the vibrations originally ~w111•1111t•d wtll lw HI 11d1111ll y d11111111•d 11111
Systems with One Degree of Freedom
28
Chap. 2
Sec. 2.4
u
Sine and cosine functions appearing in the parentheses of Eq. (6) are of the same general form that we found before for free vibrations without damping [see Eq. (2.2-5)]. However, now they have the angular frequency wd
= V w2 -
n2
= w ~1 -
(
~)2
29
Effects of Damping
(7)
which is called the angular frequency of damped free vibrations. The corresponding period of damped free vibrations is (8)
If the damping term n is small, then wd = wand~ = T. The constants C 1 and C 2 in Eq. (6) must be determined from known initial conditions. Assume that at time t = 0 we have initial displacement Uo and initial velocity u0 . Substituting these quantities into Eq. (6) and its first derivative with respect to time, we find that C2
_ uo -
+
nuo
(9)
wd
-A
.-- ---
---
--- --- --
Putting these constants into Eq. (6) yields (10)
The first harmonic term in this equation depends only on Uo, but the second dl.lpends on both u0 and uo. Equation (10) can also be written in the equivalent form u
= Ae-"
1
Vcf + c~ =
(15)
In this case the solution is not periodic, so it does not represent a vibratory 2 Uo
+
(uo
+ nuo) 2 WJ
2
(12)
und ar1
In the foregoing discussion of Eq. (2), we assumed that n < w. For the opposite case of_n > w, both of the roots in Eq. (5) become real and negative. Substituting them into Eq. (4), we obtain two solutions of Eq. (2), and the general solution becomes
(11)
cos (wdt - ad)
111 this expression the maximum value is
A=
Figure 2.10 Damped_free vibration.
= tan - I -C2 = tan _,uo +
nuo
motion. The viscous resistance is so large that when the mass is displaced from its equilibrium position, it merely creeps back to that position.Jn such a case the system is said to be overdamped, anq the motion is called aperiodic. The constants C 1 and C2 in Eq. (15) may be evaluated by substituting the initial conditions at time t = 0 into the equation and its first derivative, yielding
(13)
Wr1Uo
C 1
We may regard Eq. ( 11) as a pseudoharmonic motion having an exponentially dei.:rcasing amplitude Ae 'U, an angular frequency Wr1, and a phase angle a,1. Figure 2. 10 shows this damped free vibration with u lag time of
n,,
,,,,,
from which (16)
Thus, Eq. ( 15) becomes ( 14) II
111di1•111t•d Ill 1111' 11ppt' I li•II
1iu ,1' 1
,\', llo
"''
t•\I
I
(I 7)
Systems with One Degree of Freedom
10
Chap. 2
'l'ht· general appearance of a graph of Eq. (17) depends on the parameters n, uo ,
Sec. 2.4
from which the solution is
1111d 1.i0 .
Between the underdamped and overdamped cases lies the special case of 11 w, which is the level of damping where the motion first loses its vibratory l hn rncler. Using Eq . (3) for this condition , we have Ccr
(18)
= 2nm = 2wm = 2'Vkm
ln which the symbol cc, denotes critical damping . For the critically damped SDOF system, Eqs. (5) and (7) show that s1 = s2 = -wand wd = 0. Neither l\q . (6) nor Eq. (17) constitutes the solution, which in this particular case of 11 pealed roots takes the form
u = e-w1(C1
+
C2t)
(19)
Suhst ituling the initial conditions into Eq. ( 19) and its first derivative, we find lhnt nuo
(20)
(uo + nuo)t]
(21)
C2 = uo
+
(w2 - fi2)Pm Q - (w2 - fi2)2 + 4n2fi2
= e-w [u 0 + 1
(25)
By substituting these constants into Eq. (24), we can obtain the particular solution of Eq. (23). The total solution of Eq. (23) consists of the sum of the particular solution and the general solution derived previously as Eq. (6). Thus , considering only subcritical damping, we have u = e- "1(C 1 cos wdt
+ C 2 sin
+
wdt)
Q cos fit
+ R sin
u = A cos (fit - 0)
A
= y Q 2 + R2 =
\!(w 2
Pm
iV) 2
-
(27)
+
(2nfi) 2
Pmf w 2
ht•t'OlllCS
+
cu
+
= P cos
ku
+ 2nu +
Y[l - (fi/w)2]2 + (2nfi/w 2)2
(22)
fit
I >ivlding this equation by m produces
ii
w u
= Pm cos
(23)
fit
() =
wlw11· () und R are constants. To determine these constants, we substitute Eq . ()•I) 1111d its derivatives into Eq. (23) and obtain
( U'() I 211fiR + w 2 Q - Pm) cos fit
+
(-fi 2R - 2nfiQ
+
w 2R) sin fit
=0
'l'hiH t·quation can be solved for all values oft only if the expressions in the p11rn1Hhcscs vttnish. Thus , for calculating Q and R we have two linear algebraic ,•q1111llo11s, as follows : ()
'2.11 HU I ,,,i (J
oi Ii
1 1,1 Ii
l 11 O(! I
fi 111 II
R 2nfi = tan- 1 ---,.----,, Q w2 - fi2
tan- 1 -
_
_
- tan
(24)
+ R sin fit
u = Q cos fit
H
(28)
und 2
llll' which all of the notation has been defined previously. A particular solution 111 Hq . (23) can be taken in the form
1
(26)
where
which ugain depends on n, u0 , and Uo. Turning now to damped forced vibrations , we take the applied forcing h111ttion in Fig. 2.9(a) to be P cos fit. Then the equation of motion for the mass mii
fit
Because of the factor e-nt in the first part, the free vibrations gradually subside, leaving only the steady-state forced vibrations represented by the last two terms. The harmonic force maintains these latter vibrations indefinitely, so they are of great practical importance. The steady-state response in Eq. (24) may be written in the equivalent phase-angle form as
'l'hl·ll the general solution becomes u
31
Effects of Damping
1
2nfi/w 2 1 - (fi/ w )2
(29)
Thus, we see that steady-state forced vibration with viscous damping is a simple hunnonic motion having constant amplitude A , phase angle (), and period lj 2n/!l. Using the values of w2 = k/m and Pm = P /m and introducing the symbol 'Y for the damping ratio ,
n w
C
(30)
y =-=w1 11111y
Ccr
s11hst itull· l~q. (28) into Eq. (27) to obtain
" t1 I',. l'Os m,
o)
(3 1)
Systems with One Degree of Freedom
J2
In which the magnification factor
f3 for damped forced 1 (0/ w)2]2
/3 = Y[l -
Chap. 2
Sec. 2.4
33
Effects of Damping 4.0 ,--,----,--,--..--.-rT--ir--.-----,---,
vibrations is
y= O
(32)
+ (2y0/ w)2
f3
Also , Eq. (29) for the phase angle becomes _
e-
-I
tan
2y0/w 1 - (O/w)2
(33)
Thus, lhe amplitude and the phase angle both depend on the damping ratio y as wd l as the frequency ratio 0/ w. Figure 2 .11 (a) shows the magnification factor f3 plotted against the ratio U/
o
1.0
1.0 (a)
1.5
2.0
n
w
(34)
- = V1 - 2y 2
/
w
Fm small damping ratios ( y :5 0.20) the maximum value of {3 occurs very near to tl sonunce, and taking the value of {3 at resonance as the maximum provides 1,111 lldcnl accuracy for engineering analysis. Then, from Eq. (31) we have
y=O
0
1
P
Amax
1
P
= /3,esk = 2y k =
w
P
2n w2 m
P
= cw
(35)
'l'hus , wh ile damping has only a minor effect when the system is remote from 1N10111111ee, il has a dramatic effect at or near resonance. In structural dynamics thl' inl1uencc of damping is crucial for this case and represents its most important 1q1pl k 11tion . In metal structures the damping ratio y is usually in the range 0.01 to 0 ,0); whereas its range for reinforced concrete structures is about 0.05 to 0, I0 . ' l'hc value of y would always be less than 0.20 for practical structures. While the phase angle is of less consequence, we also show in Fig. 2.1 l(b) plots of () versus the frequency ratio 0/ w for different values of the damping llltlo. The phys ical meaning of the phase angle in damped forced vibrations is thnt the respo nse of the system lags the forci ng function by the angle 0. For the c11Sl' of zero dHmping, the forced vibrations are exactly in phase (0 = 0) with the d1st11rhi11g force I'm nil vn lues of H/ I . Also, f'or zero damping thr pl111Hl' llll)' ll• iH l11dl•t(1111ii11111l' 11! t l'Nlllllllll'l', whl'tl' U ,,,. Wlwn d111nping is 11011 tl' 111 , Wl' t1t1ll' 111 ,1111111111111 h11111w 111 /J11Mtlw 11111110 / ,,1 v11il1•H llut , tl'p11rdkss
Fll(urc 2.11 (a) Magni fication factors; th) phnsc unglcs.
2 (b)
3
D. w
of the amount of damping, the phase angle is always equal to 1r/2 at resonance. '1'11111 is, ut resonance the response lags the force by a quarter cycle.
ll,x111111,h• 2.5 I It till' 11•vt11111t11l11r pl11nl' l'ru1m• 111 Fig. 2. 8 hnvc u hnrmonic force P cos D.t in place of 1111 loll'<' 11N1•d 111 l~x11111pl1• l. \, AINo, ln1111tim• lhal lhi: ground has an acceleration
Systems with One Degree of Freedom
34
Chap. 2
1/M a cos flt instead of the sine function in Example 2.4. Calculate separately the 11h·111ly -state forced responses of the mass m due to these innuenccs, assuming that y 0.02 and fl = 0.9w. For this example the magnification factor in Eq. (32) becomes
1
5 17 f3 = \/[1 - (0.9)2] + [(2)(0.02)(0.9)]2 = · 2
Sec. 2.5
where (h)
and _
(a)
llsing this value and k = 24E// l 3 in Eq. (31), we calculate the response to the force as
35
Periodic Forcing Functions
ad=
tan
1
Qn
+ RO
(i)
Q
Wd
In the last expression the symbol ad represents the phase angle for the damped system.
3
PL u = (5.17) 24£/ cos (flt - 8) 2.5 PERIODIC FORCING FUNCTIONS
Pl3 = 0.215£1 cos (flt - 8)
(b)
<'lmNldcring the ground acceleration, we need only replace Pin Eq. (31) with - ma to ol,111111 the relative response as mal u* = -0.215 El cos (flt - 8)
(c)
1,:x111111>IC 2.6 1 IH· 111111sil!11t response of a SDOF system with subcritical damping may be found by N11hst1t11t ing initial conditions into Eq. (26). Determine the free-vibrational response of NIil I, 11 system due to the forcing function P cos flt. Using u = u 0 and u = iio (at time t = 0) in Eq. (26) and its first derivative with 11•Np1·r t to time, we evaluate the constants of integration as
C1 =
C2
Q
Uo -
_ -
uo+ n(uo -
Q) - R il
(d)
Wd
S11hNl1t11t111g these constants into Eq. (26) gives
u= e-"'(uocos wdt + :dnuosin wdt) 10 '
+ Q[cos
flt -
+ R(sin
flt - e_,,,
e- '"(cos
I
ll jlll'N~ III!'
!Iii \
1111n ~i1•11t
(' "' ( Q cos
in
phUNl'
P (t)
= ao + a 1 cos flt + a 2 cos 2flt + . + bi sin flt + b2 sin 2flt + . .. = ao +
L (ai cos iflt +
b; sin iflt)
(e)
~ sin wd9
t,i11
t
~
11111!11• 101111 ,
1,0
= 0, the transient portion u1r of the
Qn I RH .
,,,.,
Wl'
)
sin w,,t
(f)
hnw
(. l/1,
111
,,,.,
l II ~ (
t1l,,f
ti ,/ )
(1)
The period of the applied force is Tr = 27r/fl, and the symbols a0 , a; , and b; represent constants to be determined. To calculate these constants , we may use lhc procedure described next. p
wdt + ;dsin wd1 J
II th1· 1111t1ul conditions arc taken to be u0 = 11·11111111111g tl'sponsc is
11,,
In previous discussions of forced vibrations , we assumed simple harmonic functions proportional to sin fit or cos fit. It is also possible to encounter general periodic functions that are more complicated. For example, Fig. 2.12 shows a repetitive triangular function with period Tr= 27r/ fi. In this section we learn how to calculate the response of a SDOF system to such a function. We can represent a periodic dynamic load of any kind by decomposing it into a trigonometric (or Fourier) series, as follows:
(g) lo'IMtt1·1•
.2. 1.2 l'ri l11dl1 ltm 11111 f 111w1in11 .
i,y11111111N wllh U1111 l>11111"'1 11f
lfl
I II nd11111
Chllft, ')
Assuming thal a 1 in Eq. (I) is dcsir<.!d , post11111ll1ply hnth sidl'S ol' th<.! 1•q1111t 1011 hy t.:os i Ht dt and integral<.! from t O to t ·1;. 'l'hl'II W<.! S<.!C thul
L''/' a
0
cos iOt dt
J(1'/
=0
Ju
=
(1i
Jo a; cos2 iOt dt
0
wl11•1t• / und j arc integer numbers I , 2, 3, 1111d thnt a;
= -2 l1j P(t) I;
, oo.
=
a·
jI;
By using these results , we
cos iflt dt
(3a)
0
~ (1i P (t) sin iOt dt 7; Jo
(3b)
.!.
·r,
(3c)
(1i P(t) dt
7; Jo
is simply the area under the sawtooth diagram for one cycle. Since this area is zero, the constant ao in Eq. (3c) is also equal to zero. Next, Eq. (3a) involves the multiplication of the forcing function by cos iflt dt and integration from t = 0 tot = 7; = 277/0. From the antisymmetry of P(t) and the symmetry of cos iOt with respect tot = 'TT/fl, we conclude again that the integral in Eq. (3a) is zero, so that a; = 0. Finally, considering Eq. (3b), it is apparent that P (t) from t = 0 to t = 'TT/0 is symmetric about t = 'TT/20 , while from t = 'TT/0 to t = 27T/O it is symmetric about t = 37T/20. However, when i is an even integer, corresponding parts of sin iOt are antisymmetric with respect to t = 7T/2fl and t = 37T/20. Thus, for all even values of i, we conclude that b; = 0. When i is an odd integer, both P(t) and sin iOt are antisymmetric with respect to t = 'TT/0, and Eq. (3b) gives 7T
'l'l111s, l!y using Bqs. (3), we can calculate the Fourier coefficients a;, b;, and ao 1111 1111y known periodic function. . . Assuming that the forcing function has been decomposed mto ~ Fo~ner H1•iit•N, W<.! t.:1111 now write the equation of motion for damped forced v1brat1ons I
cu + ku
= a0 + a 1 cos Ot +
+ . + b 1 sin flt + b2 sin 2flt + . . . a2 cos 2flt
'l'ht• 101111 solution of this equation consists of the sum of free and forced vihrnfio11s assot.:iated with initial conditions and the terms on the right-hand side. ll1•c11111w of damping, the free vibrations will disappear with time, leaving only thr lort·t·d vibrations. Each harmonic term in the Fourier series contributes a 1,lt•111ly sfnt<.! forced vibration of the type described in the preceding section. 'l'lm1t'1111·t· , we conclude that forced vibrations with large amplitudes can occur wh1•11t·vcr th<.! period of any term in the series is the same as (or close to) the 11111111111 pcfiod of the SDOP system. This always happens when the period 7; of tlw lrnl'ing function is equal to , or an even mu ltiple of, the natural period -r, of lht• dn111pl'd syst<.!111. I .t·l 11s now apply the 1111: thod of this st·t·lio11 to lht• piccl•wise-linear forcing llrm 11011 ill11slrntl'd i11 Fil\ , 2. 12. To
O
40
-
7T
l1r/20 P(t) sin iOt dt 0
2 P(t) = -P1 0t
(5)
(6)
7T
where P, is the maximum value of the forcing function. Substituting this expression into Eq. (5) yields b;
(4)
l21r/OP(t) sm. iOt dt =
In the time interval O ::s t ::s 'TT/20, the function in Fig. 2.12 has the formula
IIN 111fi
37
r
fi b; = -
1111111lly , 111ultiplying Eq. (1) by dt and integrating yields
a0 =
the i 111t•grnl
0
111 11 si111ilar manner, multiplication of Eq. (1) by sin iOt dt and integrating pt rnhlt'l'S
b; =
111111111111 I 1111:l1111 I 11111Jtl1111•
P (t) dt
a1 cos jOt cos iHt di =
(2)
('' h1 sin j!lt cos iOt dt
tine , J.l,
2 BP, 0 l1r/20 . . 8P1 li1r/2 . =-2t sm 1 Ot dt = Tz v sm 7T
o
v dv
17To
in which the new variable is v = iOt. Integrating this expression by parts and substituting the limits produces • i'TT = BP, (- )Ci-IJ/2 b'. = ·2BP1z sm 2 ·2 2 I l 7T l 7T
(7)
In this formula the odd values of i are 1, 3, 5, 7, and so on. Thus, the Fourier series that approximates the periodic function in Fig. 2.12 is 8P1 ( sin • 1 sm . 30t + 1 sm . 50t - ...) P (t) = -:;;i Ot - 32 52
(8)
und we need only superimpose the sine curves with odd numbers of periods in 1he intervul t O to t 27T/ H . If we omit do111ping und use II frcq11<.! ncy ratio 0/w = 0.9 , the magnifirntio11 tal·to, for tlw 111 st lt:1111 in Eq . (H) is
Syste m s with One Degree of Freedom
.Ill
Sec. 2.6
Chap. 2
Arbitrary Forci ng Functions
39
p
/3, =
1 1 - (O/w)2
= 5.26
whereas that for the second term becomes
/33
1 =
1 - (30/ w)2 = -0.159
1<'1111hermore , a multiplier of 1/3 2 appears in the second term ofEq. (8). Therefm1.·, we conclude that using only the first term for the response U
8 -P1 /3 Slll . :Ht = -7T2 k t
p
(9)
0
l'llltscs an error of less than 0.4% in the solution. Suppose now that the applied force in Fig. 2.12 is replaced by a periodic Ht'm111d acceleration of the same form. If the peak acceleration a replaces the 11111xi111um force Pi, we need only use -ma instead of Pi in the response calculnlions. For example, the result in Eq. (9) would be changed to
. at 0 u = - - 8 -ma/3 , sm 7T2
Figure 2.13 Arbitrary forcing function.
(10)
k
(3)
whk h is only the first term of the series.
2,8 ARBITRARY FORCING FUNCTIONS Wl' now consider dynamic loads that have no periodic character and may v~y
which is drawn from Eq. (2.4-10). Because each incremental impulse between t' = 0 and t' = t has such an effect, we obtain the total displacement due to the applied force as the integral
111 1111y nianner with time. Such arbitrary forcing functions must be handled m 11 11pl•ci11I way , as described in this section. , . . Fi gure 2. 13 shows a general forcing funct~on P (t ) that 1s express~d ~n It 1111s of a new ti me variable t'. The value of t' 1s less than that of t, which 1s thl• time when the response is to be calculated. If the function is applied to a tl111111wd SDOF system, the differential equation of motion becomes
mu + cu + ku =
P(t')
(1)
/\1 11uy instant o f time t' , we may calculate an fncreme~tal impulse P dt' , ll'j)ll'Scntcd by the hatched strip in Fig. 2.13. That impulse 1m~arts to the mass 11, 1111 instantaneous increase in velocity (or incremental velocity) equal to
d,:i
p dt' =- = p,,, dt,
u
= -e-nt Wd
(4)
(2)
m
l'llllrlt1d1• 111111 lhl• /11r11•1111•t1(/// IIIIY 11111'1 111111· f wtll lw WI'
em' Pm sin wit - t') dt'
O
This mathematical form is known as Duhamel' s integral. Equation (4) gives the complete displacement produced by the forcing ~u~ction P(t'), acting during the time interval from Oto t. For periodic forces 1t includes both steady-state and transient terms and is especially useful in studying the response of a SDOF system to any kind of short-term (or transient) load. If the function P (t ') cannot be expressed analytically, we can always evaluate the integral in Eq. (4) by some numerical method. To include initial l'Onditions for the damped system, we need only add their effects to Eq. (4), as lollows:
'l'his l'X pr1.·ssion is vu lid rcgnrdlcss of what other forces (such us the spring f?rcc) 11111y hl• nl'ling on tlw mnsi., nnd n:g11rdlcss of i( i,; displ11c.·tmcnt t~n~I . velocity . at thl• 1l11w 1 1 'l'H•at\111-1 1lw i11t·n·111l·11t11l Vl'lol'lty us ii 11 Wl'll' 1111 1111t111I velocity (111 till' lillll' ( 1 ),
l'
,//,1'/lftll 't'lllt'III
of till' lllllSH
Ill
t I htN l'q11111io11 11°pll'li1' 11I N till'
lol11l 1wlulio11
I
<-11,,
l' e"', o
p,,, sin w"(t - t') dt'
or l!q . {I) .
]
(5)
Systems with One Degree of Freedom
40
Chap. 2
If damping is neglected, we haven == 0 and wd == w, and Eq. (4) reduces
Sec. 2.6
Arbitrary Forcing Functions
41
Pm == P1/m is constant and acts for an indefinite time. For the case of zero damping, Eq. (6) gives
to
l' . ( ')
u == -1
W O
Pm sm w t - t
(6)
dt ,
. u = Uo cos wt+ -u0 sm wt+ -1 W
W
l'
W O
l11duding initial conditions without damping also gives a simpler form of Eq. ('I), which becomes
u == Pm -
l' .
, ,
Pm sm w(t - t ) dt
Integration of this expression yields u
= p'"(l 2
(7)
O
sin w(t - t') dt'
w
COS
wt)=
Pi(l k
COS
wt)
(8)
From this result we see that a step force P1 produces free vibrations of amplitude
'l'l11s briefer equation for the total response can be used whenever damping is not
Pi / k superimposed on a static displacement of the same magnitude, as depicted
Nigni !icant. To demonstrate use of the Duhamel integral, assume that a constant force / 1 Isec Fig. 2.14(a)] is suddenly applied to the mass of the SDOF system in Fig. 1 J, <)(u) . This condition of dynamic loading is called a step function, where
in Fig. 2. 14(b). Thus, the maximum displacement due to a suddenly applied force is twice as large as that caused by the same force acting staticaJly. Next, let us consider a step load that acts only for a period of time t 1 • This type of forcing function is called a rectangular impulse and is illustrated in Fig. 2. lS(a). During the time when the force is nonzero, the response of an undamped SDOF system is the same as that given by Eq. (8). On the other hand, the response after time t, may be found by evaluating the Duhamel integral for two ranges: 0 to t, and t 1 to t. Only the integration over the first range will produce nonzero results, because the forcing function is zero in the second range. Altogether, the solution for this case is summarized as follows:
p
u = u 0
kPi (1
(8)
- cos wt)
= Pi k[cos
w(t - t 1)
-
cos wt ]
(9)
t'
The same results may be obtained by considering the rectangular impulse in Fig. 2.15(a) to consist of the sum of the two step functions, as indicated in Fig. 2.lS(b). The first step function (of magnitude P1) begins at time t == 0, while the second step function (of magnitude -P1) begins at time t = t 1• A third method for determining the result given as Eq. (9) involves finding the displacement and velocity of the SDOF system at time t,, using Eq. (8) and its first derivative with respect to time. Thus,
(a)
u
(10) II these two quantities are treated as initial conditions at time t 1, the ensuing I1t'l' vihrali onal response may be calculated from II
M111111 .1,1,1
1111 "1t, 11111111111111. 1111 ,,
fil"'""
11,, cos w(I -
11)
u,,
+ - sin w
w(t - t 1)
(11)
S111I1-111111ti111111l 1·x pn·Msio11:, ( 10) 11110 Eq. ( 11 ), followed by trigonometric manip11111111111, y1rld, thr ~11111r 11·,1111 11, lhnt 111 hi , ( 11)
v11111111• with 01111 l>11u1nn "' I 1n11!111111
<.111111 'J
/l
,111 ,
in
A1ltl1tury I 111111111I1111111111111
43
d1~pl11t•t·111t·11t 11>1< ol tlw 11111ss relative to grnund . Thus, f'or a step acceleration a, ol llw ground, l:q. (8) bccomcs
u"' - - ~( I - cos wt) = - ma , (1 - cos wt) 2
w
k
(12)
111 addition, the second part of the response to a rectangular function for ground 1wt·eleration is
ma, u* = - k[cos w(t - t 1) 0
t'
t,
-
cos wt]
(13)
which is a modification of Eq. (9).
(a)
1r.1mmple 2. 7 l•l11d the undamped response of a SDOF system to the linearly increasing force, known mmp function, given in Fig. 2. l 6(a). The rate of increase of the force P (t ') per unit 111 lime is P,/t1. As indicated in the figure , the forcing function for this example is
p
IIN 11
/ P=P, P,
(a) ~11
lhc force per unit of mass becomes (b)
Applyi ng Eq. (6) to this case, we have
l'
u = -P,- t' sin w(t - t') dt ' mt1W o l111l11,11·11ting this expression by parts yields
/ P, - - - - - - - - -
P= - P,
(b)
Figure 2.15
(a) Rectangular impulse; (b) step-load simulation.
Depending on the duration time t 1 for the rectangular impulse, the response SDOF system may take different forms. However, the maximum response l'II II ,wvcr be more than the value 2P1/ k. Instructions for determining loci of rt•1ipo11st· muxima due to various types of forcing functions appear in Sec. 2.8. I I the dumped SDOF system in Fig. 2.9(a) is subjt•ctcd to arbitrary wound ,u·,·,•l,•mtio11.1·, we need only rcplucc lhc f11111;tio11 p111 (t ') in liqs. (4) through (7) wilh tht· l11nl·tion fl~ (t ') ii~ (11 ). In imch t'lllWH tht•st· t·quutions yield the
111
11
. wt~ u = -P1 ( t - -l sm kt, w
-'-------"'-------
(c)
11111N, we sec that the response to a ramp function consists of the sum of a linearly l111 w11Ning static displacement P1 t/ kt1 and a negative sinusoidal free vibration of ampli111d1• l'1 /kt1 w, as shown in Fig. 2.16(b). The velocity at any time tis equal to the first derivative of Eq. (c) with respect to llllll• That is, Li
= .!'.!_( 1kt,
cos wt)
= .!'kt,.!_ (1
- cos 2T7Tt)
(d)
I 111111 !his 1·xprcssion we conclude that the velocity is zero at times t = 0, T, 2T, 3T, and 11 1111 'l'lll'wf'on.', the slope of the displacement curve in Fig. 2. 16(b) is zero at t = 0, 1 I 1, 11; 1111<1 .~o on . Furthermore, the velocity is always positive and has a maximum v11l111 111 1/ '1/At1 11t tinws r T/2, 3'/'/2, 57'/2 , und so on rscc Fig. 2.16(b)].
II w1• linw 11 11111:p µround 11,t'l•k'rution i11sll•ud of an applied force, it may be
Systems with One Degree of Freedom
44
Sec. 2.7
Chap. 2
45
Step-by-Step Response Calculations
Similarly, the velocity in Eq. (d) is replaced by p
u*
(g)
= - mai (1 - cos wt)
kt1
The term - ma 1 appears in these results instead of A.
Example 2.8 Suppose that the ramp function in Example 2. 7 is terminated at time 11, as shown in Fig. 2.17. Determine the response of the undamped SDOF system in the time ranges 0 :S t :S t1 and 11 :S t. p
p = P,t'
(a)
t, P,
u
:IP,T
0
kt,
P,t kt,~
Figure 2.17 Triangular impulse.
/
The function appearing in Fig. 2.17 is called a triangular impulse. For the first lime interval (O :S t :S t 1), the response is the same as for the ramp function, as given by Eq. (c). In the second time interval, we must integrate over two ranges, as for the rectangular impulse. Doing this and summarizing the results produces
/ '/I 2P, ~ kt, 1
I', I
~
(c)
U = -A~t - -1 Sill . Wt
kt,
kt,
2T
3T
(b)
.!. sin wtl
(t, :S t)
(h)
If the triangular impulse is a ground acceleration with maximum value equal to Il'Sponscs in Eqs. (c) and (h) have u* in place of u and -ma, in place of P1.
a,, the
u = 0
.fl [t,
w
kt,
cos w(t - t,)
+ .!. sin w(t - t,) w
w
J
Figure 2.16 (a) Ramp function; (b) response.
'J.,7 STEP-BY-STEP RESPONSE CALCULATIONS l'II J>l\'NNl'd LIS
(c) wh1•11· 11 IH thl' f11'01lnd lll'l'l' lo l'llli11111111lnw 1, . Thon 1l1l' n.isponHC in llq . (c) is converted 1
tn lh1• 11•l111l w 11111111111 1('
I ~Ill 1111)
,11
(I}
111 11111t1y practical problems the forcing functions are not analytical expressions !tut urc represented by a series of points on a diagram or a list of numbers in a l11lill•, 111 such cases it may be feasible to replace the data with certain formulas hy l'ttt Vl' titting 111ethods und then lo use those formulas in the Duhamel integral. I l11wrw1 11 11101l' l,ll'lll' l'III m~thod fol' t•vnlunting the response consists of using 111 11111· 1 u111pll· 111ll' 1p11l11lim1 1'1111ctl1111 in II rnpt1titiw 1wdcs of calculations. The 1
Systems with One Degree of Freedom
46
Chap. 2
Sec. 2.7
47
Step-by-Step Response Calculations 2
latter approach is discussed in this section for piecewise-linear interpolation .fimctions. Figure 2.18 shows a forcing function of general shape that is approximated by a series of straight lines. For a particular interpolating line in the time interval 1 -c;; t :s tj+ 1, the response of a damped SDOF system may be written as the sum 1 of three parts, as follows: U
Using the definition t'
=
= U1 +
U2
+
U3
tiI'j [ w2 t' - 2n + e- nr ·(2n cos wdt' - w~ wd - n sm . = !itjkw wdt ' ) ] (Zc) 2
This formula is available as the solution of Prob. 2.6-3 . By differentiating Eqs. (1) and (2) with respect to time, we can also find velocity expressions in three parts. Thus, (3)
(1)
U3
where
t - tj, we have for the first part
(4a) (2a)
'l'his equation contains the free-vibrational motion of the system due to the displacement uj and the velocity uj at time t = tj (the beginning of the interval). The formula for this portion of the response is drawn from Eq. (2.4-10).
and (4b) Also ,
p
ih = ~[1
u~k
[1 -
e-ni'
(cos wdt' +~sin WJt')] ~
(4c)
At the end of the time interval !itj, the displacement expressions in Eqs. (2) become
1'11
( u 1) i+ 1
=
e
_ 11 !!.,.(
"
\u.j cos WJutj +
(u 2)i+1 = l1 k [1 - e-11!!.1-1 ( cos
I 'I
Uj + nuj . " \ WJ sm wdutj)
" n sm . wdut " 1\) wdutj + WJ
(Sa)
J
(Sb)
(Sc)
111 addition, the velocity expressions in Eqs. (4) are rewritten as
Figure 2.18
= e- "l!. 1 [-
(u2)1, 1
= ljw
(,i 1)1 1 1
= ~~i [1
1
(ujwd + n ui :dnui) sin wd!iti + u1 cos wd!iti]
(6a)
Piecewise-linear interpolation of forcing function.
The other two parts of the response in Eq. (1) are associated with the 11t111ight line rorcing function in Fig. 2.18. The one caused by the rectangular l!llpulse of magnitude ~ is "i
(u1 )1+ 1
I
ljk I
, , 111'
( -os
\
II
'""
sin
,,,,,1 ')
I
2
kwt1
e - "tJ.'1
(6b)
sin wd!it-1
- e ntJ.r1(cos w,,titj +
;d sin wd!itj)
J
(6c)
(2b)
wl11l'11 c1m11·11 1111111 1111· s11l11ti111111t 1'111h ) ,, ) < >11 tht· nth1~1 h1111d , till' t•llt·cl of th,• II 11111pul111 l111p11l111• 111 11111v111t11d1· Ali 'i 1 1 , ; 1!1·1·111111 11
h111111i1111s (5) and (6) <.·. rnistitute rec111T<'IIC(' formulas that may be used to calcu11111' tlw d11111pvd n•s p1111 Nl' tit the cud of stt•p j and to provide initial conditions at
th1 hr i,ti1111i111-1 ol st,•p ./ I I ,
48
Systems with One Degree of Freedom
Chap. 2
Sec. 2.7
Step-by-Step Response Calculations
If damping is neglected, Eqs. (5) for displacements simplify as follows: (u1)i+t = ui cos w!l.ti p t(l -
(u2)j+I =
COS
+ !!i. sin w!l.ti
(7a)
w!:J..tj)
(7b)
w
(u3\+1 = A!l.Pki (w!l.ti - sin w!l.tj)
49
p
P,
P, sin flt
(7c)
utj W
und Eqs. (6) for velocities become (u1}j+1 = - ui w sin w!l.ti
+ ui cos w!l.ti
(8a) (8b)
- P,
(8c)
(a)
Equations (7) and (8) are simple enough for hand calculations to obtain approxi111utc results. Of course, we need not take the shaded impulse in Fig. 2.18 as the sum ol' n rectangle and a triangle. Alternatively , it could be divided into two triangles, ns indicated by the dashed diagonal line in the figure. Then it would be possible to express the second and third parts of the response in terms of ~ and ~ +1. Furthermore, if the time step !l.ti is constant, the coefficients of uj, uj, I'j, and~ (or 0, ,) all become constants for both the displacement and the velocity expressions. I lcnce, these coefficients need be computed only once and then used tl'pclitively throughout the numerical solution [3].
- 1
Ex111111>lc 2. 9
- 2
Fi1tt1t l' 2. t 9(a) shows a forcing function P = P, sin flt that is applied to an undamped Sl>Oli system. The function is discretized by piecewise-linear interpolation into 20 equal tl111l' steps of duration !::.t = 'Fj-/20. Using the method of this section, calculate approxi11111tely the response of the system. Assume that the initial conditions are uo = Lio = 0 ,
u 2
0
T,
T,
4
2
- 3 4
thl' vulucs of P1 and k arc both unity, and the frequency ratio is 0 /w = 0.9. From Prob. 2.6- 1, the exact solution to this problem (with zero damping) is
u
r. = kP, (.sm a/ -
111 which the nrngnificution fuctor
/J •
I1
A1111pp111x1111111r ~11!11111111 IN 111111111
){3
(b)
0. . wt w sm
(a) Figure 2.19
f3 hus the value
(~l/,,,)J
I · -1- !-0-,1-,)l. ~.) <, I
hy 11pply111p ht~ ( /) 111111 (H) 11'1 III NIVl'ly
(h) Ill I()
HJll'l'ifll'd
(a) Sinusoidal forcing function; (b) approximate responses.
tl11w stl•ps. Results of such calculations (by hand or computer) are summarized in Table 1 I Also riven in the table arc exact displacements obtained from Eq. (a) . As expected, till' 11pprnx1111Ull' displut'l' t11entH urc slightly less than their exact counterparts, because 11111·111 111h•1 pul11t1011 ol' llw siiw t·111vc is impt·rf'ect. Decreasi ng the step size would, of 111111~1·, lrncl to 1•xm·1 v11lm·N (l'Xl'l'J)I Im 1rn11HI ofl l'crms) .
Systems with One Degree of Freedom
110 TABLE 2.1
P;
Uj
0.309 0.588 0.809 0. 951 1.000 0.95 1 0.809 0.588 0.309 0
I '},
I
·1 ~
(1
I K 1)
10
0.006 0.048 0.154 0.338 0.593 0.896 1.203 1.461 1.613 1.607
Exact 0.006 0.049 0.156 0.341 0.598 0.903 1.213 1.473 1.626 l.620
Chap. 2
u
Solution for Example 2.9 j
11 12 13 14 15 16 17 18 19 20
Uj
Exact
1.407 1.000 0.404 - 0.338 - 1.151 -1.945 - 2.616 -3.068 - 3.220 - 3.020
1.418 1.001 0.407 -0.341 -1.161 -1.961 -2.634 -3.094 -3.246 -3.045
P; -0.309 -0.588 -0.809 -0.951 -1.000 -0.951 - 0.809 -0.588 -0.309 0
51
Response Spectra
Sec. 2.8
0.8
0.6
0.4
0.2
The approximate undamped response in Table 2.1 is plotted as the solid curve in H u, 2. I 9(b) . Also shown in this figure is a dashed curve that results from recursively 11ppl y111g Eqs . (5) and (6) with the damping ratio y = n/ w = 0.05.
t,.
t,o
0
1,:,1m1111>hl 2. lO
(b)
l 1lp1111· }.. 20(a) gives a series of plotted points simulating a blast load that impinges on 1111 1111d11 mpcd SDOF structure, such as the building frame in Fig. 2.l(c). Note that the hlnNI lon:c rises quickly to the maximum value P1 and then diminishes more slowly (and 1·v1~11 ht•comcs negative for awhile). In this case we have 16 equal time steps, each of wltl\'11 hns the value /:J,,t = T /30. Apply the method of this section to find the approximate 11•Npo11st· of the structure. Let the values of both Pi and k be unity, and assume that the l11it11il rnnditi ons are uo = uo = 0. A I lcr using Eqs. (7) and (8) recursively to obtain the time history of the response , wu 1h1111 list the calculated displacements in Table 2.2. For this problem there are no exact
Figure 2.20 (cont.) TABLE 2.2 Solution for Example 2.10 j
1 2 3 4 5 6 7 8
/l
P;
Uj
1.000 0.850 0.720 0.590 0.475 0.360 0.250 0.155
0.007 0.050 0.127 0.230 0.350 0.474 0.594 0.699
j
9 10 11 12 13 14 15 16
P; 0.070 0 -0.050 - 0.080 -0.100 -0.080 -0.050 0
Uj
0.780 0. 830 0.844 0.819 0.755 0.654 0.521 0.363
111~11l ts with which to compare these numbers. Nevertheless, we have great confidence 111 thl'ir validity. Figure 2.20(b) shows a plot of the time history of undamped response 11111 Nolid curve) , which appears to be quite reasonable. Also appearing in the figure is 11 1l11Nh1·d curve that gives the effect of using Eqs. (5) and (6) with 'Y = 0.05.
,II R SPONSE SPECTRA
()
dy11u111k louds discussed in this chapter cause vibrational responses in I 111tli iiyilll' lltS, nnd the maximum values of these responses may be less than, , 111111 1111 , 111 g n •nl l' I' 1h1111 the corresponding static responses. For a SDOF system, 1111
1
11 11 11111111 11 1 pt" ind
1111
11'1111111•
10
(II) i l1N1 l 1 il11 d iilllNI 11111d , t i,) ll)ll•t11,l11t1tl1
II
~1'1111"1I
11,
,1
(rn Ill' l[lll' lll'Y ) is llw ch11rnc1cristic that dclermines its response ptv, 11 1111, 1111 h111l·l11111 111 11ddit1l111. !Ill' slrnpl' tllld durntinn of' the forcing
Systems with One Degree of Freedom
62
Chap. 2
function itself play important roles in the response. Plots (or loci) of maximum icsponse values against selected parameters of the structure or the forcing funclion are called response spectra. Such diagrams are useful in design because they provide the means for predicting the ratio of the maximum dynamic stress in a slructure to the corresponding static stress. The time at which the maximum 1l.lsponse occurs is also of interest, and plots of this variable usually will accompnny the response spectra discussed in this article. Let us· reconsider the rectangular impulse shown in Fig. 2.2l(a), which wns applied to an undamped SDOF system in Sec. 2.6. If the duration ti of the i 111pulse exceeds the value T / 2 (half the natural period), the maximum response is always equal to 2. On the other hand, if the time ti varies from zero to T/2, lhc maximum response goes from zero to 2. Figure 2.2l(b) shows the early singes of a series of such responses for ti = O. IT, 0.2T, 0.3T, 0.4T, and 0.5T. Note that the maximum response always occurs in the first excursion, which is lhc only part for each curve exhibited in the figure . In all such cases , the muximum displacement occurs after the impulsive action terminates, because lhc velocity at time ti is positive. Thus, to find the maximum value of the response and the time when it Ot'l'lll'S, we must examine Eq . (2.6-9), for which ti s t. That equation may be w1 illcn in dimensionless form as
u
-
= cos w(t - t 1)
-
cos wt
(1)
Sec. 2.8
63
Roaponoo Spoctro
p
P, l - - - - - - - - ,
0 (a)
I
2
Ust
whl.lrl.l the static displacement due to P1 is Ust
I
o.Jr I
P,
k
=
(2)
I I
l>ilfcrcntiating Eq. (1) with respect to time, we obtain
~=
--.~o.2r
lly setting the term in brackets equal to zero, we find an expression involving lht• time t,,,, at which the maximum response occurs. That is ,
= sin w(t,n
sin wtrn
-
I
I
ti)
I I
'l'hl• following relationship satisfies this equation:
wt,,,
7r
Wf1
2+ 2
(3)
I lt•nl'l' . t,,, is II lillL'Ur fun<: tion of t 1• Substituting lhc formula for wt,,, from Eq. (3) 11110 Eq . (I) yil•lds l S Iii
111/ 1
1
\ )( l
I I I I I
w[sin wt - sin w(t - t 1)]
Ust
II
o.s7
.L,,.-"r---.10.47
(4)
7
7 4
0
2 (b)
1,·1Murl.' 2.21
(11) Rl'Clllnl!uli11 impulse; (b) response curves.
ll4
fly•1111111 wlllt Ottn l>nu11tn 11I I 1nnclo111
1;111,p. 'J
ll11npo11111 ! ,p111 I111
55
tl~ing Hq11 . ( 1) 111HI (•I), w,• c11111011111111111:1.c the rcspo11111· Np1•1 l111111 lot the rcct 1111~111111· i111pulsc us follows : Um -us, = 2 sm. -7Tf1 T
t,,,
I
t
T
4
2T
(5a)
I ~
-=-+-1
(5b)
HIid
1 II
(5c) t/11
- =T 2
I >l11gl'll111s of' these dimensionless expressions appear in Fig. 2.22 , where u,,,/us, 1111d t,,,/'J' nre plotted against t 1/T. From Eq. (5a) we see that if the time t 1 is less llu111 '/ '/6, the dynamic response is less than that caused by applying the load Pi Nl11lk11lly. On the other hand, if the time t 1 is between T/6 and T/2, the value ol 11,, / 11H1 is between 1 and 2. Al this point we note that plots of magnification factors for forced vil>rnllous constitute response spectra, as defined in this section. Figure 2.1 l(a) 1·011t11ins u series of curves for /3 = um/ U 51 plotted against the frequency ratio U/ ,,,. We recall that these curves represent only the steady-state part of the 11•sponsc nnd that a different curve is obtained for each level of damping. If the l11111sk·n1 rurls of the forced vibrations were to be included, the response spectra 111 Fi1', 2. 11 (ii) would be somewhat higher, but this effect is of little significance. u,,,
-
4
2
4 (b)
Figure 2.22 (cont.)
I 1111h,•1111ore, while damping is of great importance in the problem of forced
, 1l1111flo11s, it is often omitted as a consideration in response spectra due to f111pt1INiVt' excitations. Small values of damping have littl~ ef~ec~ on such re111111111• 111nxima, which usually occur before much energy 1s d1ss1pated. How• ,., 1, 11 group of damped response spectra can always be co~structe? for any 1t111 l11u function, with a different curve for each level of dampmg. This may be ,11 111111plishcd for simple cases by deriving the appropriate analytical functions , 11111 Im rn1nplicated situations a numerical approach must be used. 1,11111ph• 2. 11
u.,
')
t, o'---- -~,----~,----3:------:---- T
(5d)
---
- - - .;;-------------
I IJ'llll' ) 23(u) shows a ramp-step Junction that increases linearly from zero to Pi in the 11 11 11 11 1111<1 is constant thereafter. From Prob. 2.6-7, the response of an undamped SDOF v•t• 111 In lhis excitation is
u
= !:2_ (.!. _ sin wt) k
II -
(O ::;; t ::;; t,)
wt1
t,
Pi [ I + ---'-----sin w(t - !1) - sin wt] k
(a)
(b)
wt1
11 11 11hji•1•1lvc of this example is to determine the response spectrum and the correspondt11i• 111111• l1111c1ion. . . 0
---,----"',----3 ._____ ....____ t,T
ll y i11spcction of' Eqs. (a) and (b), we see tha~ th~ max_1mum response will always
11l1t•1 111111: 11 • Therefore, only the latter equation 1s of mterest here, and 1t may be , •111, MN1•d 111 llw dinwnsionless form 111,111
4
2
II (11)
1/ 1
1'111111·1• 2.22
(11) Rt•NpttONl' Nl)l'L'l111111 fm 11•1•11111~111111 l111p11IN1•: (h) tlim· ol' 11111xl 11111m
I
l
/1) -
11>(1
ll' NPIIIIM' .
1tllh 1, 11111111111{ l\q .
(1•)
with 11•sp1•1•f lo 1111w, w,• oh111i11
sin wt]
(c)
56
Systems with One Degree of Freedom
Chap. 2
Soc. 2.7
Response Spectra
57
u
p
1
- = -[cos w(t - ti) - cos wt] Ust ti
Setting the term in brackets equal to zero gives an expression containing the time tm. Thus, COS Wtm = COS w (tm - t1) This equation is satisfied by wt,,, = (a)
wt1
1r
+2
(d)
As for the rectangular impulse, t,,. is linearly related to ti. Substitution of wtm from Eq. (d ) into Eq. (c) produces
1
Um = l + ~ sin wti = 1 ± - - V 2(1 - cos wti ) u,. wt, 2 wt,
(e)
'l'hese expressions represent both maxima and minima, which depend on the value of wt1. Summarizing for maxima, we have
2
u,,,
Ust =
lm
T ' - - ~ ~~--'-~~~~'--~~~-'-~~ ~~.l.._~~ o 2 3 4
t,
r
(b)
l
+
T
I sm-T I .
1Tl1
1rt.
(f)
= .!_ + ~ 2
(g)
2T
Figure 2.23(b) and (c) contain dimensionless plots of the response spectrum from liq. (f) and the time of maximum response from Eq. (g) . We see that the response Hpcctrum has its highest value of u,,,/u,, = 2 when t, = 0, for which the input becomes 11 step function. Whenever t1 ~ Tthe value of umdoes not exceed u, , very much, and with 11 large rise time the loading is essentially static. 11:xumple 2.12
,,,
A dumped SDOF system with harmonic ground displacement u8 = d cos flt appears in 1111-1. 2.24(a). Let us derive an expression for the response spectrum of the steady-state frn cc in the spring due to that influence. For this purpose, we need only find the Nl1°11dy-state relative displacement u* = u - ug and multiply it by the spring stiffness k. By taking the second derivative of the ground displacement with respect to time, Wl' obtain the absolute ground acceleration as
T
u8
1
2
1 '1
l 1 'J:
3
(c:)
Mw1111• 1.z.1 ll' ~pttllHI'
(h)
I h1·1cfore, the equation of motion in the relative coordinate u* becomes mu*
0
= -fl2 d cos flt
( 111 H11111p Hh'p 111111111111,
t, ----'----- T
"
(hl 11·~p1111 H1· Hiu·, 11,1111 , (, > 111111 " ' 11 11i,d 11111111
+
cu*
+
ku* = -mug = m02 d cos flt
(i)
'11111pudng lhis relationship with Eq . (2.4- 22), we see that u* has replaced u and mfl2 d h11Htl'pluccd P. llence, the steady-state response [see Eq. (2.4-31)) becomes
I
11•
en k cos ~£1 /3 111!l2d
-
0)
(j)
111 wh11'11lh1• 11111~111hl'lllior1 Im Im /I 111111 lhl· ph11s1· 11111-1k· 0 111c given hy Eqs. (2 .4- 32) and ( I •I I\)
68
Systems w ith One Degree of Freedom
Chap. 2
Chap. 2
59
Problems
Dividing this equation by F,, = kd gives F - = {3* cos (Ot - 6)
(n)
F,,
<)f course,
the sine function used in place of the cosine function would produce similar l'SUlts. Figure 2.24(b) shows dimensionless plots of the response spectrum {3* = F,,./ F,,, 11g11inst 0 / w for various levels of damping. When the frequency ratio 0 / w is zero, the Npring force is zero; and when the ratio is large, the spring force approaches the static vuluc, where {3* = l. On the other hand, at resonance (0 = w ), the magnification factor W becomes the same as {3. I
~ uu
= dcosnt
u (a)
The forcing functions treated in this section lead to explicit expressions for 11111 / 4.0
1-tt•ncral, it is difficult to identify the time range within which the maximum ll'Nponse occurs . In addition, the equation containing tmis usually transcendental 1111tl cannot be solved explicitly. Under these circumstances , values of unJus, and t,,,j'r must be found by exhaustive calculations, using a series of values for the tt111c ratio t 1/T. For each ratio the expression for u/u,, in terms of t/ T may be plotlcd, and a value of u,./ us, as well as tm/T may be obtained therefrom.
I
1..-- 'I
r,
=0
.-- 0.10
3.0 ..- 0.15
13*
·"'~
2.0
-~ \
Jl'-'
1.0
0.25
~
REFERENCES
~0.375
~~
0.50 _.-,;:: .__....
1.00
1
~
0
us, and tm/T. but it should be mentioned that these are exceptional cases. In
1.0
2.0
3.0
4.0
Ti moshenko, S. P., Young, D. H. , and Weaver, W. , Jr., Vibration Problems in l:'11Rineering, 4th ed., Wiley, New York, 1974. Wilson, J. S. , "Performance Characteristics and Selection of Accelerometers," J. Sound Vib., Vol. 12, 1978, pp . 24-29 . C'rnig, R. R., Structural Dynamics, Wiley, New York, 1981.
,!l w
PROBLEMS
(b)
l•'l~ure 2.24 (a) Dam~d SDOF system with harmonic ground displacement; (b) 1cspo11sc spectra fo r spring force.
To improve the form of Eq. U), we use w2 = k/ m and define the new 11111g111fkntion factor, {3 *, as follows : {3* -
{3(~)2 = Vl
I
(0/w) 2 (0/w)r lr I (2'Y0 / wP
,i I. For the prismatic cantilever beam in Fig. 2.l(a), find approximate expressions for the angular frequency w, the natural frequency f, and the natural period T :t,~•.2. D1:lcrmine approximate expressions for the values of w,f, and T for the simply ~upportcd prismatic beam in Fig. 2. l(b). :t,l• I , Assume that the overhanging beam shown in Fig. P2.2-3 has constant flexural 11gidity W , and find approximate values of w, f , and T.
(k)
D(t)
lh1•11 11•w11k Eq . (j) in the n·dut·cd form II*'
' ll11·
~h·11dy
N( illl'
1
fl+ ,I l'ON({ lt
lrnn· Ill 1111· ~1111111• 11111y
llllW
fl)
{{')
/ fi A,/ , IIH ( 111
//)
~ ----1 ...~ - - ~ ~, - - ·Im
---- ----+ - - -l - j
Ill" tllkl'II IIN
(111)
11'111111·1• l'l.J. .I
60
Systems with One Degree of Freedom
Chap. 2
Chap. 2
61
Problems
2.2-4. Figure P2.2-4 sho~s a body of mass m hanging from a massless cable having length L, c~oss-sect1onal area A, and effective modulus of elasticity E, . What are the approximate values of w, f, and T for this SDOF system?
r
Figure P2.2-6
2.3-1. Suppose that the SDOF system in Fig. 2.6(a) is subjected to the harmonic ground displacement u8 = d cos flt . What will be the steady-state forced vibration of the mass m due to this influence? 2.3-2. If the SDOF system in Fig. 2.6(a) experiences the harmonic ground acceleration ii8 = a cos flt, determine the steady-state forced response of the mass m relative to the ground. 2.3-3. A harmonic force P sin flt is applied to the mass m at the center of the fixed beam shown in Fig. P2.3-3. Find the steady-state forced vibration of the mass, assuming that the flexural rigidity El of the beam is constant along its length and that fl= w/2.
L
Psin Ot
Figure P2.2-4
center tt 1s connected to a flexible, massless rod that has a circular cross section w!th radius rand len_gth L, as indicated in Fig. P2.2-5. Find approximate values ol w, J, and T for this problem, assuming that the bar can only rotate about the uxis of the rod.
tm
~
l,2•5. A rigi~ b_ar AB of mass density p has cross-sectional area A and length L. At its Figure P2.3-3
1·
~
• L 2
.1.
L 2
.1
2.3-4. Figure P2.3-4 shows a rigid bar AB attached at its center to a flexible rod of length L, having a circular cross section with radius r. Determine the steadystate rotational response of the bar to the harmonic moment M cos flt, where fl= 2w. A
L
-2r
-
M cos Ot
2r
I
I
')
')
-~
Flflurc P2.2-5
l,l•<,. For lhl'. 11,wd (ll''.1111 Nhow11 111 111H , P.2.} I! , dt1h11111l1w 1lpproxl1111,tl' cxpn:ssions lot 1,1, / , 1111d / , l•w lhl Np11q111H1' ll•I 1111' ll11x 11111I ilHldlty /o'I hr 1•011Nt1111t ulonH the l1111Hlh ol llu- lu-11111
8
62
Systems with One Degree of Freedom
Chap. 2
2.3-5. Let the prismatic cantilever beam shown in Fig. P2.3-5 be subjected to a ground translation Da = d sin Ot, as indicated in the figure. Calculate the steady-state response of the mass m for O = 2w/3 .
m••- ------------~ f0
9
Chap. 2
Problems
63
2.3-10. If the left-hand support of the overhanging beam shown in Fig . P2.3-10 has the harmonic displacement d cos Ot, what is the response of the mass m? Assume that the beam has constant flexural rigidity and that O = 6w.
td
cos Ot
(t),6, (t)
m
i
-L----.l
I• t + - - -
~
f - - - - - L - - --
Figure P2.3-S
2.3-6. Repeat Prob. 2.3-5 with ground acceleration Da = a cos Ot and .0 = 4w. 2.3-7. The rigid disk shown in Fig. P2.3-7 is attached at its center to a flexible rod of length L with a circular cross section of radius r. Find the steady-state response of the disk caused by a rotational ground displacement Dg = 88g sin .Ot if .0 = 3w/4.
#;, L-----+i
Figure P2.3-10
2.4-1. Considering a damped SDOF system subjected to the forcing function P sin .Ot, derive the expression for the steady-state response in phase-angle form. 2.4-2. Derive the steady-state response (in phase-angle form) of a damped SDOF system that experiences the ground acceleration ii8 = a sin Ot. 2.4-3. For a SDOF system with subcritical damping, determine the transient response due to the forcing function P sin Ot. Give the solution in a form similar to Eq. (2.4-f).
2.4-4. Repeat Prob. 2.3-3, assuming that the damping ratio is -y = 0.01 and that the frequency ratio is .0/ w = 0.8. 2.4-5. Repeat Prob. 2.3-4, but assume that the damping ratio is -y = 0.03 and that the frequency ratio is .0/ w = 1.1. 2.4-6. Repeat Prob. 2.3-6, but let the damping ratio be -y = 0.02 and take the frequency ratio as 0 / w = 0.95. 2.4-7. Calculate the magnification factor {3 at resonance for values of the damping ratio -y equal to O.Gl, 0.02, ... , 0.20. 2.5-1. Expand the square-wave function shown in Fig. P2.5-1 into a Fourier (or trigonometric) series. p
Figure P2.3-7
Da = 88g cos Ot
2.J-8. Repeat Prob. 2.3-7 with rotational ground acceleration
n = 5w, assuming that the mass moment of inertia of the disk is !,.
and
-
P,
2.J-9. Let the right-hand support of the simple beam shown in Fig. P2.3-9 oscillate in accordance with the harmonic displacement d sin Ot. Determine the steadyslate response of the mass m, assuming that the beam is prismatic and 7w/8.
n
0
,r
:n
2,r
n
d sin Ht
t
W,~,- -·-, ----9»}1 m
I~-
~
- ~~
1<1111111• I' l, I 11
p1
~
·I l•'l111m• 1'2.~- 1
3,r
n
4,r
n
Systems with One Degree of Freedom
64
Chap. 2
2.5-2. In place of the periodic forcing function shown in Fig. P2.5-2, determine a Fourier series. p
,~
p
0
I
I
I
I
'IT
2'lT
3'lT
4'!T
n - P,
n
n
n
I--
Figure P2.5-2
Chap. 2
65
Problems
2.5-5. Derive the general expression for steady-state forced vibrations of a damped SDOF system due to the forcing function given as the Fourier series in Eq. (2 .5-1) . 2.5-6. Using a frequency ratio of D,/ w = 0.9, find the undamped response of a SDOF system to the first term in the series from Prob. 2.5-1. 2.5-7. Determine the undamped response of a SDOF system to the first term in the series from Prob. 2.5-2, assuming a frequency ratio of 0/w = 0.95. 2.5-8. For a frequency ratio of D,/w = 1.05 , calculate the undamped response of a SDOF system to the first term in the series from Prob. 2.5-3. 2.6-1. Rederive the expression for undamped forced vibrations of a SDOF system subjected to the harmonic function P = Pi sin fit'. 2.6-2. Find the damped response of a SDOF system to the step function P = A. 2 .6-3. Derive the expression for damped response of a SDOF system to the ramp function P = Pit' / ti. 2.6-4. Determine expressions for the undamped response of a SDOF system to the forcing function shown in Fig. P2.6-4.
l.5-3. Decompose the piecewise-linear periodic function shown in Fig. P2.5-3 into a
p
Pourier series. p
- P, -
-
-
-
-
-
-
-'------
Figure P2.6-4 ,(1°!i.
Figure P2.5-3
2.!i-4. 11or the sawtooth function shown in Fig. P2.5-4, substitute a Fourier series.
Assuming that the forcing function shown in Fig. P2.6-5 acts on a SDOF system, find expressions for the undamped response. p
p
P,
()
/11 II
1111 II
IJ11 II
1111 II '111411111 I' L. t, ~
1111
1..,,.,,.
Cht1p. 'I
I )llllVC l!Xl)l llSShHIH 1111 till' 1111tl11111pcd response o l HSI)( )I • N)'Nlll lll to the f'o1·ci11g function shown in Fig . l'J. (1 (1 . p
l 1111hl11111 11
67
f1 1) , l>l•tlw L' Xptcss l1111s for lht undumpL·d ,csponsc of a SDOF system to the paral>ulil' g1ou11d 11ccclcrntio11 i/H a,(1 1 /1,/ given in Fig. P2.6-9.
iig
Figure P2.6-6
2.6-7. Let the ground acceleration shown in Fig. P2.6-7 be imposed on a SDOF system, and find expressions for its undamped response.
Figure P2.6-9
iig
o,
-
-
-
-
--,,...------------
l , I• 1. For u step function of magnitude P = Pi , calculate the undamped response of ti SDOF system, using a recursive procedure with 10 equal time steps of duration t.1 T/ 10. I I l, Assume that a ramp function P = P, t/t, is applied to an undamped SDOF system . With a step-by-step procedure, find the response of the system for 10 l'qual lime steps of duration !::.t = T /10 = t 1. , 1 ,. ( 'on firm the approximate results of Example 2.9 in Table 2.1. ~. f 4, ( 'onflrm the results of Example 2.10 in Table 2.2. , f.lfi. Using 20 equaJ time steps with !::.t = T/20, determine the undamped response of u SDOF system to the forcing function shown in Fig. P2.7-5.
Figure P2.6-7
.l.(,-H. For the ground acceleration shown in Fig. P2.6-8 , determine expressions for the undumped response of a SDOF system.
p
iig P,
a, Qi£-_ _ __L_ _ _
__;ll.----~----~---t,o
0
--------~~ ~------(
,.,
--------- -
1,
1,·111111·11 p.1.7-~
""J..7.(1.
Systems w ith One Degree of Freedom
Chap. 2
Calculate the undamped response of a SDOF system to the forcing function shown in Fig. P2.7-6, using 20 equal time steps with l:!.t = T/20 .
Chap. 2
2.7-8. Determine the undamped response of a SDOF system to the parabolic forcing function P = P,t 2 / do shown in Fig. P2.7-8, using 10 equal time steps of duration At = T/30.
p
p
P,
69
Problems
lo---------.
P,
QP--~~~~~~~--t~~~~~~~~--r~~~~
t,o
P,
-
-
-
-
-
-
t20
o .....""""'~~~~~~~~--''--~~~~~
_JI--------~
t,o
Figure P2.7 -8 Figure P2.7-6
p p
Apex P,
0
t,o
Figure P2.7-7 Figure P2.7-9
J..7 7. J)i vldc lhl' 1rl1111gulnr i111pulSl' shown in Fig. P2.7 7 iulo 10 cquul time steps
At
'/'/ 10 , 1111d ll11d llw
syshl111 .
1111cl11111pN I ws pollMl' wh\111 it Is Hpplicd to n SDOF
J..7-1). The parabolic forcing function shown in Fig. P2.7-9 has the formula
P 1( 1 12 /do), With 10 equal time steps of duration l:!.t = T/25 , obtain llw u11d111111wd 1l•Npo11sc of' 11 SDOF system.
P
Systems with One Degree of Freedom
10
Chap. 2
2.7-10. Find the undamped response of a SDOF system to the parabolic forcing function P = P1[1 - (t - t1 0)2 / do] shown in Fig. P2. 7-10, using 10 equal time steps of duration 6.t = T /25 .
Chap.2
71
Problems p
p
Apex P,
Figure P2.8-2
2.8-3. Repeat Prob. 2.8-1 for the forcing function shown in Fig. P2.8-3. (See Prob. 2.6-6 for response formulas.) 0
p
Figure P2.7-10
2.H-1. For the forcing function shown in Fig. P2. 8-1 , plot the response spectrum um/ Us, and the time for maximum response t,,,/ t, against the time ratio ti/T (See Prob. 2.6-4 for response formulas.) p
QL----~~.....1-~~~~~--'L-~~~t, 2t,
P,-------
Figure P2.8-3
2.8-4. Repeat Prob. 2.8-1 for the forcing function shown in Fig. P2.8-4. (See Prob. 2. 6-7 for response formulas.) p
P, P, '-- -
-
-
-
__,......__ _ _ _ __,
l<'lgm·l' P2.ll· 1
l,H•2. lfop1•11t P, oh ) .H I 101 th,• lw d 111' ) '1 "I Im 11·~pn11N1' 111111111111H )
i111ll't lo11
Nhow11 111 Fl1,1. I' ) .H l. Ukc l'mh.
72
Systems with One Degree of Freedom
3
Chap. 2
2.8-5. Repeat Prob. 2.8- 1 for the forcing function shown in Fig. P2.8-5. (See Prob. 2.6-8 for response formulas.) p
2P,
Finite Elements and Vibrational Analysis
P,
0
t,
Figure P2.8-S
2.8-6. Repeat Prob. 2.8-1 for the forcing function shown in Fig. P2.8-6. (See Prob. 2.6-9 for response formulas.) p
3.1 INTRODUCTION P,
0
t,
Figure P2.8-6
In this book we use the method of.finite elements [1-3] to discretize solids and structures for dynamic analysis. The basic concept is to divide a continuum into subregions having simpler geometries than the original problem. Each subregion (or finite element) is of finite size (not infinitesimal) and has a number of key points, called nodes, that control the behavior of the element. By making the displacements or stresses at any point in an element dependent on those at the nodes, we need only write a finite number of differential equations of motion for such nodes. This approach enables us to convert a problem with an infinite number of degrees of freedom to one with a finite number, thereby simplifying the sol ution process . For good accuracy in the solution, the number of nodal degrees of freedom usually must be fairl y large; and the details of element lormulations are rather complicated. Therefore, it becomes necessary to program this method on a digital computer. Figure 3. I shows various examples of solids and structures that are disrrclized by finite elements , with dots indicating the nodes. In Fig. 3. l (a) we see 11 rnntinuous beam that is divided into several flexural elements of the type to l>L· described in Sec. 3.4. The space frame with curved members in Fig. 3. l (b) 1111, uxi nl , flexural, and tors ional deformations in each of its subdivided memhL•1s . Figure 3.1 (c) depicts a two-dimensional slice of unit thickness, represent111~ tht· rnns lunt slutc known us plnnc strain on the cross section of a long, p11s11111lk solid . On lht· olht·1 hand, llw disnclizcd general solid in Fig. 3. l(d) h111, 11111,1wlt 1l'~liil'l11111 II lltt• lhiu pl111t• in Fiu. I l(t•) lt11s lort·t·s npp lil:d in its
74
Finite Elements and Vibrational Analysis
•
Chap. 3
(a)
/
/
/
/
/
/
/
/
/ /
(c)
/
/
(d)
(e) (/)
Jt'lgure 3.1
75
1. Divide the continuum into a .finite number__of subregions (or elements) of simple geometry, such as lines, quadrilaterals, or hexahedra. 2. Select key points on the elements to serve as nodes, where conditions of dynamic equilibrium and compatibility with other elements are to be enforced. 3. Assume displacement shape_functions within each_element so that displacements, velocities, and accelerations at any point are dependent on nodal values. 4. Satisfy strain-displacement and stress-strain relationships within a typical element for a specific type of problem. 5. Determine equivalent stiffnesses, masses, and nodal loads for each finite element, using a work or energy principle. 6. Develop differential equations of motion for the nodes of the discretized continuum by assembling the_finite-element contributions.
/
-
Stresses and Strains
infinitesimal size. The shape functions make generic displacements at any point completely dependent upon nodal displacements. Similarly, the local velocities and accelerations are also dependent on the nodal values. With these dependencies in mind, we can devise a procedure for writing differential equations of motion, as follows:
t •
•
Sec. 3.2
From the homogeneous form of the equations of motion, we can perform a vibrational analysis for any linearly elastic structure . This type of analysis consists of finding undamped frequencies and corresponding mode shapesJ or the discretized analytical model. Such information is often useful by itself, and it is essential for the normal-mode method of dynamic analysis described in Chapter 4. In the present chapter we develop one-dimensional elements that are to be used in subsequent work (especially Chapter 6) for analyzing framed structures. Other discretized continua will be discussed in Chapters 7 and 8, where the applications include two- and three-dimensional solids, plates in bending, and shell structures .
Structures modeled by finite elements: (a) beam; (b) space frame; (c)
plunc strain; (d) solid; (e) plate; (f) shell.
own pltme, it experiences a condition known as plane stress. But if the forces nrc normal to the plane, it is in a state of flexure, or bending. Finally, a general shell of the type shown in Fig. 3.1 (f) can resist uny kind of loading. All of the discretized structures in Fig. 3. 1 have mult.iple d1·g1t·l'N of freedom and will be 1l•lc11'l'd lo us MOOF .l'y.1·t,•111s. Tht· liniw l' k11w111 111l•lhod to be uNl'd in lhlNhook 111v11lws thl· ussumption ol r/i,1'/ lfrll't'l/lt'III .1'1111/11' /tllll'll//11,\' withi11 l'lll'h 1•h·1111· 11I ' ('ht•w lt111t•lio11s giVl' 11pp111x111rnlr 11•1111II N whr11 1111' 1•lt•1111•11t 111 111 1111111 - 111· 111111 1•x11l'I 11•111111~ 111
3 .2 STRESSES AND STRAINS lt1 I his book we assume that the continuum to be analyzed consists of a linearly l'lustic material with small strains and small displacements. In any case, strains 1111d their corresponding stresses may be expressed with respect to some rightl11111d orthogonal coordinate system. For example, in a (rectangular) Cartesian 11111, llu; coordinates would be x, y, and z. On the other hand, in a cylindrical 1 omdlnntl· systl'rn , the symbols r, 0, and z would serve as the coordinates. Figun· \ , , shows 1111 inrl11itcsi111ul l'ic1m•nt in Cartesian coordinates, where llw 1•d)'1'H 1111• ol lt·HHths r/1 . di', ,11ut d·. Normal nnd s/11•r11·/111{ .1·1r<•sse.1· arc
76
Finite Elements and Vibrational Analysis
Chap. 3
Sec. 3.2
77
St resses and Strains
For convenience, the six independent stresses and the corresponding strains usually will be represented as column matrices (or vectors). Thus,
erx ery u, Txy
0-1
I I I I
dy
=
U4
I
ri, )----- -
er6
T,,
-----
/
/
EX
/
indicuted by arrows on the faces of the element . The normal stresses are labeled ,rv, and er,, whereas the shearing stresses are named rxy , Ty,, and so on. From l'q11ilibrium of the element, the following relationships are known:
Verx
1 E (- verx
E, =
11 "
'l'hus, only three independent components of shearing stresses need be con11idl·1cd. Corresponding to the stresses shown in Fig. 3 .2 are normal and shearing ,1•/ro/11.1·. Normal strains Ex , Ey, and E, are defined as
ax
E y
=
ay
(2)
o, and ware translations in the x, y, and z directions. Shearing strains, Y, 1., 'Y1•11 un<.I so on , are given by 'Y.
av
= ay + ax = Yyx
av + -aw yy: =CIZ ily <'. IV ! i)11 dI ii '
I 1!•1111• ,
011l y (h11•1• 111
lh1·
~l11•111 IIIH ~(1111111, Ill
1111l1·111·111l1•11t
E4 Es
Yxy Yy,
E6
'Yzx
- very
+
er,)
Yyz
Ty,
=G
1'zx 'Yzx = G
E
=
2(1
+ v)
In these expressions Eis Young' s modulus of elasticity, G the shearing modulus, and vis Poisson's ratio. With matrix format, the relationships in Eqs. (5) may be written as e=C
-v
-v
-v -v E 0 0 0 0
C = l_
0
y,
+ ery - Ver,)
G
Whl1l'l' 11 ,
au
E,
where
(l)
aw E =z az
E3
X
Ey = - ( -
Stresses on an infinitesimal element.
av -
Ey
1
l,W
au Ex= -
E2
= -E (CT - vay - ver,) 1 E
dx
Figure 3.2
=
Ex
where the boldfaced symbols denote the vectors shown. Strain-stress relationships for an isotropic material are drawn from the theory of elasticity [4], as follows:
/
f-'·"
E
Tyz Tzx
ers
! )-.,. y,v
(J'
er2 er3
E1
0
-v -v 1
0 0 0
0 0 0 0
0 0 0 2(1
+ v) 0 0
2(1
+ v) 0
0 0 0 0 0 2( 1 + v)
Matrix C is an operator that relates the strain vector E to the stress vector CJ'. By the process of inversion (or simultaneous solution), we can also obtain stressstmin rt'lotio11ships from Eq. (7) , ns follows: (9)
78
Finite Elements and Vibrational Analysis
Chap. 3
where
E
= c-
(1
+
E v)(l - 2v)
J)
J)
V
J)
l-v
V
J)
J)
0
0
1-
0
0 0 0
0 0 0
0 0 0
1 - 2v 2
0
0
J)
--
ooct.e.s., q (t)
D..Q)
0
0
0
--
0
0
0
0
0
2
1 - 2v
--
2
3 .3 EQUATIONS OF MOTION FOR FINITE ELEMENTS
Wu shall no w introduce definitions and notations that pertain to the finite eleinunts to be studied throughout the book. By using the principle of virtual work , we can develop equations of motion for any finite element. Such equations inc lude energy-equivalent stiffnesses, masses, and nodal loads for a typical d cmcnl. T hese terms are treated in detail for one-dimensional elements in the next section . Assume that a three-dimensional finite element with zero damping exists in Cartesian coordinates x, y, and z. Let the time-varying generic displacements 11(1) at
= {u, v , w}
(1)
whcru u, D, and w are translations in the x, y , and z directions, respectively .* If the cle ment is subjected to time-varying body forces, such fo rces may lw placed into a vector b(t) , as follo ws: (2)
= l, 2, ... , n n) 0
(3)
11111 l11N1•tl hy h1 111•1•N (
l ,111111
wilh
(4)
However, other types of displacements, such as small rotations (av/ iJx, and so on) and curvatures (a 2v/ ax 2, and so on) will be used later. Similarly, time-varying nodal actions p(t) will temporarily be taken as only forces in the x , y , and z directions at the nodes. That is,
p (t)
=
{p; (t)}
(i
= 1, 2, ... , n.n)
(5)
in which p;(t)
= {px; , py;, p,;}
(6)
Time functions for Pxi , py;, and p,; at each node may be independent and arbitrary . Other types of nodal actions, such as moments, and so on, will be considered later. For the type of finite-element method to which this book is devoted, certain assumed disvlacement shave functions relate generic displacements to nodal displacements, as follows:
=
u (t)
f g (t)
ru
ln this expression the symbol f denotes a rectangular matrix containing the functions that make u(t) completely dependent on q(t). Strain-displacement relationships are obtained by differentiation of the generic displacements. This process may be expressed by forming a matrix d, called a linear differential operator. and applying it with the rules of matrix · multiplication. Thus,
E(t) = d u (t) In this equation the operator d expresses the time-varying strain vector e(t) in terms of generic displacements in the vector u(t) [see Eqs. (3 .2-2) and (3.2-3)]. Substitution of Eq. (7) into Eq. (8) yields
I lc rl' the symbols b, , h,., and b: represent co mponents of force (per unit of volu111u, arua, or length) acting in the ru forunce direl'lions nt n gcnuri c point. The lll W
(i
q;(t) = {qx;, qy;, q,;} = {u; , V; , w;}
0
Matrix E is an operator that relates the stress vector CT to the strain vector e . For the elements in this chapter, we will not need the 6 X 6 stress-strain matrix given by Eq. (10). With one-dimensional elements, only one term , such us E or G, is required. Later, when we deal with two- and three-dimensional t:lcments , larger matrices, such as that in Eq. (10) , will be needed.
*'1'11 NIIVl' Np ill'I', rnl1111111 Vl'l ' IIIIN 11111y lw wt1 11,111 111 11 rUIIIIIIIIN ~11p11111l lt1f1 lh1• 11•1111~
= {q;(t)}
where 1 - 2v
0
u (t)
79
Equations of Motion for Finite Elements
time variation for each component of body force is assumed to be the same throughout the element. That is , we may have one time function for bx, another for by, and a third for b,. Time-varying nodal displacements g(t) will at first be considered as only translations in the x, y, and z directions. Thus , if n. 0 = number of element
1
1-
Sec. 3.3
e (t)
= B q.(f2_
whurc
n
dr
110
I l11llu I l11111u1111111111 I Vll11n11111111I A1111ly1ln
Chop. 3
I q1111llm 111 ul Motlo11 fw 11111111 I 1111111111111
m ,1.,1
81
Muttix B gives struins 111 uny pol111 within the clemcnl d11t lo 1111i( vnlucs of' nodal displacements. From Eq. (3.2-9) we have the matrix form of' .1·rre.1·.1·-.l'lrai11 relationships. That is, o(t) = E e(t) ( l I) b,(t) dV
where E is a matrix relating time-varying stresses in the vector o(t) to strains in E(t). Substitution of Eq. (9) into Eq. (11) produces
o(t)
= E B g (t)
1,odv LJdv
(12).
in which the matrix product E B gives stresses at a generic point due to unit values_of nodal displacements. Virtual Work Principle: If a general structure in dynamic equilibrium is ,l'ubjected to a system of small virtual displacements within a compatible state of dt:f'ormation, the virtual work of external actions is equal to the virtual strain l'lll'l'RY of internal stresses. When applying this principle to a finite element, we huve
J--/
/
dx
)-x"·'
z, w,W
(14)
Figure 3.3 Applied and inertial body forces.
(15)
Substitution of Eqs. (17) and (18) into Eq. (13) produces
llsinp lhc strain-displacement relationships in Eq. (9), we obtain ae = B aq
L
aeTo(t) dV = aqTp(t)
(16)
Now lhe infernal virtual strain energy can be written as
+
L
8uTb(t) dV -
where integration is over the volume of the element. For the <'Xterna/ virtual work we turn to Fig. 3.3, which shows an infinilt•si 11111I clement with components of applied body forces bx(t) dV, by(t) dV, and !,, (I) dV. The figure also indicates inertial body forces pu cl\/, pv dV, and pw dV d11t• lo lhc accelerations u, i5, and w. The symbol pin these expressions repreNt•11ts !he 1110.1·.1· density of the material, which is defined as the inertial force per 111111 HL'L'clcration per unit volume. Note that the incl'liul forces act in directions lh111 me opposite to the positive senses of' lhc ut·n•ll•f'11lio11s . Thus, we add the t'Xlt111111I vlrtunl work of' nodul nnd clislribu(L·d hody loll't'N us follows: h11 'p (I) I ( Im 'h (I) 111'
Iv
" r'ln 11111 rlV
L
8uTpii dV
(19)
Now assume that
q
ii= f
l>W,.
piidV - - - - - -
/
'l'ht111 lhe resulting virtual generic displacements become [see Eq. (7)] au= f aq
b,(t) dV
/
y, V, ii
1, 2, . . . , nen)
//
I b,(t) dV
when: fJU, is the virtual strain energy of internal stresses and aw. is the virtual wmk of external actions on the element. To develop both of these quantities in dt•lllil , we assume a vector aq of small virtual displacements. Thus,
=
-·-
/
au.= aw.
(i
dy
, ( 18)
'l'ltcn we can substitute Eqs. (12) and (20) into Eq. (19) and use the transposes nf Eqs. ( 15) and (16) to obtain J>ql'
LBTE
B dV q = 8qTp(t)
+ aqT
L
fTb(t) dV - SqT
L
pfTf dV q (21)
<'1111ccllution of aq rand rearrangement of the resulting equations ofmotion gives
Mq+ Kq
= p (t) +
Pb(t)
Wltl' ll'
K
r II
I
14:
n
Finite Elements and Vibrational Analysis
82
Chap. 3
Sec. 3.4
83
One-Dimensional Elements u
and
q,
M =
2 q, ------,!.------------x,u
Iv pfT f dV
-
Also, Ph(t) =
Iv er b (t) dV
(a)
Matrix K in Eq. (23) is the element stiffness matrix, which contains stiffness coefficients that are fictitious actions at nodes due to unit values oL nodal displacements. Equation (24) gives the form of the consistent-mass matrix. in
1 ~ (b)
which the terms are energy-equivalent actions at nodes due to unit- values of nodal accelerations. Finally, the vector pb(t) in Eq. (25) consists of equivalent nodal loads due to body forcesi n the vector b(t). Other equivalent nodal loads due to initial strains (or stresses) could be derived [l], but analyses for such influences are considered to be statics problems.
1
(c)
Figure 3.4
3.4 ONE-DIMENSIONAL ELEMENTS
In this section we develop properties of one-dimensional elements subjected to n,dul, torsional , and flexural deformations, starting with the axial element in Fig. 1.4(11). The figure indicates a single generic translation u in the direction of x. 'l'hus, from Eq . (3.3-1) we have u(t) = u
Axial element.
The first function diminishes linearly from 1 to 0, whereas the second increases linearly from O to 1. From Fig. 3.5 we see that the single strain-displacement relationship du / dx for the axial element is constant on the cross section. Thus, Eqs. (3.3-8), (3.3-9), and (3.3-10) yield du df E(t) = Ex= - = -q(t) = B q(t) dx dx Therefore, we have
'l'hc corresponding body force is a single component bx (force per unit length) , 11cting in the x direction. Therefore, Eq. (3.3-2) gives
B = d f = df = .!. [- 1 dx L
1]
b(t) = bx Nodul displacements q 1 and q2 consist of translations in the x direction at nodes I nnd 2 Isee Fig. 3.4(a)]. Hence, Eq. (3.3-3) becomes q(t) ( ' oi responding
= {q1, q2} = {u1,
y
u2}
nodal forces at points 1 and 2 are given by Eq. (3. 3-5) as
P (t)
= {pi, P2} = {Pxi, pxi}
Figure 3.4(b) and (c) show linear displacement shape functions/1 and,f; that we ussumc for this element. That is, Eq. (3.3-7) gives
u
t' q (t)
wlwn• I'
(I) Flw1111• I,!!
i\~ i11l
1h'lt11t111tlilll1N
(2)
84
Finite Elements and Vibrational Analysis
Chap. 3
Sec. 3.4
which expresses the strain Ex.in terms of the nodal displacements. Similarly, the single stress-strain relationship [see Eqs. (3 .3-11) and (3 .3-12)] becomes merely
O'(t)
=
CTx
85
One-Dimensional Elements
q, 1 -0. 2 q, - - - - - -.......- - - - - - - - - - - - - X
= E E(t) = EEx = EB q (t)
Hence,
E=E
and
EB=
(a)
f [-1 1]
(3) 1
The latter expression gives the stress crx in terms of the nodal displacements. The element stiffness matrix K can now be evaluated from Eq. (3.3-23), as follows:
K
=
Iv BT EB dV = i
= EA [ L
1 -1
[-!
1
Jr-1
-!J
(c)
(4)
assuming that the cross-sectional area A is constant. Similarly, the consistentmass matrix M is found from Eq. (3.3-24) to be
M
= Jv pfT f dV = { 2 =
pAL[2 6
f L[
L : x ]EL - x
x] dA dx
=
l]
(5)
1 2
f
fT bx dx
=
if[
L :
X
J dx = bx ~ [ !J
(6)
which shows that the equivalent nodal loads are equal forces at the two ends. Turning now to the torsional element in Fig. 3.6(a), we use a single generic displacement Ox, which is a small rotation about the x axis (indicated by a double-headed arrow). Thus , u (t)
=
Figure 3.6 Torsional element.
which is a moment per unit length acting in the positive x sense. Nodal displacements in the figure consist of small axial rotations at nodes 1 and 2. llence,
q(t)
assuming also that the mass density p is constant. We see that the stiffness matrix K and the consistent-mass matrix M are unique for a prismatic axial element of uniform mass density . However, an infinite number of equivalent nodal load vectors Pb (t) may be derived, depending on the distribution of body forces. For the simplest case, we assume that a uniformly distributed axial load bx (force per unit length) is suddenly applied to the axial element. Then Eq. (3.3-25) produces
Pb (t)
(b)
f),
= {q1, q2} = {()xi, Ox2}
In addition, the corresponding nodal actions at points 1 and 2 are
p(t)
= {p ,, pz} = {Mx1, Mxi}
which are moments (or torques) acting in the x direction. As for the axial clement, we assume the linear displacement shape functions /1 and !2 shown in Fig. 3.6(b) and (c). Therefore,
ex = f q (t) in which the matrix f is again given by Eq. (1). Strain-displacement relationships can be inferred for a torsional element with a circular cross section by examining Fig. 3. 7. Assuming that radii remain 11lraight during torsional deformation, we conclude that the shearing strain 'Y vurics linearly with the radial distance r, as follows: 'Y
wlwrc
1/1
d()x
= r - = rijJ dx
is the twist , or rate of change of angular displacement. Thus,
Corresponding to this displacement is u single body action b(t)
Ill,
dO, 1/1
ffi
86
Finite Elements and Vibrational Analysis
Chap. 3
Sec. 3.4
87
One-Dimensional Elements
y
1]
= GJ[ L
(L r21r (R
Jo Jo Jo (r 2 )r dr dO dx
1 -l
(12)
where GJ is constant. The polar moment of inertia J for a circular cross section is defined as 4 21T 11R J = r 3 dr dO = 0 0 2
z
L
Figure 3.7
Torsional deformations.
Equation (7) shows that the maximum value of the shearing strain occurs at the surface. That is, 'Ymax
= Rt/;
where R is the radius of the cross section (see Fig. 3. 7). Also, we see from Eq. (7) that the linear differential operator d relating 'Y to Ox is
d
d
=
df
dx
r
= L[-1
1]
(10)
which is the same as for the axial element, except for the presence of r. Shearing stress r (see Fig. 3.7) is related to shearing strain in a torsional element by r
= Gy
where the symbol G denotes the shearing modulus of the material. Hence,
E
G
and
If the cross section of a torsional element is not circular, it will warp. Such warping is mosLsevere for elements of open cross sections, such as channel or wide-flanged sections. For most practical cases, the theory of uniform torsion described here may be used by substituting the appropriate torsion constant [5] for J. If a more precise analysis is desired, the theory of nonuniform torsion [6] may be applied. To obtain the consistent-mass matrix M for a torsional element, we will first integrate over the cross section and then over the length. Due to the small rotation Ox, the translation of a point on the cross section at distance r from the center is rOx. Also, the acceleration of the same point is r8x. By integration over the cross section, we find the inertial momentper unit length to be -p 18,., where J is again the polar moment of inertia given in Eq. (13). Use of this inertial moment in conjunction with the corresponding virtual rotation Mx leads to
= r-
Thus, the strain-displacement matrix B becomes
B
LR
GU
Grl /,
11
( 11 b)
Thcsl' 1l'111tionships llll' 111111Io»ous lo Eqs . ( 1) 101 till' 11x111 l l'll•1rn·111 , W,• 1111,y 110w 1111d th,• 1<11si111111l stll l111•11N 11111111, K hy 11pply111» l\q. ( I I I I), 11~ lollow~
which is a specialized version of Eq. (3.3-24). Integration of Eq. (14) over the length yields
M = pl2 (L [L - x ][L - x L Jo x
=
pJL 6
x ] dx
[2 1]
(15)
1 2
'f'his array is the consistent-mass matrix for a torsional element of constant cross Hcction and uniform mass density. The simplest case of a body force applied to a torsional element consists of u uniformly distributed axial torque (or moment) mx per unit length. For this l1111ding Gq . (3 .3-25) gives
(' f"'m, dx
Jn
1'11,·'lt' l'q111v11l,•111 1111d11I londs
llll'
m1 ('·
,, Ju
I1L -, xl dx = m)::.2 [1] 1
(16)
l'q1111l 1111111w11111 nl lhl' two ends or the clement.
Hnlto E:lomontt1 nnd Vll1111tl111111I A1111lynl11
88
Chup. 3
. F_igure 3.8(a) show~ a ~traii!U}l~ elert1J.'III , for whlt:h the x-y plane is a principal ~Ian~ of bending. Indicated in the figure is a single generic displacc111cnt v, which 1s a translation in they direction . Thus,
=V
U (t)
The corresponding body force is a single component by (force per unit length),
nc
I i1
01111
89
lll11111111lm111I I 111111111111
1111111-1 111 tht· y direction . I loncc, b(t)
= by
I 1111dc I !sec Fig. 3.8(a)] the two nodal displacements q1 and q2 are a transl,1111111 111 the y direction and a small rotation in the z sense. The former is h11lk 11tt·d by a single-headed arrow, while the latter is shown as a double-headed 111111w. Similarly , at node 2 the displacements numbered 3 and 4 are a translation ,111d II small rotation, respectively. Therefore, the vector of nodal displacements 111 l'tllllCS
111 which fJ zl
=
dv, dx
fJ z '
= dv2 dx
I ht'Sl' derivatives (or slopes) may be considered to be small rotations even though they are actually rates of changes of translations at the nodes. Corre-111111ding nodal actions at points 1 and 2 are
P (t)
lb)
~
,-
f,
·y=:==-- ----- =__.~
= {p1, Pi, p3 , p4} = {py1, M,1, Py2 ,
I ht• terms Py1 and py2 denote forces in they direction at nodes 1 and 2, and the -y111hols M, 1 and M,2 represent moments in the z sense at those points . For the flexural element we assume cubic displacement shape functions in 111111ri x f, as follows:
l/1 Ji f3 J:i] I /}
- [2x 3
-
3x 2L + L 3 x 3L - 2x 2 L2 + xL 3 - 2x 3 + 3x 2L
(17)
x 3 L-x 2 L 2]
(c)
Id)
M,2}
l'hl'sc four shape functions appear in Fig. 3.8(b)-(e). They represent the vari11t1011s of v along the length due to unit values of the four nodal displacements ,,, through q4. Strain-displacement relationships can be developed for the flexural ele1m111t by assuming that plane sections remain plane during deformation, as lll11strutccl in Fig. 3.9. The translation u in the x direction at any point on the 1 rnss section is dv u = - ydx 1llling thi s relationship , we obtain the following expression for flexural strain:
(11)
y
(19)
90
f In Ito Elomont11 nncl Vlhr ntl1111nl A1111lyuls
Chaµ. 3
I
111,
3.4
One-Dimensional Elements
91
" L
BTE B dV
y,v
ri- J Ey6
-ry
~--
z
........
Jo
+---\
A
L
~ :t
2
2 [ ~:;
-
l 2x
]
+ 6L
[12x - 6L
. . .
6xL - 2L 2 ] dA dx
6xL - 2L 2
-_;;;-""--~--
Mu ltiplication and integration (with El constant) yields
......-"
I M, /
\
10
\_,,
_ dv dx
I '.,........J----
\ \
I
+---
----------1~-v,....
K = 2El [3: L3 -6 3L
\
-:;,I
................
-dx~
[=
~~2
-3L L2
L
:3: 6 -3L
~~
] -3L 2L2
(24)
2
y dA
Figure 3.9 Flexural defonnations.
in which represents the curvature:
d 2v
= dx2
(20)
I:rom Eq · ( 19) we see that the linear differential operator d relating
Ex
to v is
d2
d
= -y dx2
(21)
rrptl'Scnts the moment of inertia (second moment of area) of the cross section wi th respect to the neutral axis. Additional contributions to matrix K due to • lll'11ring deformations are given in Ref. 5. The consistent-mass matrix M for a flexural element will be developed in 1wo parts. A typical cross section of this type of member translates in the y dlr1•l'ti on, as indicated in Fig. 3 .8(a). However, the section also rotates about its m·utru l axis, as shown in Fig. 3. 9. The translational inertia terms are much more lrupmtant than the rotational tenns, so they will be considered first. Using matrix I lwm Eq. (17) in Eq. (3.3-24) , we find that
M, = fv pfTf dV =
The n Eq. (3.3-10) gives the strain-displacement matrix B as
U
d f
= _ 1'._ [l 2x L3
6L
6xL - 4L 2
-l2x
+ 6L
[ 156 22L 420 54 - 13L
6xL - 2L 2]
= pAL
(22) In addition, flexural stress
er.. in Fig. 3.9 is related to flexural strain Ex simply by (23a)
l lcnce,
E =E
f
pAfTf dx
22L 4L 2 13L -3L 2
EB
IW
(23b)
Element stiffnesses muy now be ohlnirwd 1111111 11•11• (,, l .,l, 21, l ) ' us, ,.() ows: 11
l
-13L -3£2
- 22L 4L 2
(26)
whkh is the consistent-mass matrix for translational inertia in a prismatic beam. Notational inertia (or rotary inertia) terms for a beam can be deduced from I •IH I . 9, where the translation u in the x direction of a point on the cross section 111
and
54 I3L 156 -22L
II
111 thi11 1·xprl1Hsion ,
- y f)z
92
Finite Elements and Vibrational Analysis
(), = V,x = f,xq (t)
Chap. 3
(28)
where the symbols v,x and f,x represent differentiation with respect to x. Similarly, the acceleration of the same point in the x direction is
u = -yo, 0, = V,x = f,xQ (t)
(30)
By integrating the moment of inertial force over the cross section, we find the inertial moment per unit length to be - pl 0,, where I is again the moment of inertia given in Eq. (25). Use of this inertial moment in conjunction with the corresponding virtual rotation 80, leads to the formula
=
f
93
One-Dimensional Elements
and flexural elements. Let us reconsider the torsional element in Fig. 3.7 and integrate the moment of the shearing stress T about the x axis. Thus, we generate the torque Mx , as follows:
Mx
(29)
where
M,
Sec. 3.4
=
fr
2
dr d()
Substitution of the stress-strain and strain-displacement relationships from Eqs. (l la) and (7) produces
Mx
= Gt{,
fr 1r
r 3 dr d()
= GJ t{,
(36)
1
lf we take Mx as generalized (or integrated) stress and as generalized strain, the generalized stress-strain (or torque-twist) operator G becomes G
p!f;xf,x dx
Tr
1T
= GJ
(37)
which is the torsional rigidity of the cross section . Hence, from Eq. (36) we have
which is a modified version of Eq. (3.3-24). Differentiating matrix f [see Eq. ( 17)] with respect to x, we find that
Mx = Gt{,
(38)
(32)
By this method the operator d in ~. (9) does not include .the mu.ltiplier r. Furthermore, the generalized matrix B [Eq. (10) devoid of r] is used mstead of matrix B. That is, (39) B = rB
Suhslilulion of this matrix into Eq. (31), followed by integration over the length, pmduccs
Prom this point we can conclude that evaluations of th~ terms in the stiffness matrix K do not require integrations over the cross section. Therefore,
f,1
= ~3 [6(x
2
-
2
3
3x L - 4xL + L
xL)
M,
2
=
36 pl 3L 30L -36 [ 3L
3L 4L2 -3L -L2
-6(x
-36 -3L 36 -3L
2
-
xL)
I
-3L L2
-3L 4£2
2
3x L - 2xL
2 ]
K = JoL BT G
which is the consistent-mass matrix for rotational inertia in a prismatic beam. Additional contributions to matrix M due to shearing deformations have also been developed and are given in Ref. 7. We now consider the simple loading case of a uniformly distributed body force by (per unit length) applied to a flexural element. Equivalent nodal loads ul points I and 2 [see Fig. 3.8(a)] may be calculated from Eq. (3.3-25) as
( - Jorl f
p,, t) -
T
by dx -- by L { 6, L, 6, - L}
12
B dx
(33)
(34)
For this integration the displacement shape functions / 1 through [4 were drawn frolll Eq . ( 17). By using f./.l't11'mll: l'd ,\•tr,·.1·.w·.1· wul .1·trol11.1·, wt• c1111 uvoid n.ipctitious inlc µ111110111, ow, llll' nrn,1, "l'l'tio11s o l om· d1111l•11si111111l l·lt·111l' nl s Allhou~h this rom·,•pt 1, 111thl'I t11vi11l l111 111111x111l l·k11u•111, 11 l'llll hr 1111111• II Mt•lul Im 1t11 s101111I
This expression for K is equivalent to Eq. (3 ._3-23) used previously. Turning now to the flexural element in Fig. 3.9, we integrate the moment of the normal stress r:r., about the neutral axis to obtain M ,, as follows: M,
=
L
~ U xY
dA
Then substitute the stress-strain and strain-displacement relationships from Eqs. (23a) and ( 19) to find 2 (42) M , = E y dA = EI
L
hu this clement we can take M, as generalized (or integrated) stress and as l' l'nc,·tilizcd strain . Then the generalized stress-strain (or moment-curvature) lljll'l'll(OI' /: is (43) F
m
94
Finite Elements and Vibrational Analysis
Chap. 3
which is the flexural rigidity of the cross section. Thus, from Eq. (42) we have
M,
= E
(45)
Thus, Then the shearing stress at any point on the cross section can be found with Eq. (49). For a flexural element, the relationship between the nonnal stress CTx and lhe bending moment M, is CTx
which is analogous to Eq. (40). In Sec. 3.5 we will see how to assemble finite elements, and in later chapters we shall learn methods for calculating the dynamic response q (t) of the nodes . After those steps, we can find the time-varying stresses C1(t) within each clement, using the equation
(3.3-12)
This ex pression may be converted to a special fonnula for generalized stress, depending on the application. For example, an axial element has the simple .~lrcss force relationship
l'x A
= -
(47)
where
=
.EB q(t)
(48)
where E=EA is the axial rigidity of the cross section. After the axial force has been determined from Eq. (48), we can find the normal stress using Eq. (47). A torsional element with a circular cross section bears the following relutionship between the shearing stress T and the torque M.,: T
M,, .I
-My -' = I
-yEBq(t)
I knee,
M,
= EB q(t)
(52)
I'quation (52) is a formula for the bending moment at any point along the length nexural element in tenns of the nodal displacements q (t). After this calcululion, we can find the normal stress at any point on the cross section from l•q (51). If we wish to find actions only at the ends of a member, we can simply 111111iiply its stiffness matrix__K and the vector of nodal displacements q(t). While 1h1H calculation is valid for framed structures, it has no such physical meaning liu two- and three-dimensional finite elements, where the resulting actions are th lllious.
111 11
11 /oni/ axes fo r a finite element are not parallel to global axes for the whole ~,, m lure, rotation-of-axes transformations must be used for nodal loads, dis-
Therefore,
Suhslilulinµ lhi s 1l•1111 1" wl'II m, I•~
I
,(S TRANSFORMATION AND ASSEMBLAGE OF ELEMENTS
l'x = EB q(t) l'x
= _ M,y
'l'his term as well as E = E and Eq. (45) can be substituted into Eq. (3.3-12), yie lding
K= fiJTEBdx
(T X
(50)
Mx =GB q(t)
Then integration over the cross section for terms in matrix K becomes unnecessary. Hence,
C1(t) = E B q (t)
95
Transformation and Assemblage of Elements
(44)
With this approach the operator din Eq. (21) is devoid of the multiplier -y. In addition, the generalized matrix B[Eq. (22) without the factor -y] may be used in place of matrix B. That is, B = -yB
Sec. 3.5
M,r .I
(42.}
<; 111ul l\q ( l'J) 11110 l\q . (J .J 12) prnduc:es
11l11l"l'll\Cnts~ accelerations, stiffnesses, and consistent masses. Thus, when the , I, 111l'llts are assembled, the resulting equations of motion will pertain to the 1,1 l11h11I directions at each node. The concept of rotation of axes applies to a force. a moment, a translation, 11 , 111ull rntatio.n.. velocit~celerations, orthogonal coordinates, and so on. I 1v.1111· \ . IO(a) shows a two-dimensional force vector F and its components E1111111~ 111 the I and y directions. The figure also gives the components E-· and F;,· 111 1111· d11cl'tions of inclined axes x' and y'. The scalar values of the components 111 11111•, 1u111s of the primed axes can be computed from those for the unprimed IKI 11, IIN follows: (1a) /•:, (11: 10 · I ' A11 I·: A11 f ~
I\•
( I•:
I.;,) • .I '
Aq / •;
A11 I•:
( I h)
96
Finite Elements and Vibrational Analysis
Transformation and Assemblage of Elements
Chap. 3
97
y
(2)
y'
or
.F' =RF The 2 x 2 matrix R is called a rotation matrix, consisting of direction cosines of the primed axes (with respect to the unprimed axes) listed row-wise. Hence, it is orthogonal, and the reverse transformation becomes
(a)
y y'
Similarly, a three-dimensional force vector F appears in Fig. 3. 1O(b) . Also shown in the figure are the force components E, Fy, and F, in the x , y, and z directions. Components E·, F;,·, and F,, in the directions of inclined axes x', y ', nnd z' are given as well. For this case Eqs . (1) can be expanded, as follows:
f'x· = (E + Fy + F,) · i' = A1,f'x + A12F, + Al3F, F,· = (E + Fy + F,) . j = A2, f'x + A22F, + A23F, F,, = (E + F;, + E) · k' = A31f'x + A32f', + A33F, I
(5a) (5b) (5c)
where k' is a unit vector in the z' direction. In matrix form, Eqs. (5) become A,2 Al3] A22 A23 A32 A33
[F,] F,
(6)
F,
for which the additional direction cosines are associated with axes z and z'. nquation (6) may be stated more concisely as in Eq. (3), but the rotation matrix R is of size 3 x 3. In addition , the reverse transformation given by Eq. (4) pertains to the three-dimensional case as well. Simultaneous transformation of a force vector F and a moment vector M lllay be accomplished by
A'= RA= [: :][~]
z
" 111 this expression Risa rotation-of-axes transformation matrix containing two Identical rotation matrices in diagonal positions. The reverse transformation is
z' (b)
Figure 3.10 Rotation of axes: (a) two dimensions; (b) three dimensions.
In_th1cse cxp;cs~ions the symbols I' and J' denote unit \ll'ctor.1· in the directions of -~ nndy . h .1rllwrn1ort.·, llw cm•f'fkit.•nls A 11 , A 1J, und so on, ore dir,•ction '.'(/,\' /lit',\ ' thl' p1 llllt'd IIXl'S wil lt ll'N Pl'l't lo lhl' 1111p1 lllll'd IIXl'N, ( Fo1 l'Xlllllple , i\ 1 IN lhl• dlll'rl1011 l·o.~ 1m• ol 11~11, 1 willt ll'~ Pl'l' I 10 1i,us 1•, 11 111 1 ~11 011 ) ' I ht• 11111 1r/~ 1111111 of h1, t I J 1,
of
0 J[F'J A= R'A'=RTA'=[RT 0 RT M' As menlioned before, displacements and other types of vectors also can be l11111 Nl111111t•d to and from local and global di rections. Therefore, we shall convert 111,• 1•q11111lo11s or motion for a finite clement from local axes to global axes. For 111111 p111111>Ml', ll•I us rt·wdtl' Eq . (1 . l ?2) rm incl ined axes, as follows:
M 'c1' I K'c1'
p '( t) I p /,(t)
(9)
Finite Elements and Vibrational Analysis
98
Chap. 3
Sec. 3.5
[DF] + [SFF SRF
In acc~rdance with Eqs. (4) and (8), premultiply Eq. (9) with RT and substitute I .. " " ' q = R q and q' = R q to obtain
M q + K q = p (t) + Pb (t)
(3.3-22)
in which
K = RTK'R
(10)
MFF MFR] [ MRF MRR DR
[DFJ = [AF(t)J AR (t)
SFR] SRR DR
parts gives
..
..
..
..
MFFDF + MFRDR + SFFDF + SFRDR (11)
and
MRFDF + MRRDR + SRFDF + SRRDR
Also,
p(t)
= RTp'(t)
(12)
MFFDF + SFFDF
= R_TpHt)
(13)
= AF(t)
(18a)
= AR(t)
(18b)
If support motions (at restraints) are zero, Eqs. (18) simplify to
and
Pb(t)
= AF(t)
(19a)
and
I !ere the matrix Rcontains rotation submatrices for all the nodes of the element. After stiffnesses, masses, and nodal.Joads for 'ndividual elements have h~~~ transformed to global directions, we can assemble them by the direct sqf./ness method [5]. With this approach we need only add the contributions from all the elements to obtain stiffnesses, masses, and nodal loads for the whole structure. Thus , by summation* we have
(19b) These equations will be used in subsequent work for calculating free displacements DE and support reactions A 8 (t). In many problems it is sufficiently accurate merely to lump tributary masses at the nodes of a discretized continuum [8]. When doing so , we form the lumped mass matrix Me for the whole structure as
lie
Ms= LM;
(14)
i= I
M, 0
0 M2
0 0
0 0
0
0
MJ
0
0
0
0
Mn,,
and Me=
n,
As(t)
= L p;(t) i=I
(15)
(20)
i=I
where n, is the n~mber of elements. In Eqs . (14) the symbols Ss and Ms represent l!~e ·~·tructural stif.(ness matrix and the structural mass matrix for all the nodes. S11rnlarly , the action vectors As(t) and Asb(t) inEqs. (15) are actual and equivalent nodal loads for the whole structure. Then the undamped equations of , notion for the assembled structure become
MsDs + SsDs = A,(t) + A .,b(t)
(16)
i~ wh!ch Ds an~ Ds are vectors of structural displacements and accelerations. Equation ( 16) gives the structural equations of motion for all nodal displacements, regardless of whether they are free or restrained . . In pre paration fo r solving Eq . (16) , we can rearrange and partition it, as lollows:
*
(17)
In this equation actual and equivalent nodal loads have been combined into a single action vector. The subscript F refers to free nodal displacements, while the subscript R denotes restrained nodal displacements. Writing Eq. (17) in two
and
M = RTM'R
99
Transformation and Assemblage of Elements
flm till' 111w111tlonNIll 111,~. ( Iii ) 111111 (, , i. 1111' 11111!1 IX 111 Vl'l'llll 1111 lh1• 1l11ht lllll Nl Ill' 1•x.p11111k•d , With /l'IIIN Ill 111111'1• II 1111• Hllllll' Nill IIM till' llllllii~ 111 Vt'I 1111 1111 1111• h II
where n,, is the number of nodes. The typical submatrix M1 in Eq. (20) signifies 11 small diagonal array defined to be (21)
lu this expression Mi is the tributary mass lumped at node;, and Io is an identit~ 11111lrix with l replaced by O wherever a nontranslational displacement occurs. ' l'hus, the lumped-mass approach has the advantage that the mass matrix Me is 11lw11ys diagonal, although not always positive-definite. As an b.umpid of transformation and assemblage of element properties, we 11111111 l'Onsider the plane truss in Fig. 3.1 l(a). For this structure let us find the 11ltll11L'l'IS matrix S, and the consiste nt-mass matrix Ms in rearranged and part ll Hllll'd fo1111s. Ass11llle th111 the cross sL·clional area of member 1 is equal to A, 111111 ol ,11L·111lwl' J is l'q111d lo O,Cl/\, 1111d th11l ol' tlll!lllber 3 is equal to 0.8A.
Finite Elements and Vibrational Analysis
100
2
Chap. 3
r
--2
0
=
}2
(22)
2}
Those for a typical joint j are illustrated at the left of Fig. 3.1 l(a). For this structure we have three degrees of freedom (numbered 1, 3, and 4) and three restrained displacements (numbered 2, 5, and 6). Note that the arrows for restrained displacements have small diagonal cross lines to distinguish them from free displacements . Figure 3. ll(b) shows a typical member i in the plane truss. Its ends are numbered j and k, and local axes x ' and y ' have their origin at point j. Axis x' lies along the member, and its positive direction is from j to k, whereas the direction of y' is perpendicular to the member. These axes are inclined at the angle y from the global axes x and y. Using the length of the member as
j2
t
101
Transformation and Assemblage of Elements
jl = 2} - 1
y
j O - j1
Sec. 3.5
0.8L
L
= \/x;k + Yjk
(23)
we find the direction cosines of the inclined axes to be
(a)
y
x' q~
y'
qk2
\tY
x,,
----- ---
=
cos )'
A 21
=
-sin )'
L = Cx
Xjk
=
\ /\J2 -
= -cy
A22
.
-
sm '}' -
=
cos )'
=
/
TABLE 3.1
Plane truss: (a) structure; (b) member.
An arbitrary system for numbering the members and joints of the truss nppcurs in Fig . 3. 11 (a). Member numbers are enclosed in c ircles adjacent to the mc1ubcrs, and joinl numbers arc placed adjacent to the joints. The numbering Hystr111 for joint displuccm<.: nt s is shown hy nu111bcrl·d arrows that depict the posi ti w dirl't'lio11s of the poss lhk· displ11l'l'IIIL'llts, ind11di111~ thoi,c ut restruincd polut11. Thl'Sl' displlll'l'llll'lll 111111thl' l'S llll' ol>tul11L·d l'n1111 llll' joint 1111111bcrs, us lollows :
Cy
(24) Cx
Direction Cosines Member i
Joint j
Joint k
I
1 1 2
2 3 3
3
(b)
L-
Member Information for Truss of Fig. 3.ll(a)
~qk1
k
YJk -
In these expressions the symbols Cx and Cy denote direction cosines of the member axis itself, with respect to axes x and y. Table 3.1 summarizes member information for the truss , including the arbitrarily chosen joint numbers} and k, the cross-sectional areas, the lengths, and the direction cosines cx and Cy.
2
Figure 3. U
A11
Length
Area
c, L 0.6L 0.8L
A 0 .6A
0.8A
0.6 1.0 0
Cy
0.8 0 - 1.0
By considering two nodal displacements in the x' and y' directions at each end of the member, we can write the element stiffness matrix K' for local axes as follows:
K'
= EA L
[
~~
-l 0
O 0
-~
~i
1 0 0 0
(25)
This matrix is the same as that for the nxial clement in Eq. (3.4-4), except that tht zuros arl' insl'lll•d for st iffm·ssl'H 111 the y' direction. Thus, the size of the
102
11111111 f- hm11111t11 111111 Vll11 nll1111nl A1111ly1l1
Cho p.
a
matrix is expanded from 2 x 2 to 4 x 4. Equation ( I 0) l'nnhks us to trunsl'or111 the stiffness matrix from local to global axes. !Jenee,
-c; K
= RTK'R
= EA L
[c:!Y -c;
(26)
[~
Noll' that tht\SC indexes uppeur us subscripts for the .t and y translations at joints Fig. J. I l(h). As un aid in the transferring process, the numerical values 111 1 t through k.2 arc listed down the right-hand side and across the bottom of r11d1 mutrix in Eqs. (29). After assembling the structural stiffness matrix using 1111' first of Eqs. ( 14), we obtain / 1111d kin
R is
Ss
(27)
~]
where the submatrix R can be written as R
=[
l'x
-cy
Cy]
(28)
Cx
Ki
=
0.48 0.64 -0 .48 -0.64
-0.36 -0.48 0.36 0.48
-0.48J -0.64 0.48 0.64
3
4
2
1.36 0.48 = EA - 0.36 L -0.48 -1 0
0.48 0.64 -0.48 -0.64 0 0
-0.36 -0.48 0.36 0.48 0 0
-0.48 -0.64 0.48 1.64 0 -1
2
3
4
Ss =
[SFF SRF
SFR SRR
2
=
EA
L
[
I 0 -1 0
K, -EA[~ L
0 0 3
00 0 0
-]0 1 0
2
5
0 1 0 0 0 -1 0 4
5
00 J 0 0
2
6
3
(3 1a)
4
5 6
0 1.36 -0.36 -0.48 ' 0.48 -1 ' -0.36 0.36 0.48 :-0.48 0 0 EA -0.48 0.48 1.64 : -0.64 0 -1 = - -------------------------~-----------------L 0 0 0.48 - 0.48 -0.64 : 0.64 ' -1 1 0 0 ' 0 0 -1 0 0 0 0 1
2
3
I
4
6
3
4
2
5
4
(31b)
2
5 6
6
wltl'rc the indexes down the right side and across the bottom show the rear11111gcment. As with stiffnesses, we can generate the consistent-mass matrix M ' for lrn·al axes of a typical member to be
6
M'
3
-1OJ 0
5
1
6
4
6
AJ.
2 O 6 1 [ 0
= pAL
0 2 0 1
1 OJ 0 1 2 0 0 2
(32)
111 lhis instance, the terms in Eq. (3.4-5) for the axial element are repeated for 1hr 11' direction, because accelerations in that sense also give rise to inertial
The terms in each of these matrices may be transferred to the appropriate locations in the structural stiffness matrix S, by culculating the following joint displacement indexes: )2 2/ JI
J. I
5
2
I
(29)
5
0 0 0 -1 0
J
3
K
-1 0 0 0 1 0
lfr11rranging and partitioning this matrix in the form shown by Eq. (17) produces
The latter matrix contains only the direction cosines of the member axis. When Eq. (26) and the data from Table 3 .1 are applied to each of the members, we obtain the following stiffness matrices for global axes: 0.36 EA 0.48 L -0.36 [ -0.48
103
l 111111,l1111n11!11111 111111 A11111nhln1111 111 I ln1111111lM
-c;
For this type of element the 4 x 4 rotation-of-axes transformation matrix
R=
.u,
s.
-cxCy
-cxCy
nl'
• 1.J.
,111ro11s. Thus, the consistent-mass matrix for inclined axes is of size 4 x 4. It 11111y he lrnnsformcd to global axes using Eq. (11). However, when that equation 1- 11pplil.'d, we !ind the resu lting matrix M to be the same as matrix M ' in
ht ( \l). Thc1cforc , wt· rnnl'luclt· that for II plnne truss 1111u111
111111tix ii!
i11v111 lunl
element the consistent-
with rotation of uxcs.
/\pply1111:, Eq . ( \l) 1111d Oil' <111111 ltu111 Tuhlt· \. I, we find the consistent-
104
I lnltn I lnmn1111 nnd Vlhr1tl11r111I A11nly•I•
Chnp. J
mass matrices for the members to be M1
=M
M2
M = pAL s 6
0 0.36 0
M,
0 1 2.72 0 0 3.28 1 0 0 0.64 0.36 0 2
O.MM
(JJ)
0 1 0 3.28 0 0.64
0.36 0 0.64 0 2 0
0 0.36 0 0.64 0 2
4
5
6
3
2
3
Vlhrnllonnl A11nly1l1
105
to the diagonaLten:n_s.
j11 1·11lw1 M, o, M ,. Al this point we shou ld observe that considering the masses for the three-
1111111lll'1 truss lo be associated only with jQint translations is a poor approxi11111t1011 to the truth . In fcw.L. there often will be flexure in the membeis thl!.t hl~11if11.·1111tly influences the dynamic characteristics of the structure. For examl'k, ti II time-varying force is applied in the x direction at the middle of member I l'lt't' Jlig. 3. 1 l(a)I, the flexural deformation in that member will dominate the 11 ~p1111se. We shall deal with this type of problem for trusses by the component11111111• method in Sec. 10.6. In the meantime, we will use Ms or Me in the forms l111w11 here, even though the numbers of members and joints are small.
(34a)
4
5
It VIBRATIONAL ANALYSIS
6
111 ~kc. 3.5 we derived undamped equations of motion for free nodal displace111r11l'> of a discretized structure [see Eq. (3.5-19a)]. If there are no applied ,1111011s, these equations can be written in homogeneous form as
Rearranging and partitioning this matrix as shown in Eq. (17) gives M,
111
1111'11 111 lh1.· Joi111~ of 11 1111ss, tlwy 111ny IK· mld~·d directly
= 0 .36M
In these expressions the matrix M is the same as M' in Eq. (32). Terms in matrices M 1, M 2, and M3 may be transferred to the consistent-mass malrix M, for the whole structure using again the displacement indexes j 1 through k2 . Thus , when M, is assembled in accordance with the second of Eqs. ( 14), we find that 2.72 0
11
= [::: ::]
MD+SD=O
(1)
I
2.72 1 1 3.28 pAL O O - 6 0 0 0.36 0.64
0
0 3
0 : 0 : 3.28 :
0 0
I I
1 1 :I 2.72
0 : 0 0.64: 0.36 I
2
4
0.36 0.64 0 0 2 0
0 0 0.64 0.36 0 2
5
6
I lt•1l' the subscript F (for free displacements) is omitted to simplify the notation. Equation (1) has a known solution [9] that may be stated as follows:
3 4
2
I I
4
'
2
(i
= I, 2, ... , n)
(2)
wh1•1c II is the number of degrees of freedom. In this harmonic expression , 4>; 1 11 vector of nodal amplitudes (the mode shape) for the ith mode of vibration. I h1 symbol w1 represents the angular frequency of mode i, and a; denotes the 11/t,111• a11gle. By differentiating Eq. (2) twice with respect to time t, we also find
5
6
where the indexes at the right and bottom again show the pattern of rearrangement. Alternatively, the consistent mass matrix M, in Eq. (34b) could be replaced by a lumped-mass matrix Me, as follows: 0.68 0 0 0 0 0 0 0.82 0 0 0 0 0 0.82: 0 0 0 0 Me= pAL ------------ -- -----L----------------0 0 0 0 : 0.68 0 0 0 0 ' 0 0.5 0 ' 0 0 0 '' 0 0 0.5
D; = 4>; sin (w;t + a;)
(34b)
111111
D;
= - wT4>; sin (w;t +
a 1)
(3)
'111h1,litution of Eqs. (2) and (3) into Eq. (1) allows cancellation of the term 111 (1,,,1 f- a1), which leaves
(4) 3 4
2
(35)
5 6
I Ills ruanipulation has the effect of separating the variable time from those of ,pun· , und we are left with a set of n homogeneous algebraic equations. Ll4uution (4) hus thc..form oLthe algebraic eigenvalue problem From the 1hr111 y of homogeneous equations 1101, nontrivial solutions exist only if the d1'11•1111i11unt of the coefficient mutrix is equal to zero. Thus,
JS
"
For L'IIL'h muss matrix M, and Ml', the totul muss of llll' trt1ss is 2p/\/, for hoth llw I and 11 dirL'rtions, us it shou ld hL· . ll 11011til1ur1u1.il nHIL'L'Utrntcd 11rnss1.·:,; also
cu f MJ
0
(5)
I ,p1111:-;i1111 nl this d1•ll'll11i1111111 yi1• ld~ 11 pol y11n111111l of order II called the charac'lh1• 1111111l~ ,o: ol llm poly1111111i11 l 1111• thl' 1·ht11'<11·1,·ri.1·fic values,
1,·1/1ft11•1111,1f11111
11111111 ~ ln111nnt1 nnd Vll11 nll1111nl Annly•I•
100
or eigenvalues. Subslilulion of lhese rools (one at n tillll') 11110 lhc homogeneous equations LEq. (4)J produces the characteristic v,•cwrs. or l'i>:envectors <1>1, within arbitrary constants. Alternatively [9), each eigenvector may be found as any column of the adjoint matrix Hf of the characteristic matrix H;, obtained from Eq. (4), as follows: H;; = 0 (6) where (7)
The methods implied by Eqs. (5), (6), and (7) are conducive to hand calculations for problems having small numbers of degrees of freedom. Examples at the end of this section demonstrate such calculations. However, a structure with a large number of degrees of freedom must be handled by a computer program, as described in Sec. 3.8. Such a program would include a subprogram (or subroutine) for calculating eigenvalues and eigenvectors. The most efficient type of subroutine for structural vibrations accepts the eigenvalue problem only in the following standard. symmetric form: (A - A;l )X;
=0
!11
Ch11p. 3
(8)
in which A is a symmetric matrix and I is an identity matrix. The symbol Ai dcnoles the ith eigenvalue, and X; is the corresponding eigenvector for a new syslcm of n homogeneous equations. We can put Eq. (4) into the form of Eq. (8) hy fm;toring either matrix Sor matrix M, using the Cholesky square-root method 11 , 51. We choose to factor S for an important reason that will soon be apparent. Thu.~,
111
107
Vll11111l111111I A11nly• I•
I 'Iii ( I I) I, wl11ch is gunruntccd to be symmetric. In addition, we see that the , 1p1•11v11l11t' Ai is cquul lo lhe reciprocal of the square of the angular frequency. l ht~ 11, tlw t'(.>ll~<:quencq of ~hu; in Eq. (10) is related to •I•, hy lhc last of Eqs. ( 11 ). This expression constitutes a change of coordinates 111 11 IIL'W set where the stiffness matrix is equal to I. After the eigenvalues and 1,.1•11vcctors have been found from Eq. (10), the angular frequencies and mode h11pt·s (i n the original coordinates) can be determined as 1
W· = - -
'
VA;
Because the matrix Mu in the new coordinates is symmetric, all of its 1lpr11vectors are linearly independent [10]. In addition, two eigenvectors ~,; 111111c1,,IJ corresponding to distinct eigenvalues A; and A1 will be orthogonal with ll ~Pl'L't lo each other. Thus,
(i
* i)
l luwcvcr. the back-transformation giyeilJl.Uhe second expression_in Eq£. (12)
,loc~ 1101 necessarily preserve orthogonality among the eigenvectors, Instead, the , l~l·nvcctors ; and <1>1 in the original coordinates are orthogonal with respect IU umltix...S.. as follows: TS j = b;u- Ts
u-1«1>uj = ~;uj = 0
lli2
111 ~L'C. 4.2 we will show that the eigenvectors are also orthogonal with respect 111111rix M. From vibrational analysis of structures, we may find repeated frequencies 1111 r,•peated eigenvalues) as roots of the characteristic equation. If an eigenvalue I tl'peated m times, it is said to be of multiplicity m. We can find m linearly 1111ll•pendent eigenvectors corresponding to the repeated eigenvalue, but such \'l'l'tors are not unique. A new set of eigenvectors can always be formed as linear 111111binations of the original set and still satisfy the eigenvalue problem. That is , 111
(9) where lhe factor U is an upper triangular matrix. Substitute Eq. (9) into Eq. (4) to obtain
(UTU - w7M);
=0
Then premultiply this equation by u-T and insert I = u - 1 U after matrix M, which yields
m
i=I
u - T(UTU - w7M u - u); = 0 1
lkwriting terms in reverse order, we find that Wlll'll'
M11
U 'l'M U
1
,,,1
l llJ
Eq11ntio11 ( 10) is now in thl' sln11d111d , ~y 111111l•t1 ll' 1111111 of till' l'igl·11val11l' p111hll 111 g1vt·11 hy 11q (H) TIil' 1111111h A i'4 ll'Pll'~t·11tt•d hy M11 ltlll' lu st of 0
(15)
i= I
wht•ic ci is a scalar multiplier of the ith modal vector and Am is the repeated lu1•11vulue. ll is always advantageous t0 form a new set of eigenvectors that are 111 thogonal with respect to each other. They will automatically be orthogonal to 1 l l,ll 11vectors corresponding to distinct eigenvalues. But there is still an infinity 111 d1oil'es for nonunique orthogonal vcclors corresponding to a repeated eigen\'llh11·. The (iram Sd1111idt 1>1·tlrow malizutio11 procedure [ 11] is a formal mathe11111l 1t·11 l 11pproad1 rn111111011ly usL'd lo l'o11st111c1 1111 orthogonal set of eigenvectors 1111111 11 l11ll'11rly i1uk•p1•11dt•11t Ht•I II llll' 'tllflt1l'"" 11ml11~ ts s1:111tlk·hnill', tt rn1111ol hl' fnl'torl'd as dcslTibcd 1
(Mu - A;I)u1 = 0
m
M u L c;u; = · AmLC;u;
1111110 1:-lomontN nnd Vlh111ll1111nl Annly•h•
108
Chnp. 3
Soc. 3.0
109
Vlbrntlu1111I A1111ly11I•
hen:. bccausc ut h:ust one rig,id-body modG (with
where M 1l 2 contains diagonal terms equal to the square roots of those in M. In this case the transformation to standard, symmetric form yields (17)
where A;=
WT
(18)
In the first of Eqs. ( 18) the matrix M- 112 contains the reciprocals of the diagonal terms in M 1l 2 • After finding the eigenvalues and eigenvectors from Eq. (17), we can obtain the angular frequencies and mode shapes from W;
= \IA;
(19)
Note that the numerical advantage mentioned before is not present in this )!Pproach. In general , the process of directly extracting the roots of the characteristic equation for Eq. (8)' must be done iteratively and is not efficient for large problems. If the eigenvalues and eigenvectors of all the modes are to be found, it is best to use Householder transformations (see Sec. B.2) and convert matrix A to tridiagonal form. Then the final values of A; and the vectors X; can be determined by iteration with the QR algorithm. On the other hand, if only a few modes are desired, the method of inverse iteration with spectral shifting is more efficient (see Sec. B. l). Example 3 .1 Figure 3. l 2(a) shows a plane truss with two degrees of freedom at joint 1. Assume that the cross-sectional areas of members 1 and 2 are equal to 0.8A and A. Using the consistent-mass approach, find the angular frequencies and mode shapes for thi s structure. The 2 x 2 stiffness matrix S for the free displacements in this problem is
S = EA [ 0.36 L - 0.48
- 0.48] 1.64
11 Wl' ll•I .I'
!•:/\ / /, 1111d Ill
\,} HriA I
(a)
i
I I I
I I I I
(b)
I
pAl, , .1 .2H O (1 0 1,2 H
/ 11 , lhl• l111111111•l' lll' IHI N l' lllllllillllN in
0.6L----
(a)
and the consistent-mass matrix has the diago nal form
M
1..,__ _ _
(b)
1-'.q. ('1) hl l' lH1W 1
Figure 3.12
(c)
(a) Plane truss; (b) mode l; (c) mode 2.
Finite Elements and Vibrational Analysis
110
0. 36s - mw'f [
-0.48s
-0.48s 1.64s - mw'f
Chap. 3
=
111
Vibrational Ana lysis
y
J[ 1;J [OJ
Sec. 3.6
(c)
0
in which the 2 x 2 array of coefficients is the characteristic matrix H;. Setting the determinant of H; equal to zero in accordance with Eq. (5) produces the characteristic equation, as follows:
m 2 w1 - 2mswt + 0.36s 2 = 0
G)
(d)
s
= 0.2m
w
2
1.s.:. m
=
2
(e)
from which we find the angular frequencies to be Wi
=
(oTs = \j-;;-
0.6049 L
@.
\jp
W
2
=
_ 1.815 --
L
lo -
p
z
/
'
•
1~
The roots of this quadratic equation are
wf
t
0
2
X
o/.l
L (a)
(f)
S ubstitution of these values (one at a time) into the homogeneous equations (c) produces the mode shapes (g)
q>I = [~]
(b)
which arc scaled arbitrarily. These shapes are depicted in Fig. 3. l 2(b) and (c) .
!001.uiue 3..Z A t·nntilc vcr beam consisting of one fle xural element appears in Fig. 3.13(a). It is fixed nl nodc I but has two degrees of freedom at node 2 . Assuming that the beam is prismatic, dlltllrn1i ne the angular frequencies and mode shapes, using translational consistent mass ll'IIIIS.
From
Eq. (3 .4-24) , we find the 2 x 2 stiffness matrix for the free displacements
to he
(c)
S = 2EI [ 6 L 3 -3L
- 3L] 2L2
Figure 3.13
(h)
wr =
Also, Eq. (3.4-26) gives the consistent-mass matrix as
pAL [ 78 M, = 210 - Ill
Lettings
-
lll]
(i)
2L2
_ 3.533 W1 -
uhlul n the ho mogeneous equations
I
6(s - 13mw;)
L (3.1· - I lmwr)
-L(3s - t tmwn ] [ <1>11] = 2L2(s mwr) <1>21
[OJ 0
(j)
whc1l' Ilic 2 x 2 cocfli cicnt matrix is the c hnructeristic matri x 11,. As indicated by l\q. (~). Wl' Sl't tlw dctc1111in11nt ol' 11, cqunl to Zl'l'O , produci ng ( k)
whil'h IK 1111' 1 l1111 m h·iiNll1
1•q1111111111
'lh,•11 1hr 1111 ,tM111 l
1
q (~ ) 111 1• lrn1111l to lw
s m
0.02971 -
2
W2
s m
= 2 .885 -
(C)
' uhstituting the known values of sand m and taking square roots , we find the angular li1•quc ncics, as follows: ·
2EI /C and m = pAL /2 10, we substitute matrices S and M, into Eq . (4) to
r
(a) Cantilever beam; (b) mode l; (c) mode 2.
~
/El \/pA
Wz
= 34.81
L2
/El
\jpi4.
(m)
Wl11111 these formulas arc compared with exact values [9] , the errors are found to be I 0.48% and e 2 = +58%. Thus , the first-mode frequency is a good approxi111111 hu1, hut the second-mode frequency is very poor. In this cxumplc we obtain the mode shapes by using the first column of the adjoint
, 1
11111t ilx
II ,', which is
II '/,
. I
21.2(.1·
/, ( \,\
lll(rJl ) I I 1111,,f )
l
(n)
f lnltu I l11111nn1 11 nncl Vih111tl111111I A11nly1l1
112
Substitution of cu ~ and
wl from (1)1 =
Chnp. 3
113
!iy111111111tlr 111111 A11t l1ym111nltlt: Mod111
Eqs . (f) into thi s column yil'ld~
L ] [ 1.378
I
(())
1- sym.
Of course, the second column of Hf would serve equally well. Figures 3. l 3(b) and (<:) show the mode shapes in Eqs. (o), which are both scaled so that the translation iN numerically equal to L.
I
3. 7 SYMMETRIC AND ANTISYMMETRIC MODES
Figure 3.14 shows two examples of symmetric structures. The plane frame in part (a) of the figure has one plane of symmetry, while the discretized plate in part (b) has two such planes , as indicated by the centerlines. When a structure has one or more planes of symmetry, the natural mode shapes for vibrations all will be either symmetric or antisymmetric with respect to those planes [12]. In problems of this type we need only analyze a portion of the original structure . The reduction to a smaller sized problem may be accomplished by introducing artificial restraints at joints located on planes of symmetry. In addition, the properties of members that lie in those planes must be altered. These changes may be incorporated into the structural data for a computer program and do not require any additional coding. If there is one plane of symmetry [as in Fig . 3.14(a)], only half of the structure need be analyzed. If two planes of symmetry exist [as in Fig. 3.14(b)], only a quarter need be analyzed, and so on. When a vibrational mode is symmetric with respect to a plane of structural symmetry. the nodal displacements. strains. stresses. and reactions will also be symmetric with respect to the same plane. Therefore, nodes located on a plane of symmetry must be restrained in such a manner that the structure deforms symmetrically with respect to that plane. Figure 3 .15(a) illustrates schematically a typical node j located on a plane of symmetry that is normal to the x axis. The figure also shows nodes k and k' that are symmetrically located on opposite sides of the plane. Displacement vectors at each of these nodes indicate a symmetric pattern of deformation. Note that translations in the y and z directions and rotations in the x sense are all in positive directions at both points k and k' . Therefore, we conclude that the same displacements on the plane of symmetry must be free to occur. These displacement vectors at point j are labeled j 2, j 3, and }4. On the other hand , the translations in th~ x direction and the rotations in they and z senses are in opposite directions at points k and k'. Thus , the same displacements on the plane of symmetry must be set equal to zero. Hence, the vectors labeled} 1,} 5 , and j6 at point} need to be restrained, as indicated by the small slashes on their arrows. In general , the component of nodal translati<)Jl normal to a plane of__symmetry and the C<)111po11l'nts of rotnt ion in the plane 11111:,t he prcvc11t1.:d in order to cnforcL' a i:;ymnwt1 k p11ltt·1n of distortion.
y
I
7,i 77, -
X 7} 7,7,
7,i 7,7,
7,i 77,
7} 7,7,
(a) /
/
/--sym.
(bl
1,·11111n1 J.14 Sym111c1ric i,11 u~·tl11~·s: (u) frumc; (b) plate.
114
Finite Elements end Vlbrntlonol Analysis
Chap.3
y
t
/A
I
Plane of / symmetry~
k·--
//
/ +5
/ -t.1 / ( ,· t r----~-~------+--- •
j2
Y,k,
I/
V
z1,,
/
Jl'J3 /.
I/
/4----x
z
I
,,-4
-1
i1i4) /
i6
/
//
V
z1•
/
/L:. ---y'
1
,
~
t
I I I /I 1/ I I / Y;,
I / I /
y
/,1
-·/i
/
I I Yw
(/
t i5
j
tT
j2
~ - ---+----,- -- ~ · i4) '/
r-----+--r-----~--. -
/.i1 3
I ,/ 1 I , j6
7~/
z
/
I
/
t t
I I I k•-I / I 1/ I I I Y;,
// /
/
7/:.~ /
I ,
V
115
Pl'CI to
the plane.
If a member of a framed structure or an element in a discretized continuum ltt 111 a plane of symmetry. we must divide its rigidities by two in order to cut llll ~tn1cture into egllill.parts, In the case where a member lies in two P-lanes of \ 111111ctry, we need to divide its rigidities bs...fuut, and so on Jf a member ar 11 ll111h: element is normal to and bisected by a plane of symmetry, we must ,llvldt it into two equal parts and introduce new nodes on the bisecting plane th.at llll H'Strained as described aboYe.,
(a)
Plane of ---. / symmetry 7
Symmetric and Antisymmetric Modes
If a vibrational mode is antisymmetric with respect to a plane of structural \·111111ctry, the nodal displacements, strains, stresses, and reactions will also be 111111symmetric with respect to the same plane. For this case Fig. 3.15(b) depicts ill!iplucements at points k and k' that represent an antisymmetric pattern of ,Ii l111111ation. That is, the translations in the x direction and the rotations in the 1· 1111d z senses are all in positive directions at both points k and k'. From this we , 1111dude that the same displacements on the plane of symmetry must be allowed h I nl'nir freely. They are the displacements labeled j I, j 5, and j 6 at point j. On 1111 olhcr hand , the translations in they and z directions and the rotations in the , 1•11sc are in opposite directions at point k and k'. Therefore, the same displace111r 11ts on the plane of symmetry need to be set equal to zero. This may be ,11111111plished by introducing restraints corresponding to }2, }3, and }4, as 111d1t·ntcd by the slashes on their vectors. In summary, the components of nodal 1111111,lation in a glane of sxmmetry and..Jbe com12.onent of rotation normal to the 11li111t· must be prevented to give a pattern of distortion that is antisymmetric with I\
,/
k'i/
n, 3.7
z1,
I , 11111plc 3.3 I 11•1111• \. I 6(a) shows a simply supported beam composed of two flexural elements. This 111 11111 hus four degrees of freedom and is symmetric with respect to its centerline, as 1111111 ntt·d in the figure. To take advantage of symmetry, we shall analyze only the 111t111 l111nd half, using restraints at node 2 for symmetric and antisymmetric deformations. I 11•1111• 1. 16(b) illustrates the symmetric case, for which the rotation D 3 in the plane of 11111111•1ty is restrained. In addition, the antisymmetric case is given in Fig. 3.16(c), "111 11· lhc translation D 2 in the plane of symmetry is restrained. For each of these cases ", lluw only two degrees of freedom instead of the four degrees of freedom in the 1111.,111111 problem. Now let us find two angular frequencies and mode shapes from each 111 1111• NUhsidiary problems, using translational consistent mass terms. Fm lhc symmetric case in Fig. 3. 16(b), the 2 x 2 stiffness matrix for the free 1ll•11l11n•mcnts (D2 and D4) is
1// v (b)
2£1 [ 6 3e 3e u 2
s=7 \\hlo h INtl111w11 from Eq. (3.4-24).
Figure 3.15 Restraints on planes of sym111ct1 y: (11) syn11m·1ric modes; (b) anti· symmetric mmks.
M,
" 11v, 11 hy hi ( I •I
J
(a)
Also, the 2 x 2 consistent-mass matrix becomes r>A
fr
420
156
I H'
)(!) Thl' l h11111l 'h•11NI 11 1111111 ix
l.3el 4f2 II, tnr 1hi~
(b) L'IISL' hns
1lw I'm m
Finite Elements and Vibrational Analysis
116
Chap. 3
117
Symmetric and Antisymmetric Modes
Sec. 3.7
WT =
2
0.007305 !_
W3
m
y
s
(e)
= 0.9026 m
Substituting the known values of s and m and taking square roots , we find that I
, D'/4
Ja,
CD
/Ji-,
Wt
0
DJ/~ e--------~
(ii
= 9.909
3
D
. 4/ ~ -
W
\JpA
l2
= 110.}
3
l2
(ii
\JpA
(f)
where L = 2£. When these formulas for the angular frequencies are compared with exact values [9], the errors are found to bee, = +0.40% and e3 = +24%. We obtain mode shapes corresponding to w, and W 3 using the first column of the adjoint matrix Hf, as follows:
--x
e~ , j
z
2
2£ (s - 2mwT)
H1 = [ ' -£(3s + 13mwf)
(a)
Substitution of
J
(g)
wr and w5 from Eqs. (e) into this column produces (h)
Mode 1
These mode shapes appear in the left-hand portion of Fig. 3.16(b) . Of course, each of them represents half of a symmetric mode shape for the whole beam. Considering now the antisymmetric case in Fig. 3.16(c) , we form the stiffness 11mtrix for the free displacements D3 and D4 as
S 2£/e [2 1] =
lb)
Mode 3
1
(i)
2
1111d the consistent mass matrix is 3
_ pAC M, - 420 Mode 2
[
4
I hl.ln the characteristic matrix becomes 2
3
H, =
[2(s - 2mwf) s + 3mw f
Mode 4
where .1· giws
2/i//
1m 2 wt - 22mswf
(c)
(a) Symmetric beam; (b) symmetric modes; (c) antisymmetric modes.
HI =
s-
e nnd 111 1
2
w, M, =
[
6(s - 26mwn e(3s + l 3111c,l )
e(3s + I 3mwl) 2 (s 2mwn
ze
l
11111 11
2
W2
111m 'l'l
0
(k)
1 h111 111 h il Mih 1·q 1111tl1111
,ll ,11111111; I I.I ~
1111 11111(~ 111 l'q
0
(ti) 111r
+ 3s 2
(£)
= 0
1 s
= - -
1m
2
W4
=
s 3-
(m)
m
ding us before, we find the angular frequencies to be ,,, 1
I" 1111
J
which the roots are
(l' )
pA (' / 420 . /'k'lling thl· dctc, mi111111t or 11111tri x 11 1 equal to ~.l'1t1 •l"l ',111 ' 111,'
Whh h
s + 3mw f 2(s - 2mwf)
whl·re s and m are the same as before. Expanding the determinant of H, and setting it 1 qunl to zero gives the characteristic equation
-.....::_:::;.. Figure 3.16
(j)
-3
(d) lt11 w li1d1
th1• 1·11111 ~
111r
43. 82 /,J
I 11 'Y.,
J/i/ 11A
111111
I J. / 17,,
(1) ,1
2001.8 I,
J!J·I 11/\
(n)
118
Finite Elem ents and Vibrational Analysis
Mode shapes are given by the first column of H 1, which is a
_
Hi; -
[
2(s - 2mwt) -(s + 3mwt)
J
119
Program VIB for Vibrational Analysis
Chap. 3
Flowchart 3.1
Main program for VIB*
IOI
(o)
SDATt
I. Read and write structural data.
Substituting w~ and w~ from Eqs. (m) into this vector yields 2. Generate structural stiffness matrix.
(p)
These mode shapes arc displayed in the left-hand part of Fig. 3. 16(c), where each of them depicts half of an antisymmetric mode for the whole beam. The error calculated for the first angular frequency w1 is acceptable, while those for the other modes are not. Better accuracy for these modes could be obtained by using more finite elements with more nodal degrees of freedom.
3. Generate structural consistent mass matrix.
4. Convert eigenvalue problem to standard, symmetric fonn.
3.8 PROGRAM VIB FOR VIBRATIONAL ANALYSIS
In this section we discuss a computer program named VIB for vibrational analysis of any type of linearly elastic framed structure or discretized continuum. Steps in the main program appear in Flowchart 3. 1, which calls seven subprograms indicated by the names in double boxes. Subprogram SDATreads and writes input data for a particular type of structure and calculates nodal displacement indexes. Subprogram STIF generates the structural stiffness matrix (for free nodal displacements only) by assembling contributions from element stiffnesses, as indicated by the first of Eqs. (3.5-14). Next, the consistent mass matrix for free nodal accelerations in the structure is assembled by Subprogram CMAS, using contributions from individual elements [see the second of Eqs. (3.5-14)] . The subprogram named STASYM then converts the eigenvalue problem to standard, symmetric form by factoring the structural stiffness matrix, as shown in Eq. (3.6-9). If the stiffness matrix is found not to be positive definite , the mass matrix is factored instead. However, if the mass matrix is also found not to be positive definite, an error message is written and calculations stop. Otherwise, the subprogram EIGEN2 solves the eigenvalue problem using Householder transformations and the QR algorithm (see Sec. B.2). Then the eigenvectors are transformed back to the original coordinates with Subprogram TRAVEC, using the second of Eqs. (3.6-12). Finally, the subprogram named RESl writes the resulting angular frequencies and mode shapes obtained from solution of the eigenvalue problem . As shown by the flowchart, several structures of the same type may be processed in one run of the program . Program VIB may be specialized to become VIBCB for continuo us beams, VIBPT for plane trusses, and so on. The main program for each specialization has four subprograms that are different for each type of structure, as indicated by the second footnote helow Flowt:harl .1. I . For l'Xa111plc, the suhpmgru111 named SDAT hl't'O llll'' SDATC'B fill II l'1111ti1111011s lw11111 , Sl>AT(Y(' fo1 n plalll'
5. Solve eigenvalue problem by Sec. B.2.
6. Transfom1 eigenvectors to original coordinates.
7. Write angular frequencies and mode shapes.
e
Go to IOI and process another structure.
END
*Applies to any type of linearly elastic structure. t Subprograms that differ for every type of structure.
tt nd so on. Notatio n for this and other programs is given as Part 5 in the II 1 111 11otatio n near the end of the book. Detailed steps in the logic for various ulipwgrnms are shown in the flowchart for Program DYNAPT, which appears 111 Appl•ndix C. '1'11hlc 1. 2 shows preparatio n of structural data for plane trusses. In the 11111111 ltm· of the table arc the number of nodes NN, the number of elements NI , lhl• 1111111bcr of restrained nodes NRN , the modulus of elasticity E, and the 11111 11 d1•11sily RIIO. Enl'h linl' of till' duta for nodal coordinates (NN lines total) \ 1111111111, 11 nmk 1111111lw1 J, tlw I l'Oo11li111ttl' X (J ) of tlw node, nnd they coordinate
1111 N,
Flnlto El omonts 1111tl Vlhr 11tlo1111I Arw lysls
120
121
l'royrnm Vll3 lo, Vlbrn1lo1111I A11ulysl11
Chop. 3
t4 ,_T
TABLE 3.2 Structural Data for Phmc Trusses
3
No. of Lines
Type of Data Problem identification Structural parameters Plane (a) (b) (c)
Items on Data Lines Descriptive title NN ,NE, NRN, E, RHO
I I
truss data Nodal coordinates Element information Nodal restraints
O.Bl
J, X(J), Y(J) I, JN(T) , KN(I), AX(l) J, NRL(2J-l), NRL(2J)
NN NE NRN
Y(J). The element information (NE lines) consists of the element number I, the j node JN(I) at one end , the k node KN(I) at the other end , and the cross-sectional area AX(I) . Each of the NRN lines in the last block of data contains a node number J and two code numbers which indicate the conditions of restraint at that node. ThL· symbol NRL(2J-1) denotes the condition of restraint against translation in the \ direction at node J , and the term NRL(2J) gives the restraint against translation in the y direction. The convention adopted in this book is that if the restraint exists, the integer 1 is assigned as the value of NRL; but if there is no restraint, a value of zero is assigned. Of course, the vector NRL must initially contain on ly zeros (by clearing it) before the restraint information is read. If desired, we could include data for external masses that are idealized to be concentrated at the nodes of the structure. Such extra masses may be con, veniently added to the data lines for nodal coordinates. When the program assembles the consistent mass matrix for the structure , these concentrated masses would be added to diagonal terms for translational accelerations . Such a procedure for handling superimposed masses could be applied to any type of framed structure or discretized continuum.
0
~r
• t J - - -- - - - - - 0 ~
~l~
\- ~ 0.6l~
/
L
5
X
(b)
(a)
_l_
Example 3.4 We will now use the specialized program VIBPT to find frequencies and mode shapes for the plane truss with three members examined previously in Sec. 3.5. This truss is reproduced in Fig . 3 .17(a), where the free and restrained nodal translations are indicated by numbered arrows. Structural data for this problem are listed in Table 3.3, in wh ich the following numerical values are assigned to the parameters E, p, L, and A: E = 3.0
x 104 k/ in. 2
L = 250 in .
p = 7.35 x 10
A
7
k-s2 / in.4
10 in. 2
where the material is steel and US units arc used. (Sec Appendi x A fo r a discussion o1 systems of units and 11111tcrinl properties.) Tobie \.'I contuins the rnmputer n:sult ~ lor rlus l'x11111pll'. In the rir st pnrt of tlw lnhle, Wl' Sl'l' 1111 "l'l'itn" pr 1111 o l lht· d11111 1l'lld hy lhl' l'0111p11lt•1 AIN11 co111p11ll'd 1111d prinll'd lll l' till' t'lt•1t \l'III l1•1111th11 l•l • 1111' d11 1•1 l11 111 l m1 111· ( ·x 111111 ( 'Y, 1111· 11111111w, 111 \lt'j,\ll'\'S 111
(c)
l•'l1111rc 3. 17
(d)
(a) Plane truss; (b) mode l ; (c) mode 2; (d) mode 3.
11,, 111 1111 Nl>I•, 11 nd the number of nodal restraints NNR. This is followed by the angular 111 q,11 Ill il'~ 11 11d 111odc shupes for cuch of the natural modes of vib.ration, which may be , 111tll1 i11rd hy h1111d c11k11h.11ions. The angul ar lh•qucnc:7 for mode .1 1s w.'... = 420.~;ad/sec '"' ) ). 111111 1111 IIHHk]. IS (t/J 11 68 s 1 '1111d th111 tor mode 3 IS W3 - 1862 s . Each 11u11lnl vi- 1 1111 11 11 ~ hl·t·11 ,1111 1111tli:n·d with H'N(ll'l I 111 its l11rgcs1 term, and the mode shapes 11• d1 ph lrd Ill l•tp \ I l(h) (d )
Finite Elements and Vibrati,Jnal Ana lysis
122
TABLE 3.4 TABLE 3.3
MODE 2 ANGULAR FREQUENCY NODE DJl 1 8 . 6725E-01 2 -l.7149E-01 3 O.OOOOE+OO
Alphanumerical Values
Problem identification Structural parameters
Example 3.4: Three-member plane truss 300(().0 0.000000735 3 3 2
Plane truss data (a) Nodal coordinates
I
0.0 150.0 150.0
2 3
2
I I
3
2
I 3
0
I
(b) Element information
(c) Nodal restraints
TABLE 3.4
(Continued) ·
Structural Data for Example 3.4
Type of Data
-
123
Program VIB for Vibrational Analysis
Chap.3
I
MODE 3 ANGULAR FREQUENCY l.8618E+03 DJ2 NODE DJl 1 1. OOOOE+OO O.OOOOE+OO 2 -6.0504E-01 -6.1068E-01 3 O.OOOOE+OO O.OOOOE+OO
0.0 200.0 0.0
2
JO.O
3 3
6.0
8.0
1.1677E+03 DJ2 O.OOOOE+OO l.OOOOE+OO O.OOOOE+OO
K11.11mple 3.5 1<1 urc 3.18(a) shows a plane truss with a larger number of members. The cross-sectional
11
I I
111• s 1 11
Computer Output for Example 3.4
PROGRAM VIBPT
of diagonal members are equal to I .SA , and those of other members are equal to \ This truss happens to be symmetric with respect to its centerline, so we need only 11111lyzc half the structure. For this purpose, we impose restraints in the plane of sym1 1111,11y for symmetric and antisymmetric deformations, as indicated in Figs. 3.18(b) and ti ), In addition, the cross-sectional area of member 12 is divided by 2 . Assuming that y
*** EXAMPLE 3.4: THREE-MEMBER PLANE TRUSS*** STRUCTURAL PARAMETERS NN NE NRN E 3 3 2 3.0000E+04 NODAL COORDINATES X NODE l 0.000 150.000 2 150.000 3
RHO 7,3500E-07 y
-
0 . 000 200.000 0.000
ELElMENT INFORMATION AX J K ELEM. 10 . 0000 1 2 1 6.0000 1 3 2 8 ..oo 00 3 2 3
ex CY 0.6000 0.8000 1 .0000 0.0000 0. 0000 -1. 0000
EL 250.0000 150 .0000 200.0000
(al
NODAL RESTRAINTS NODE NRl NR2 1 0 1 3 1 1 NUMBER OF DEGREES OF FREEDOM: NUMBER OF NODAL RESTRAINTS: MODE 1 ANGU~AR FRIDQUENCY NODIC I 2 I
OJ l
2.J137lt 01 I • OOOOICi 00
0,000!llCtOO
NDF • NNR •
3 3
4.1995m~o2 D,12
o. 0000,c, oo 2, 472 21C 0 1 o. ooomc, no
1"1)l1111• ,\. Ill
(11) 1'111111• t111NN; (h) 111111!1- 1 (Nyn1111cttlc); (c) mode 2 (antisymmetric).
-
x
124
11111111 I l11mn11t N1111d Vlh1ntl111111I A1111IYNh1
126
I 11ol!ln111a
t hnp I
l lrrirniltl•,iko , S. t>., Yo11111-1, I>. 11. , 111111 Wl' IIVN, W, , J1., , Vl/m 1tion Problems in 1,:11Hl1w1•rl11H, 4 th ed ., Wil ey, New York , 1974 . . r W Jr Mlllri. A Algebra f or Engmeers, 2nd ed.' . I W It)
-- ~-"" l
ll " \. ", I
---~--
(c)
Figure 3.18 (cont.) the truss is aluminum , we give parameters the following numerical values:
E = 6.9 X 107 kPa
p = 2.62 Mg/ m
l=5m
A = 6
X
PROBLEMS
3
F P3 4-1 shows an axial element with a linearly distribut~d load (fo~ce per u~~~~; ngth) given by the formula bx = b_i : (h2 - bi)x/ L. Fmd the eqmvalent f' {. 1; [ ,-{ {_ nodal loads Pb(t) = {pbi' Pb2} due to this mfluence.
10- 3 m2
f" [
where SI units are implied (again, see Appendix A). To process this truss with Program VIBPT, we must analyze half the structufl• twice. In the first analysis , restraints on the plane of symmetry allow only symmctrk modes of vibration; and the second analysis uses restraint data for only antisymmetrir modes. Figures 3.18(b) and (c) illustrate the mode shapes corresponding to the first and second angular frequencies w, = 79.55 s - 1 and w2 = 168.9 s - 1 . We see that the first mode is symmetric with respect to the plane of symmetry, while the second mode is antisymmetric.
•. f,~
~ b ,
xJ
' ; [ 1/ '<"'L' ri.•:.,
- I--------1 rt [ rh,. Pb,
p,,,
l\,f)"
b,
.,
L
, .... 2.
I. Weaver, W. , Jr. , and Johnston, P. R. , Finite Elements for Structural Analysi.1·, Prentice-Hall, Englewood Cliffs, N.J., 1984.
1,~'(
o
=-1 ).
Figure P3.4-1
REFERENCES
, L-X
(kb,)
-
L [(L-X)
"
l \,,+ li,._ -~')-;:-
~ ( 1,i-1-( b, - L,) I:_
A parabolical!y distributed load (force per unit length) has . th~ f~rm~~ bx = bi(x / L )2, as illustrated in Fig. P3.4-2. Determme the equiva en no loads Pb(t) = {pb,, Pb2} resulting from this body force.
2. Zienkiewicz , 0. C., The Finite Element Method, 4th ed. , McGraw-Hill, Maidenhead, Berkshire , England, 1987. 3. Cook, R. D., Concepts and Applications of Finite Element Analy sis, 2nd ed., Wiley, New York, 1981.
1
5. Weaver, W., Jr., and Gere, J.M., Matrix Analysis of Framed Structures, 2nd ed ., Van Nostrand Reinhold , New York, 1980. 6. Oden, J. T., Mechanics of Elastic Structures, McGraw-Hill , New York, 1967. 7. Archer, J . S., "Consistent Matrix Formulations for Structural Analysis Using FiniteElement Techniques," AJAA J., Vol. 3, No. 10, 1965 , pp. 1910- 19 18. 8. Clough , R . W. , "Analysis of Structural Vibrations and Dynamic Response," Rel'. Adv. Mat. M ethodsStruct . Anal. Dt•s., ed . R. 11. Gullugher, Y. Ynrnadu, a nd J. ·1: Oden, University of Alabama Press, llunt svillc, Alu ., 197 1, pp. 25 45.
~b~--x,u
p:--i.-,,-~---.1
4. Timoshenko, S. P., and Goodier, J. N., Theory of Elasticity, 3rd ed., McGraw-Hi ll , New York , 1970.
.,.....,.
p.,
Figure P3.4-2 " ss ume that an axial clement has three nodes , as shown in Fig. Pd?.41-3(a). I "te 'rms · of the coordinate x measure d from node 2 ' the quadratic . - (2x1sp - acemer L)x/ L; .I . · p·irt• (b) (d) of the figure are. f1 ' " 2 0 cn·vc thc 3 x 3 stiffness matri Jishape(I} uncti4.,ons i )/Im }, and.{, (2.i I /,) ,1· / / '.. K l'or thi s l'I Cllll'llt ii' the nxinl d /\id ity /\'; \ is constant along the length.
128
1111110 I lorrntrUN 1111d Vlltrnll111111I A1111fy11IH
--
I . ~q,
2
3
q,
L
/:
2
./ L
,K
-
1•111hl11111 11
127
----x, u
Q3
L
2
~ · ~ - - - - --x,IJ,
.I
/1 111
P.,
, - m, - - - - -
'
L------+l
(a)
Figure P3.4-6
Suppose that a concentrated moment Mx is applied to the torsional element at the distanccx from node 1, as shown in Fig. P3.4-7. Determine the equivalent nodal loads Pb (t) = {pb1, Pb2} caused by this moment.
~----____::::;=" (b)
P11,
-M.
2
-----+--------- -
------
X,
IJ,
Figure P3.4-7 (c)
I ,,,It
The flexural element shown in Fig. P3.4-8 is subjected to a triangular load by = b2x/L (force per unit length). Derive the equivalent nodal loads Pb (t) = {pb1, Pb2, Pb3, PM} indicated at points 1 and 2.
(d)
Figure P3.4-3
3.4-4. For the axial element with three nodes [Fig. P3.4-3(a)], derive the 3 x 3 consistent-mass matrix M, assuming that p and A are constant along the length .
3.4-5. Let a uniformly distributed load b., (force per unit length) be applied to the axial
Figure P3.4-8
element with three nodes [Fig. P3.4-3(a)]. Find the equivalent nodal loads Pb(t) = {Pb1, P1,2, Pb3} due to this body force .
3.4-6. For the torsional clement shown in Fig. P3.4-6, obtain the equivalent nodal loads p1, (t) {P111, p1,2} caused by a puruholicully dislribulecl moment (per uni I length) given ns 111, 111,d I (l'/ 1.)2 J,
1.4-1), Figure P3.4-9 depicts a concentrated force Py and a concentrated moment M, nppl icd to II flexural elcmcnl at the distance x rrom node 1. Obtain the equivalent nodtil londs 1>1o(I)
{P1o 1, J>1oi, JJ1,1 , p1,.1) for cuch of these actions.
128
1lnltu I lon10nta1 1111d Vlh1nll1111nl A1111ly1ltt
Chop. :l
y, V
t
y
t
P",
;·/1:
t
pr
I, PoJ
r
2
1
z
129
l 111>l>lt1111"
x
M,/,1 L
P-~,
X
II II
CD
2
CD y
l,
X
'/
I 3.4-10. In Fig. P3 .4-10 a linearly distributed load by = b1 + (b 2 - b1)x/L (force per unit length) acts on a flexural element. Find the equivalent nodal loads Ph (t) = {Phi, Pb2, P&3, Pb4} due to this influence .
0.8LI
0.6L
2
Figure P3.4-9
/
I,
0.8L
X
Figure P3.5-2
Figure P3.5-1
Repeat Prob. 3.5-1 for the plane truss shown in Fig. P3 .5-2, assuming that cross-sectional areas of members l and 2 are 0.8A and A. For the plane truss shown in Fig. P3.5-3, repeat Prob. 3 .5-1. In this case assume that the cross-sectional areas of members 1 through 4 are equal to A and that the area for member 5 is V2 A.
y,v
y
r L
l
Figure P3.4-10
3.4-11 . Rederive the 2 x 2 stiffness matrix K for a torsional element, using moment M, and twist i/1 as generalized stress and strain. Assume that the torsional rigidity GJ is constant along the length. 3.4-12. Derive again the 4 x 4 stiffness matrix K for a flexural element, with moment M, and curvature rp as generalized stress and strain . Let the flexural rigidity El be constant along the length. 3.S-1.
The plane truss shown in Fig. P3.5- I has cross-sectional areas of 0.6A and A for members I and 2. For thi s structu re !incl the stiffness mutri x S.,· and the consistent mass matrix M, in rourranged und pnrtit ioncd f'onns.
I,
CD
2
0
0 3
0 L
X
·I
Figure P3.5-3
l.~-4.
Ropcut Prob. 3.5- 1 for the plane truss shown in Fig. P3.5-4. Assume that the cmss sectional arcus of mc1nhers I 11nd 2 arc 0.8A, those of members 3 and 4 111'l' 0 .6/\. und th11t or 111c111lwr ~ Is l'qunl to/\ .
130
1lnltn ~ lnt11n11t• nnd Vlhr11tl1111nl A11n ly1l11
l'rohlnm•
Glinp. 3
y
1
I
I
l•'1gtlll' 11 I..~ I NhowNn two d ct11lllll ht•1111 1 Im whk'h lhc pnrnmctcrs E, I , A, and 11111c l'OIINtn11l 11ln11g the length . Asscnthle the stiffness matrix S., and the consisll•nt-11111ss mutrix M, (for trunslutionul inertias) in rearranged and partitioned forms .
CD
o----_..;-----o 2
0.6L
L
131
y
0
2
,.
,/
Figure P3.5-4
4.
~
,
0
~l•- ---L~----+------
L
3
~
X
.1
Figure P3.5-7
3.5-5.
For the plane truss shown in Fig. P3 .5-5, repeat Prob. 3.5-1. Assume that the cross-sectional areas of members 1 and 2 are 0.8A, those for members 3 and 4 are 0.6A, and those for members 5 and 6 are equal to A.
Repeat Prob. 3.5-7 for the two-element beam shown in Fig . P3.5-8 . y
y y
r
/,
I
,I
0
2
G
0.6L
L
4
- -x
j.
O.BL
Figure P3.5-5
3.5-6.
r
G) 1~
CD z
/ ~L
L
~
0
~3
X
L~
Figure P3.5-8
©
L
2
For the two-element beam shown in Fig. P3.5-9, repeat Prob. 3.5-7. y
- -x
3~L
1
Figure P3.S-6
Repeat Prob. 3.5-1 for the plane truss shown in Fig. P3.5-6. In this case let the cross-sectional areas for members I through 4 be equal to A, while those for members 5 and 6 are equal to V2 A.
,
/
•
G)
2
4 l
I FIMlll'C PJ.5-9
0
~3
l~
X
132
Finl to rlum1111t" 111111 Vll11 nllorml A1111ly11l11
Rl•twnt Prnh. \.6· 2 for the pl1tlll' 11 II NN show11 In Fig. P3 .6-4 , but let the crossNl'Clionul ureas of mcmbcrs I and 2 hc l'qunl to 0.8A and A.
3.5-10. Repeat Prob. 3 .5-7 for the two-element beum show11 In Fig. P3.5- 10.
,~
y
1~
G)
,/~,
0
2
~
3
•
X
l~
Figure P3.5-10
J1'l1111rc P3.6-4 3.6-1.
3.6-2.
Assuming that the mass matrix M is positive-definite and not diagonal, factor it into M = VTV. Then transform Eq. (3.6-4) to the standard, symmetric form of Eq. (3.6-8). Also, show the back-transformation of eigenvectors to the original coordinates. Figure P3.6-2 shows a plane truss with two degrees of freedom at joint I. Assume that the cross-sectional areas of members l and 2 are equal to A and A V2. By the consistent-mass method, find the angular frequencies and mode shapes for this structure.
1.,,-s.
The beam shown in Fig. P3 .6-5 consists of two prismatic flexural elements with two degrees of freedom at node 2. Member 1 has moment of inertia a~d cross-sectional area equal to I and A, but member 2 has 2/ and 2A for its properties. Determine the angular frequencies and mode shapes for this beam using translational consistent-mass terms. y
[,
.-- L--•I Figure P3.6-2
3.6-3.
l
j
1 i-,,, - - --
0.8L - ---....i Figure P3.6-3
Repeat Prob . 3.6-2 for Lhe plane truss shown in Fig. P3 .6-3, buL let the cross sectional areas of mcmhors I und 2 lw l'q111al to /\ 1111d O.M .
0.8L- - --
"l L~ I
1~
r L~_,.
133
l'whlntn•
Chnp. 3
z
/
CD /, A
~L
2
•
.1.
0 2/, 2A
~3
X
L~
Figure P3.6-5
.U-6.
Repeat Prob. 3.6-5 for the continuous beam shown in Fig. P3.6-6. In this.case the beam has constant values of I and A along its length and has two rotational degrees of freedom (at points 2 and 3). y
11111111 1111111111111111111 Vll11 11ll1111nl A1111ly1l1
134
Ch11p J
Repeat Prob. 3.6 5 for the two clement co11tin11011Nlw11111 shown in Fig. PU, l Cross-sectional properties I and A arc constunt , 1111d the free displacc111cnh 111 joints l and 2 are both rotational .
3.6-7.
"''' \, I,
1:
l'rnhl"'"'
hu thl• Ny1111m•tric rnnti11uo11s IK·11111 shown in Fig. P3.7-3, find the angul. tll'(llll' lldcs and mode shapes for (11) symmetric and (b) antisymmetric di
hu lions. Use onl y half the structure, assuming that each of the four elements h, tht· snn1c v11 lucs of I und A.
y
y
, _____;:CD:::;.____2_ _ _ _0.;:;2:;.__ _ _~ _3 - - x z
/
~
#i
I
~l,1--~~~[~~~----,-i-~~~-L---------+-j Figure P3.6-7
3.7-1.
The fixed-end beam shown in Fig. P3. 7- l consists of two prismatic flexurnl elements having the same values of I and A. Using only half the beam, find thl' angular frequencies for the (a) symmetric and (b) antisymmetric modes. y
1 7 4. I
i-Sym. 2•
CD
1~
z
Figure P3. 7 -3
0
I
/1·
e
e
·I·
3~
.,
The symmetric plane truss shown in Fig. P3.7-4 has cross-sectional areas fo members I and 2 equal to V2 A , whereas those for members 3, 4, and 5 arc equal to A. Using only half the structure, calculate the angular frequencies anc the (a) symmetric and (b) antisymmetric mode shapes.
X
r-Sym. I
L= 2e
-1
Figure P3.7-l
3.7-2.
Figure P3.7-2 shows a prismatic continuous beam composeq of four flexural elements. Determine the angular frequencies and mode shapes for (a) symmetric and (b) antisymmetric deformations, using only half the beam.
y
,-:2@
L
(Ds_l_
/J ' > o - - - - - . = - - - ~ - - - - o ( ' v ' .
y
r-Sym.
,~
CD
I z
/
l-e
2
~
,I,
0
31
~
X
/
I
®
e+e Figure P3. 7-2
~ L - 1 4 L~
4
~ ,J,
© ~s
Figure P3.7-4
\. - x e--j
L- u
1.7-5. For the symmetric plane truss shown in Fig. P3.7-5, determine the angular frequencies and the (n) symmetric and (b) antisymmetric mode shapes. Use only half the structure, and assume thnt the cross-sectional areas of members I and 2 111c O.<,A , thot for mcmhcr 1 is 0 .8A, nnd those for members 4 and 5 are equal lo/\ ,
136
Fl11 lto l: lomo11t11 1111cl Vll>111l lo1 111I A11uly11lo r
Chop. 3
13,
l 1111 l>lt11 1111
y(Sym .)
"ro6L-j-o6L--::i CD
, 0
2
'l
O.SL
_j~x Figure P3.7-5
3.7-6.
Figure P3 .7-6 shows a plane truss having cross-sectional areas equal to A for members 1 and 2, 0.6A for member 3, and 0.8A for members 4 and 5. Find the angular frequencies and the (a) symmetric and (b) antisymmetric mode shapes, using only half the truss.
(a)
rSym. 1
I
y
f
0 l.---=----0---=--------G
3
-
-• I
0.6L
____.__l- x
- o.sL- -....,, j,_.--o.sL------j Figure P3.7-6
(cl
(b)
Figure P3.7-7
3.7-7. The symmetric plane truss shown in Fig. P3.7-7(a) has cross-sectional areas of all members equal to A. Using only half the structure, calcul ate the angular frequencies and the (a) symmetric and (b) antisymmetric mode shapes. For thiH purpose, part (b) of the figure shows symmetry restraints, und part (c) showN antisymmetry restraints. (Note: For the antisymmetric case, nodes 2 a11d I translate equall y in the x direction; so llll' ll1iw 1· '\ mny h.: trenlL•d as 111 igid body.)
4 Normal-Mode Method of Dynamic Analysis
4.1 INTRODUCTION
In Chapter 3 we discussed the formulation of action equations of motion for a MDOF structure using finite elements. Then we showed how a vibrational analysis can be performed by solving the algebraic eigenvalue problem associated with the homogeneous equations. Extracting the angular frequencies and mode shapes in this manner sets the stage for the normal-mode method of dynamic analysis [l], described in the present chapter. A structu.re subjected to impulsive (time-varying) loads responds with a combination of rigid-body and flexible-body motions. If a structure is restrained (immobile), the response will involve only flexible-body motions. However, if the structure is unrestrained or partially restrained, certain rigid-body motions can occur as well. Within the linear theory of the noanal-mo.de..Jru:.th.od,_such motions may consist of small or large translations but only .smaluQtations . Thus, problems involving large rigid-body rotations are beyond the scope of this theory. Regarding the flexible-body part of the response, a linearly elastic solid or structure has an infinite number of degrees of freedom and an infinite number of natural modes of vibration. If the structure were analyzed as an elastic continuum, its flexible-body response to dynamic loads would consist of the sum of an infinite number of vibrational motions . However, if the structure is dis cretized by the finite-element method, the resulting analytical model will have only a finite number of nodal degrees of freedom and a finite number of nnturul
138
n, , 11.:l
139
l'r lrwlpnl n11d Nor mnl C:111111tl11ntn11
111rnl(•S
of vihmtion . 'l'hcrcfme, such a model has only a finite number of vil111111011ul motions contributing to ils dynamic response. Systems that are subjected to arbitrary dynamic loads be~ome extremely dlt lll'ult to analyze rigorously in their original p~ysic~l coordmates: We can nvold these difficulties by using natural modes of vibration as gener~hzed coordl11111cs . When this path is followed, the equations of undamped m~~on bec?me llllt'tlllplcd . In these coordinates each equation may be so~v~d as 1f it pertamed hr II system with only one degree of freedo~. Superposition of. ~ese SDO~ 1, "ults is accomplished through a transformation back t~ the ongmal .coordiuules. By this means we can evaluate time-vW?'ing nodal displacements, mtemal lilH•sscs, and support reactions for the analyttcal model. . . In this chapter we develop the normal-mode method and apply it to simple •ltuctures such as beams and plane trusses. In later chapters we will use it ~or 111111 c co~plicated framed structures and other continua. discreti.zed b.y fimte l'lr 111 cnts. Undamped systems are treated first, and special considerations req111rcd for damped systems are discussed in the latter parts of.the chapter. An important advantage of the normal-mode me~od is t~at onl)'....lh.e M l11,11ifa:ant modal responses need be included in a dynamic analysis.. The oth~r 11 u,dnl responses may often be omitted without much loss of accuracy. This trt•hnique, known as modal truncation, can make the normal-i:node.method more dllcicnt than the numerical integration methods to be descnbed m Chapter 5.
4,2 PRINCIPAL AND NORMAL COORDINATES 111 order to study relationships among the natural modes o~ vibratio~, let us 11 msider modes i and j of the eigenvalue problem for action equat10ns, as lollows: (la) ; = WT M ;
s
S <1>1
=
wJ M
(lb)
<1>1
l'lwsc expressions are modified versions of the algebraic eigenvalue problem µlwn previously in Eq. (3.6-4). Premultiplication ~f Eq. (la) by J and post11111ltiplication of the transpose of Eq; (lb) by ; yields
= wfJM ;
(2a)
~S~=~~M~
~
JS ;
I hl· lcfl hand sides of Eqs. (2) are equal, so that subtraction of the second 1·q1111lion from the first produces the relationship
(wr - wJ) (l>T M , 1111
0
the other hand , if Wl' divide holh silks of' Eq . (2a) by
(3)
wr and both sides of
140
Normal-Mode Method of Dynamic Analysis
Eq. (2b) by
Chap. 4
TS 1 = 0
(4)
J
*
To satisfy Eqs. (3) and (4) when i j and the eigenvalues are distinct (wr wJ), the following relationships must hold:
*
141
Principal and Normal Coordinates
Premultiplication of this equation by T and insertion of I = <1>- 1 before D and D produces
wJ, the right-hand sides become equal. Then subtraction gives - _l) ( _!_ Wf WJ
Sec. 4.2
JM ; = ;M i = 0
(5)
JS ; = ;S i = O
(6)
and
These expressions represent orthogonality relationships between the modal vectors ; and i. From Eq. (5) we see that the eigenvectors are orthogonal with respect to the mass matrix M. Equation (6) also shows that they are orthogonal with respect to the stiffness matrix S, as demonstrated in Sec. 3.6. For the case when i = j, Eqs. (3) and (4) yield
which can be restated as (13) By virtue of Eqs. (10) and (11) , the generalized mass and stiffness matrices in Eq. (13) are both diagonal. Also , the displacement and acceleration vectors in the latter equation are defined to be
DP
= <1>- 1 D
(8)
in which Mp; and Sp; are constants that depend on how the eigenvector ; is normalized. For operational efficiency, we place all of the eigenvectors columnwise into an n x n modal matrix of the form ,.]
(9)
where n is the number of degrees of freedom. Then we can state Eqs. (5) and (7) collectively as
..
MDISI>
in which the modal matrix is given by Eq. (9). The symbol w 2 in Eq. (16) represents a diagonal matrix with values of wf in diagonal positions, as follows:
wf 0 0
w2 =
(11)
0
( 12)
(15)
Here we see that the generalized displacements in vector DP operate as multipliers of the modal columns in to produce values of the actual displacements in vector D. Thus, the shape functions for the principal coordinates of a MDOF system are its natural modes of vibration. We now restate the eigenvalue problem in Eq. (la) more comprehensively as (16) S = M w 2
in which MP is a diagonal array that will be referred to as a ,vrinciDal mass matrix. Similarly , Eqs. (6) and (8) are combined into • •
where SP is another diagonal array that will be called a principal stiffness matrix. Equations (10) and (11) represent diagonalization of matrices Mand S. If either of them is already diagonal, the operations merely scale the values on the diagonal. To take advantage of the diagonalization process, let us reconsider the action equations of motion for free vibrations of an undamped MDOF system, as follows:
..
D= DP
(10)
T s = Sp
(14)
Ihe generalized displacements De given by the first of Eqs. (14) are called Jl.rincipal coordinates, for which the equations of motion [Eq. (13)] have neither inertial nor elasticity coupling.. From Eqs. (14) we find that the displacements and accelerations in the original coordinates are related to those in principal coordinates as follows:
(7) and
Op = <1>- 1 I)
0
0
0
0
(J)~
0
0
0
W5
0
0
0
(J)~
Ll1l
This array, sometimes
caHed the spectral matrix, will be referred to as the eigenvalue matrix, or matrix of characteristic values. It postmultiplies the matrix in Eq. ( 16), so that a typical modal column ; is scaled by the correspondPremultiplying Eq. (16) by T and using the relationships ing eigenvalue ( 10) and (11) , we obtain (18a)
wt.
llcncc, ( 18b)
Normal-Mode Method of Dynamic Analysis
142
Chap. 4
Thus, in principal coordinates the ith principal stiffness is equal to the ith principal mass multiplied by the ith eigenvalue. Because the modal vectors may be scaled arbitrarily, the principal coordinates are not unique. In fact, there is an infinite number of sets of such generalized displacements, but the most common choice is that for which the mass matrix is transformed to the identity matrix. We state this condition by specifying that Mp; in Eq. (7) must be equal to unity, as follows: (19) where (20) Under this condition, the scaled eigenvector N; is said to be normalized with respect to the mass matrix. The constant C; in Eq. (20) is computed as
C;
=
i j;(i Mjk)1
±VTM ; = ±
j= I
(21a)
k= I
Sec. 4.2
143
Principal and Normal Coordinates
normal coordinates. From Eqs. (14) , these vectors are related to their counterparts in physical coordinates by
DN = N1D
DN= N D 1
(25)
and the reverse transformations are
(26) The inverse of the normalized modal matrix required in ~~s. (25) ~ay be easily found . We need only postmultiply Eq. (22) by N to obtam the relationship (27) Thus, the desired inverse may always be calculated by this simple ~atrix multiplication. If we wish to include only a selected number of modes mm our analysis, the modal matrix will be of size n x m. Then Eqs . (25) must be restated as (28) This technique-. for using only a limited number of. modes s called modal
If the mass matrix is diagonal, this expression simplifies to
truncation.
II
C;
=
±
L Mj J;
(21b)
Example 4.1
j= I
When all of the vectors in the modal matrix are normalized in this manner, we use the subscript N and revise Eq. ( 10) to become
<1>iM N = MN = I
(22)
To demonstrate the use of normal coordinates for action equations, let us reconsider the three-member plane truss in Fig. 4. l(a), which we studied prev!ously in Cha~ter_3. Fro~ Hq. (3.5-3lb) the 3 x 3 stiffness matrix for the free nodal displacements md1cated m
Pig. 4. l(a) is
Thus, the principal mass matrix is now the identity matrix. Furthermore, the principal stiffness matrix, from Eqs. (11) and (18a), is seen to be (23a)
Or, for the ith mode,
- 0.48] 0.48 1.64
(a)
In addition the consistent mass matrix from Eq. (3.5-34b) is
(23b)
Therefore, when the eigenvectors are normalized with respect to M, the stiffnesses in principal coordinates are equal to the eigenvalues. This particular set of principal coordinates is known as normal coordinates. Rewriting Eq. (13) in normal coordinates, we have
'
M =
2 2 ·;
pAL[ 6
o
~J
3.~8 O
(b)
3.28
hu this structure, three eigenvectors were calculated in Example 3.4 (see Table 3.4). I hl'NC vectors become the columns of the modal matrix, as follows:
(24a) or
-0.36 0 .36 0.48
1.36 S= -0.36 L [ -0.48
EA
111 1111111111 11,l· thii. 111 111y wtth
0.2314 1.0000
0.8673 0 . 171 5
l.OOOOJ - 0 .6050
• O.2472
I.()()()()
0 .6 107
r
ll'S \ll' l' l
to thr
11111HH
111nt11x,
Wl'
(c)
compute from Eq. (2 1a) the
144
Ll111p. 11
Nn111111I Mnd11 M1111t111l 11f 11y11111 11lc l\1111lynlt1
l'1l11nlpnl 111111 N,11111111
< 11111dlt111l n•
146
, 1111~llllllN
C2 F,,
-,
,.
I ll vldl11g the columns o f'
)(
/
0.9384
1. 2114
- 0.1856
\/pAL - 0.2995
1.0820
SN = ;!::S N
/ /
=
F,,
-J I--
(a)
lb)
I
F,,
/r ----+-1I
-'.j+I
~
i1 r ,.
F 32
r,
I I I
I
I
/ / I
I
I I
/ /
I
'/
-0.7542
(e)
0.2701 0 O
0 2.088 O
0 ] 0 5.308
= w2
(f)
wi, and w~on the diagonal. Their square roots are the the angular frequencies given in Table 3.4. Of course, the eigenvalues in Eq. 11) were already available; so the advantages of transforming stiffnesses to normal , 1u11 di11atcs are not obvious. These advantages will become apparent for the response , ,iii 11lutions in later sections. 111111• 11s
/
, - 0.6 L - - -
1.2350] -0.7472
" l1 k h contains the values of wr,
2.__io----~~~ -
:2 [
p
I
/
-
(d)
111hNlilution of Eqs. (a) and (e) into Eq. (23a) yields the result
/
/
0.2803
1 [
/
C3 = 0.8097\/pAL
by these values produces
N = -
I
/
(J>
0.9242\/pAL
If a structure is statically determinate and immobile [2J, flexibility 11w/ficients are not difficult to calculate. In such a case, the action equations of 1111111011 expressed by Eq. (12) may be replaced by displacement equations oJ. 11tr>tiulL, as follows:
FMD + D=O
(29)
111 whi ch
F =
s-1
(30)
I lit•.flexibility matrix F contains values of free nodal displacements due to unit v,tlucs of the corresponding actions. We tr.ansforrn Eq. (29) to principal coordi1111lt•s by substituting Eqs. (15) for D and D. Then premultiplication by <1>- 1 and hlMt' I tion of the identity matrix I = «1>- Tcl>T before M produces
«1>- I F «1>- Tcl>T M cl> Op + Op = 0 whk h can be rewritten as (31)
I
II
(c)
l'.'lgurc 4.1
l10n 111
I.
(a) Plane truss; (b) condition A,
Id)
=
I; (c)
condit ion A2 = J; (d) condi-
1lw symbol Fp in Eq. (31) represents & orincioal flexibility maJ.Jix., correspondl11p to S,,, and is defined as
Fp
=
'F
T
= Sr'
(32)
111 ,·m11·sc, this definition applies only when S (nnd hence S,,) is positive-definite. l<'u,thl.!nllol'c, th<.· cxpflndcd frn111 ol llw t'i)'l:nvuluc problem in Eq. (16) is
Normal-Mode Method of Dynamic Analysis
146
Chap. 4
F M
= A
(33)
The eigenvalue matrix A in this expression consists of a diagonal array containing values of .Ai = 1/ in diagonal positions, as follows:
wt
A1 0 0
.A2
0 0
0
.A3
0 0 0
0
0
0
An
0
= (1) - 2
4.3 NORMAL-MODE RESPONSE TO INITIAL CONDITIONS
(35)
For a MDOF system, suppose that w~ know the initial conditions (at time t uf displacements Do and velocities Do, as follows:
Do=
(36) Thus, lhe flexibility matrix in normal coordinates becomes the eigenvalue matrix ~. which is also equal to the inverse of w 2• From this we conclude that Eq. (24b)
Hives the equations of motion in normal coordinates, regardless of the method ol formulation in the original coordinates.
wmJe U
/\s 1111 cx11mplc of the use of normal coordinates for displacement equations, we again rn11sidcr the three-member truss in Fig. 4.1 (a). Because this structure is statically deter111l1111l c nnd immobile, we can find terms in the flexibility matrix by the unit-load method Il l, Figures 4 . l(b)- (d) show the flexibilities obtained from applying unit values of A,, /\ J, and A.1 (all forces). The resulting flexibility matrix for this structure is
F =
:A
l. 000 1.000 [ 0
1. 000 5.556 - 1.333
J
0 -1.333 1.000
(g)
= <1>i M = VpAf,
0.3290 0.3945 [
0.4353
0.7089 0.05494 - 0.2026
- 0.1637] 0.5915 - 0.41 23
(h)
•I•,., I J1' •l• N1
,,, /['·70) () I
()
0
ti •I /IN
0
tl
ll
II I HH I
Do=
Doi Do2 D03
(1)
111 accordance with Eqs. (4.2-25), these initial values may be transformed to 11mmal coordinates by premultiplying them with the inverse of the normalized 1110<.lal matrix. Thus, we have
DNo
= ,v 1 Do
DNo
= ,v' Do
(2)
I he second relationship in Eqs. (2) is obtained by differentiation of the first with 11'Hpcct to time. The forms of the displacement and velocity vectors in normal 111ordinates are the same as those in physical coordinates given by Eqs. (1). From Eq. (4.2-24b) we see that a typical equation of motion for undamped l11•t· vibrations in normal coordinates is
+
wt DNi = 0
(i = 1, 2, ... , n)
(3)
I 11l'h equation of this type is uncoupled from all of the others , and we will tr~at 1111• l'Xpression as if it pertained to a SDOF system.[se~ Eq. (2.2-4)]. Knowmg 1111' rnnditions of initial displacementDNo, and velocity DNo;, we find the response 111 11 ll' i th normal coordinate as
(4)
1111'4 t•x p1'l' Ssio11 is dl'awn from Eq . (2.2 9) for un undamped one-degree system.
w,
(i)
= 0)
Don
(i = l, 2, ... , n)
With this matri x und the nex ibility matri x from Eq . (g) , we find FN using Eq . (36), as l11llow11:
Ii',.,
Doi Do2 Do3 Don
DNi
S11hslituti11g <1,i and M from Example 4.1 into Eq. (27), we obtain the inverse of
1
and .\3 = 1/wt as
At this point we should remember that working with displacement equations and flexibilities is feasible only for structures that are statically determinate nod restrained against mobilities. Upon reflection, we realize that this is a very limited class of structures that would be encountered rather infrequently. In ~cneral, the approach using action equations and stiffnesses is much more Nuitable for dynamic analysis.
When the modal matrix is normalized with respect to the mass matrix, the principal flexibility matrix from Eqs. (32) and (35) takes the form
11'
= 1/ wf, A2 = l I wt
(34)
Transforming Eq. (33) to principal coordinates as before, we find that
FpMp = A
147
Normal-Mode Response to Initial Conditions
This diagonal array contains the values of A, confirmed by Eq. (f) in Example 4.1.
replaced by
A. =
Sec. 4.3
11u111ppl y Hq. (II) rqwlitiwl y to rnlrnl11tl' till' terms in the. vector of normal111111lr tl1 hplm·t·111l'1Ils l)N \nN,l 'l'hr N1· 1t•1nilt 'I 111t· thl'II tn111Hlornwd hack to the 111lp11111I 111111d11111tr 'I, ll 'l i11p 1hr 11p1· 111111111 p1vrn hy th1• 1111,I ol 11.qs, (11. 2 Ui) .
Normal-Mode Method of Dynamic Analysis
148
Sec. 4.3
Thus,
DNo
D
=
NDN
...,, _ , = 'YN Do = •vr-;-;pAL {4.486,
=0
= DNo; + DNo;t
(7)
D =
,-1
t
o,~
[
- 0.1407
1.257 =
PoL EA
COS W1 t -
5.434 cos w,t [ -1.343
COS Wit -
3
1111d ' - - - 0 . 6L - l
M
The.initi td displacements of the joints caused by the force may be calculated as tlw produet of lli nnd the second column of the flex ibility matri x F in Eq. (4.2-g). Thus,
(u l
AI No, Ill 111111• I () tilt• 111ltl11l w lodtk•N llll' i'>11 0 IJN it11-1 tlu• tli NI ol l\qN: (l). Wl' 1111 111111111 t1111pli111·1111' llt ~ ht 1111111111l 111111dl1111h N, IIN lnllnw" :
111111Ml111111
= pAL 420
Plane truss with initial load.
1:.'.~'
0.09600
w2 t
COS W 2t
0.1738
COS W3t
1
+ 0.1051 cos w3 t + 0.1061 COS W 3t
(d)
Assume that the flexural element in Fig. 4.3(a) is initially at rest when node 2 is struck In such a manner that it suddenly acquires a velocity D03 . Determine the smalldisplacement response of the element to this impact. Nodes of the flexural element have no restraints against either translations in the I' direction or rotations in the z sense. Therefore, the stiffness and mass matrices are
L
l.:lJ.1 )
COS W2t -
In this example the response of the first mode of vibration is about one order of magnitude )lrcater than each of the other two.
3
- 6
-3L
L2 3L
-6
3L 2L2 -3L
6
- 3L
3L
L2
- 3L
2L2
K-2Et:
( 1.000, 'i. 'i:'i6,
0.08325
+ 0.01647 cos
0.8L
Ou
COS W3t
11;xample 4.4
Po
y
Figure 4.2
(c)
~:
NDN
Example 4..3 Suppose that the plane truss in Examples 4.1 and 4.2 is subjected to an initial force R in the x direction at joint 2 , as illustrated in Fig. 4.2. Let us find the free-vibrationa~ response of this structure due to suddenly releasing the load. 3
J
4.486 cos w, t - 0.08872 cos W2t
Then with the normalized modal matrix N from Eq. (4.2-e), we transform these results back to the original coordinates, using Eq. (5) as follows:
This f~rmula is used in_ place ofEq. (4) to evaluate the response of a rigid-body mode m normal coordmates, assuming that rotations are small.
D
'VpAf,
DN =
(6)
Integration of this equation twice with respect to time yields
DN;
(b)
EA
For this purpose the inverse of the normalized matrix ,v is available from Eq. (4.2-h). According to Eq. (4), we find the terms in the vector of normal mode displacements to be
wr
DN;
} PoL
- 0.08872, - 0.1407 1
This _sequence o~ operatio?s is the same regardless of whether the original equat10ns of motion are wntten as action equations or displacement equations. Howeve~, _for the action-equation approach . there exists the possibility of one or more ngid-body modes. For such a principal mode the eigenvalue is zem..._and Eq. (3) becomes
..
149
Normal-Mode Response to Initial Conditions
l
22L
54
156 22L
4L2
13L
54
13L
156
- l3L
- 3L2
-22L
1
-13l2 - 3L
(e)
1 (f)
-22L 4L2
llh itiven by Bqs. (3.4-24) and (3.4-26). In this case the stiffness matrix is positive"1'11iidefinite, so we should expect lo find two repeated zero roots when solving the 1 l1wnvnhu.: prnb lern . Thus, so luti on or Eq. (4.2- l a) yields the e igenvalues
,,,,'
'
'" '
I io fo.'/
11Af
8400 El I
pAJ,''
(g)
160
No rmal-M odo M othod ul l)yn11mlc Analysis
Chap.4
y
Sec. 4.3
Normal-Mode Response to Initial Conditions
151
l'11cse eigenvalues and eigenvectors were found using a computer program called VIBCB Im vibrational analysis of continuous beams (see Sec. 3.8). Each of the four column
D,/
t°'
CD
,/
Vl'Ctors in Eq. (h) is scaled so that the first translational displacement is numerically equal tu L. Figures 4.3(b) and (c) depict the rigid-body modes, and Figs. 4.3(d) and (e) show the vibrational modes. For the purpose of normalizing columns in the modal matrix with l\'Spect to M , we use Eq. (4.2-2la) to find the constants
tb03 D3L
o./.1
L
X
C1 =
LVm
C2 =
L~
C3 =
L~
C4 =
Li
(i)
111 which m = pAL. Dividing the columns of Eq. (h) by these values, we obtain the 11ormalized modal matrix as
(a)
L
«l>N= - l [~
L
LVm L
(b)
0
L\/3 -2\/3 - L\/3 -2\/3
LVS
LY?]
- 6V5
-12V7
LVS
- LY?
6V5
- l2V7
U)
Initial velocities in normal coordinates now may be calculated using the second in Eqs. (2). Hence,
1 1qlrcssion
L ~ (c)~ - L
. DNo =
L~
.
0
.
«l>N Do = «1>iM ~OJ D 03 1
[
~---~
~L
(k)
(d)
I q1111tions (4) and (7) give the vibrational and rigid-body responses in normal coordiL~
1111ll'S
,,----.____
~ " 1-L
as
ce)
(e)
Figure 4.3
N111t· that there is no response of the third normal mode, which has a symmetric vil1111timuil shape. Finally, with Eq. (5) we transform the results in Eq. (e) back to the 1111pi1111l coordinates to obtain
(a) Flexural element with initial velocity; (b) mode I; (c) mode 2;
(d) mode 3; (e) mode 4.
us wl· ll as the modal matri x
r~
()
L
- tL + (L sin W4t)/ w4] 12t - ( 12 sin W4t)/ w4 [)03 I ltl (L sin W4t )/w4 lOL [ 121 ( 12 Sill W,il)/ W,1
L
2
6
/,
/,
2
"
~21 /,
I)
(h)
1111'~1' ll'Npt!llNl'H Ill\' v11 )1d 011ly Im
111111111
111-dd hotly 111t111ltHIN.
(m)
152
Normal-Mode Method of Dynamic Ana lysis
Chap. 4
4.4 NORMAL-MODE RESPONSE TO APPLIED ACTIONS
Now we shall consider the case of an undamped MDOF system that is subjected to applied actions corresponding to the nodal displacements. For this situation, the action equations of motion are
MD+SD=A
(1)
Sec. 4.4
the terms in the vector of normal-mode displacements DN = {DN;}. Then the results are transformed back to the original coordinates using Eq. (4.3-5). For a normal mode corresponding to a rigid-body motion , the eigenvalue 11Jt is zero. Then Eq. (6) becomes (8) In this instance, the normal-mode response (with the system initially at rest) is
where the symbol A denotes a column vector of time-varying applied actions, as follows:
(2) and n is the number of degrees of freedom. Equation (1) is transformed to normal coordinates,.by premultiplying both sides with <1>i and substituting Eqs. (4.2-26) for D and D to produce
<1>iM<1>NriN
+ <1>is
NDN
= <1>iA
This equation may also be written as
DN+ w 2 DN = AN
(3)
The symbol AN on the right-hand side of Eq. (3) represents a vector of applied act ions in normal coordinates, computed by the operation AN= <1>iA
(4)
153
Normal-Mode Response to Applied Actions
DN; =
J:
r
(9)
AN; dt" dt'
This equation replaces Eq. (7) whenever a rigid-body mode is encountered. In summary, we calculate the dynamic response of an undamped MDO~ ~ystem to applied actions by first transforming those actions to norm~! coord111ates using Eq. (4). Then the response of each vibrational mode is ob~amed from tire integral in Eq. (7), and that for each rigid-body mode is de~enmned by t~e double integral in Eq. (9). Finally, the values of the actual displacements m physical coordinates are found with the back-transformation. operation of Eq. (I\ .3-5). If applied actions do not correspond to the nodal displacements, the 11ppropriate equivalent nodal loads discussed in Sec. 3.3 can always be found as 11 preliminary step (see Example 4.6 at the end of this section). At this point, let us examine the effect of a dynamic load A1(t), correspondIng to the jth nodal displacement, on the response of the kth nodal displacement " k· From Eq. (5) the ith normal-mode load due to A1 is
In expanded form, the results of this multiplication are
AN!] AN2
[ ~·N;, =
[N11A1 + N21A2 + , , , +
+
(10)
NnlAnl Nn2An
;;ln°~I -~ ;N~n~~ ~- ·· ·· • ~ ·;;,.,.·~n
(5)
If the system has only vibrational modes, the response of the ith mode is found lr om Eq. (7) to be W;
= 1, 2, .. . , n)
(6)
where the normal-mode load AN; is taken from the ith row in Eq. (5). Each of the normal-mode equations of motion given by Eq. (6) is uncoupled from all of the others and has the same form as a SDOF system with unit mass. Therefore, we can calculate the response of a typical vibrational motion using Du/w111el' s integral, as follows: t') rlt'
Jo
(11)
dt'
'l'nrnsformation of this result back to the original coordinates by Eq. (4.3-5) yidds the response of the kth nodal displacement as
The i th equation of motion in normal coordinates is (i
1
= Nji 1 A1 sin W;(t - t')
DNi
0
(7)
TIIIH t'Xjlll'HHio11 Is d111w11 lm111 Hq. () <> <,) 1111d w11s dl·riwd for 1111 u11d111111wd cu11• dl'l-' ll't' 'lYNll' lll 111111 ,~ l11i1l11lly 111 11·~1 W1• 11pply II H' pt•liliwly lo c11ll'lll1111•
(Dk)A ,
=
J
i
[NkiNji rAj sin W;(t - t') dt']
i= l
W;
Jo
(12)
Sl111ilarly, the response of thejth nodal displacement caused by a dynamic load , \ 1{(), corresponding to the kth nodal displacement, may be written
(0j)A1 -_ 11 /\, 1 q1111tt·
~ [<_l)Njl cpNki
L, I
I
W/
lt · (t _ ') d '] Ak
Sill W;
t
t
(13)
o
/\ 4, the right hand sides of' Eqs. (12) and (13) are equal, and we may llw kfl 111111d sides to ohtai 11
U>1h,
(/l1 ),,
(14)
154
Normal-Mode Method of Dynamic Analysis
Chap. 4
This relationship constitutes a reciprocal theorem.for dynamic Loads [3] that is similar to Maxwell's reciprocal theorem for static loads [2]. It states that the dynamic response of the kth nodal displacement due to any time-varying action corresponding to the ith displacement is equal to the response of the ith disP-lacement due to the same action applied at the kth displacement. The theorem bolds .for systems with rigid-body modes as well as vibrational modes, as can be seen by using Eq. (9) in place of Eq. (7) in Eq. (11).
we obtain the vector of normal-mode displacements, as follows: 1.2114(1 DN =
COS W1 t)/ Wi]
- 0.1856(1-cosw2t) / w! [
-0.7472(1 -
COS W3t)/w3
4.485(1 - cos w,t)
A= {O, Pi, O}
DN =
T
[
(b)
- 0.1408(1 -
P2pL2
1.000 - 1.257c, D
= NDN =
COS W3t) + 0.08341 c2 + 0. l 739c3]
AL
5.556 - 5.433c 1 - 0.01650c2 - 0.1052c3 EA [
-1.333
(e)
+ 1.343c, + 0.09618c2 - 0.1062c3
where c 1 = cos w 1t, c2 = cos w 2t, and c3 = cos W3t . Inspection of these results shows 1hat the joints of the truss vibrate about the displaced positions:
-0.7472
P2L D .. = {l.000, 5.556, -1.333} EA
where m = pAL. For this purpose the normalized modal matrix N is available from Eq . (4.2-e). From the result of the Duhamel integral for a step function, given as Eq. (2.6-8),
(d)
Transforming this solution back to the original coordinates using Eq. (4 .3-5), we find that
As required by Eq. (4), we transform vector A to normal coordinates with the operation:
AN= ((>NA=
(c)
JEVm
- 0.08889(1 - cos Wi t) [
(a)
1.2114] P2 -0 .1856 Vm
1
2 Substitutionofwi = 0.2701E/ pL2,w~ = 2 .088E/ pL , andw5 = 5.308E/ pL2intoEq. (c) yields the simpler form
Example 4.5 We shall consider again the plane truss of Examples 4.1, 4.2, and 4.3. Figure 4.4(a) shows a step force of magnitude P = A applied in the x direction at joint 2. Let us determine the response of the structure to this suddenly applied load, starting from rest. The vector of applied actions for this case is
155
Normal-Mode Response to Applied Actions
Sec. 4.4
(f)
due to the force Pi applied statically. Proceeding in a similar manner, we can also calculate the response of the truss to 11 step force of magnitude P = P, applied in they direction at joint 2, as indicated in Fig. •I.4(b). In this case, the results are
+ 0.3108c, - 0.4863c2 + O. l 755c3J AL - 1.333 + l.343c 1 + 0 .09618c2 - 0. l062c3 EA 0
P = P2
D
=
[
(g)
1.000 - 0.3321c, - 0.5607c2 - 0.1072c3
P.quation (g) shows that the joints of the truss vibrate about the displaced positions:
AL Ds, = {O, - L333, 1.000} EA
y
(h)
due to applying the force P3 statically. If we let the step force P2 be equal to ?3, th~n the 11•sponsc D 3 caused by P2 in Eq. (e) will be equal to the response D2 caused by P3 m Eq. (11). This equality confirms the reciprocal theorem for dynamic loads in Eq. (14). 1'
ANsumc that the unrestrained flexural element in Example 4.4 is subjected to a ramp force
/
(b)
~ 0.6 l ~
/'
P,t/ t, applied in they direction at its center, as indicated in Fig. 4.5. We shall 111ll·11late the response at the nodes due to this influence, beginning with the element at 11•NI
(n) ll ll(lll'l' " ·"
l'llllll'
IIIIN~ (11) Wllh Nh' Jl 11111 I'
mpl~4.6
,.,. (h)
wllh ~h·p lmn• ,.,
lJs111p I I /'2 111 ll w displnl'l'llll'lll sh11pl' lunctions f from Eq. (3.4-17) and npply, 1114 Fq ( I I )~) without inlrp,1111011 , w1• hntl th1• 1•quiv11knt nmlul actions to be
Normal-Mode Method of Dynamic Analysis
156
Chap.4
Sec. 4.5
y
157
Normal-Mode Response to Support Motions
Lt 3
ISL~ - ~ sin
-
3
90~ - ~ sin 3
ISL~ - ~ sin 3
- 90~ - ~ sin 3
W3t) / w5 W3~/
w~
(n)
w3t) / w5 W3t) /
w~ 3
llere we see that the first rigid-body mode contributes translations equal to P1t / 6mt1 at hoth nodes l and 2. Figure 4.5 Flexural element with ramp force. AP -- f T P --
{
4 , L, 4 , - L }Pit 8t1
(i)
which consists of a force and a moment at each end . Premultiplication of this vector with the transpose of the normalized modal matrix from Eq. (4. 3-j) yields T
AN = NAP =
{
• Ir
}
2, 0, -vs, 0
Pit •1 2t1 vm
(j )
wlwre 111 = pAL. The vector in Eq. (j) contains normal-mode loads of types 1 and 3 for th is example. Now we integrate the first term with Eq. (9) to obtain
P1t 3 6t1Vm
(k)
DNI = - -
whil'h is a symmetric rigid-body motion [see Fig. 4.3(b)]. Similarly, evaluation of the thinl term in Eq. (j) in accordance with Eq. (2.6-c) for a ramp function produces DN3 =
-
P1VS
• 2 1 2t1W3V m
(t
-
-1 W3
. sm
~
W 3t
ce)
4.5 NORMAL-MODE RESPONSE TO SUPPORT MOTIONS
We are often interested in the response of structures to support motions instead of applied actions. In this section we discuss problems in which either rigid-body ground displacements or accelerations are specified. In addition, independent 111otions of multiple restraints will be treated. Figure 4.6 shows six possible displacement components Dg,, Dg2, . . . , nH 6 for a point g that is assumed to be a reference point on ground. The figure 1tlso depicts a typical joint (or node) j on a structure that is connected to ground. The six possible displacement components at pointj are labeled Dj1, D j2 , . . . , n1c,. A location vector r8 j is directed from point g to point j and has scalar ,·wnponents x 8 j, Y gj , and Zgj· We may calculate the displacements at j due to t lf,!,id-body displacements of the ground at point g using the concept of translotion of axes [2, 4]. For this purpose, the rotational components of the ground tliNplacements must be small. Under this condition, the displacements at j in ti•, ms of those at g are (1)
whldt is a sy mmetric vibrational response [see Fig. 4.3(d)]. The antisymmetric rigidhody and flexural modes 2 and 4 do not respond to the centrally-placed load. Altogether,
till' vector of normal-mode responses is
where
(2) 0
-3vs( -~ 1
sin
(m)
(3)
w,,) / w5
0 1'1 1111Nlrn11111tlo11 o l thl'Nl' dlsplttL'l'lllL'lllN bnL·k to phy11k-11t L'ttonll1111tL'N using llq . (4 .J ~) fLIV!'N
1111d
I II Hq. ( I ) the transformalion matrix is
l, 0
r
(4)
Normal-Mode Method of Dynamic Analysis
168
Chap. 4
to,~ .to,, / o,6
tD 1~
o,~,I
This matrix equation has the same form as Eq. (2.3-11) and can be restated
MD + SD
--D,,
.:;\bSo \ ..t.Q..
Coo-rd,.,,..fe.i (9) (10)
Thus, the vector A on the right-hand side of Eq. (9) contains equivalent nodal 8 actions due to rigid-body ground displacements . Similarly, the accelerations at a typical joint j may be expressed in terms of rigid-body ground accelerations at point g , as follows: (11)
I :
:~ -----------r---------,,---X
-
/ /
= A8
in which
D14
95
g
159
Normal-Mode Response to Support Motions
To include the effects of ground displacements in the action equations of motion [see Eq. (4.2-12)), we write them as (8) M D + S (D - ..:1g) = 0
y
I .
Sec. 4.5
o.,
o••
I
Dg3
1I
Dg6
--- - -- . __ _____ __J/
(12)
/: /
I
where
// 91
/
llnd (13)
X gj
Then the vector of accelerations
/ I
..:i8
at all free joints becomes
..
-
..:18
Figure 4.6 Rigid-body ground displacements.
-
T ••
(14)
T 8 D8
In order to use this vector in the equations of motion, we must change to the 111 which 13 is an identity matrix of order 3 and
cL =
-c8i
relative coordinates:
~i
= C18 = [-~, 1 Y g;
Xgj
-:.;J
D* (5)
0
This skcw-sy~metric submatrix contains positive and negative values of the l'0111poncnls of vector r81• These components are arranged in a manner that p, od 11t·cs Ihe cross product of a small rotation vector at g and the location vector l'NJ · Then the time-varying displacements ..:18 at all free joints may be written as
(6)
=
D - ..:18
D* = o - ..:i
(15)
g
In these expressions the symbot D* denotes a vector of displacements relative lo the ground, and the vector D~. contains the corresponding relative acceler11tions. Substituting D - ..:18 and D from Eqs. (15) into Eq. (8) and rearranging, wc find that MD*
+ SD*
= A[
re.t"-t, ve
Cao'<~;
nt1.ti>{ l6)
111 which (17)
Whl' H'
TT _ /I -
T TI,1 [ T
112
,- , 1'
I NII/
1111d 111 I~
IIH•
11111111 11'1
ol NIil ii 10111111
(7)
'l'lll'rcforc, the vector At on the right-hand side of Eq. (16) consists of equivalent 11od11I ac tions caused by ri gid-body ground accelerations. If we compare Eq. (16) 111 illl SDOF counl crpurt in Eq . (2.3- 18), we see that both are of the same form. Afkr tlw l'q11 ivule11t nmlul actions AHor At have been found, the response l llil'11l 11tions in nhsolull' m 1t·l11l iVl' l'OOl'di1111ll'S proceed as described in Sec. 4.4 1111 11pplil·d m·lillllH 11111 IIH' 111111·1 l 11sr, lht• 11h~11l11tt• disph1l'l' 1Hl' nl s at free nodes
160
Normal-Mode Method of Dynamic Analysis
Chap.4
may be calculated from the first of Eqs. (15) as D
= D* + 4g
(18)
Sec. 4.5
The rigid-body rotation 8z, centered at point 2, causes the followi ng step-translations at point 1:
48
which is the sum of the relative displacements and the effects of the rigid-body ground displacements. Example 4.7 Figure 4. 7 shows the plane truss used previously in Example 3 .1. In this truss the cross-sectional areas of members I and 2 are equal to 0.8A and A. Now let us calculate the response of the structure to a small rigid-body step-rotation D86 = 8, about point 2. From Example 3 .1, the stiffness and mass matrices for the free displacements at point I are
S=s [
0.36 -0.48
-0.48] 1.64
(a)
s
2
w~ = 1.8..::.
m
m
= TJD8 =
A8 = S 4 8 = [
1 [3
8,
(c)
-0.288] sL8, 0.384
(d)
which result from applying Eq. (10). From this point we may proceed to calculate the response by the method shown in Sec. 4.4. Thus, Eq. (4.4-4) gives the normal-mode loads as
A= T A= [-1] 0.48sL8, '\!'To,;; 3 N
N= ~ 1
[-0~8L]
as given by Eq. (6). The 2 x 1 operator TJ in Eq. (c) contains only terms from the first and second rows and the sixth column of the general operator in Eq . (4). Equivalent nodal loads (forces) at joint 1 are
in which s = EA/Land m = 3.28 pAL/6 . We also know that
w, = 0.2-
161
Normal-Mode Response to Support Motions
(b)
N
(e)
g
Then the normal-mode responses to these step loads become
_[-(1 - cos w,t) / wf] 0.48sUJz 3(1 - W2t)/w~ '\!'To,;;
(f)
DN -
y
COS
These expressions may be simplified by substituting wf and w~ from Eqs. (b), as follows:
_[-3(1l -- cos cos w,t) ] 0.8mL8, '\!'To,;;
DN -
(g)
W2f
Transformation of this vector back to physical coordinates with Eq. (4.3-5) produces
_ _ [-10 D - NDN -
+ 9c1 +
c2 ]0.8L(J,
3c1 - 3c2
-10
(h)
where c, = cos w 1 t and c 2 = cos w 2 t . Here we see that the truss vibrates about the displaced position given by Eq. (c). 0.Bl
1,· 11mple 4.8 4.8 . We M hnll determine the steady-state response of node 2 caused by a rigid-body rotational ll('tClcration Dg6 = sin Ot of ground at node 1. Stiffness and mass matrices for node 2 are
'l'hc prismatic cantilever beam analyzed in Example 3.2 is shown again in Fig.
e,
S = /
Dno l _ /
~
0.6l
s[
6
- 3L
-3L] 2L
M=m [
2
78
-lll]
- llL
2L2
(i)
whllll' .1· 2/:'/ / 1,1 and m = pAL/2 l 0. Normalization of the modal matrix (from Examph 1,2) with respect lo the mass mulrix yields
I
(l,N
l
I I.
Ill
0. 1,\()4/,
0. 1943/, l
O I ()21
I .•181
(j)
Normal-Mode Method of Dynamic Analysis
162
Chap.4
Sec. 4.5
y
MFF MFR][~F] + [SFF SFRJ[DFJ [ MRF MRR DR SRF SRR DR
/
Dg
•
(3.5-17)
AR
produces (3.5-18a)
X
L
6
[ AF]
As in Sec. 3.5, the subscript F denotes free displacements , while the subscript R pertains to restraint displacements. Writing these equations in two parts
JD,
CD
1~
/
163
Normal-Mode Response to Support Motions
and (3.5-18b)
MRFDF + MRRDR + SRFDF + SRRDR = AR
7
Rearranging Eq. (3.5-18a) gives Figure 4.8
Cantilever beam with rigid-body ground acceleration.
MFFDF + SFFDF = AF - SFRDR - MFR DR In this form we can see that the terms
Due to the ground acceleration at node 1, the accelerations at node 2 are
.
. [L] 0,
"18 = T;'D8 =
1
AFR (k)
sin fit
ns given by Eq. (14). In this case, the 2 x 1 operator T; in Eq. (k) has only terms from rows 2 and 6 and column 6 of the general operator in Eq. (4). Equivalent nodal loads (a force and a moment) at node 2 become
Ai
=
.. = - [ 67 ] mLO, sin fit -M .1 8 -9L
(C)
* NAg T *= AN=
[
7.613 ] L ,v -0.3109
.. . me, sm
fit
(m)
F,om these actions we find the steady-state normal-mode responses to be
D,t
= -
7.613{3,/wf] . , · · . [ - 0. 3109/32/ W22 L v m (}, sm fit
D
* -_ N D N* -_
-
Example 4.9 We shall now reconsider the plane truss in Example 4.7 , which appears in Fig. 4.7. Let the support point 3 have a sudden independent step translation DR 3 = din the x direction. Then determine the response of the free displacements at joint 1. By including terms for the restraint displacement at point 3, we can extend the Nliffness matrix in Eqs. (a) to become
(},
0.36
SFR s = [sRF sRJ = s [ ~~::: SFF
Sill fit
(o)
I
--~:::
i-~:::
3.28 0 i lJ M [M . MRR] m, [- ~1_ _ }_:_~l~ i2 =
When u stru cture has multiple connections to ground , it is also possible to l'nku lalc the respon se to i11rl<'fJ<'tult•11t 11101io11s t?f'support rl'.1·1rai11t.1· by generntinp lltl' 11ppropri11tL· Ntiff11n 1s 1111d 11111ss l'ocffkiL'llts I 1, 'i I, 111 such II t'llsc, thl' rcl11tiVl' diMpli1cl'llll' llls 111 thl' ~11pp111 h 11111st Ill' s 11111II i11 111dl•1 to H•t11i11 liiwur lwlwvlrn I 1•t II~ 11•w1it1• 1111• 11rnl 111 11p1•d l'q11111lo11~ ol 1111,111111 1111 1111 prn11, tl>l1• 11od11l displ11n•
-0.48 i -0.36J (p)
Slil'fnesses for unmoving restraints are omitted from Eq. (p). Similarly, the mass matrix 11 0111 Eqs. (a) is extended to MFF
llll" III N Ill 11·11 1111111•1•d 1111t l pill titlo111•d 1111111 , 11~ 111111 ,w~
(20)
DE
(n)
)L] . .
[ (l.06 lb , - 0.0604lb2 l.46 lb1 - 0.4604b2
-SFR DR - MFR DR
on the right-hand side of Eq. (19) are equivalent nodal loads. They are caused by the independent restrai1:.t displacements in the vector DR and their ~orresponding accelerations in DR .. After the displacements DE and the accelerations are found b y normal-mode analysis , the reactions Ae at su1mort goints may be obtained from Eg. (3.5-18b) if desired. Of course, .@Y..._restraints without motions may be represented with zeros in vectors D,uind D.R (or omitted al-
llpon transforming this vector back to physical coordinates, we find the relative displacelltLlllls:
=
Logether).
whkh is dictated by Eq. (17). Then the normal-mode loads are 1
(19)
I
MFR
=
(q)
Q
RI
,,M,/6. Thon l'rom Eq . (20), the equivalent nodal loads due to the step dl~p(lll'l ' llll' lli J>R I r/ lll l ' w hi•il' 111
1
() \/, I O ·IH
rl
f
0(),,tH .361,I'(/
(r)
164
Normal-Mode Method of Dynamic Analysis
Chap. 4
Next, the normal-mode loads may be calculated as T AN= cf>NAFR =
[
0.6] -sd- 1.8 V10m
(s)
From these actions, we find the normal-mode responses to be
t)]
- [ DN-
0.6(1 -cosw,t) / wT] -sd - - [3(I - cosw 1 - 1.8(1 - cos W2t)/ w~ V10m - - (1 - cos w 2t) Finally, the transformation back to physical coordinates yields D
= cf>NDN =
[10 -- 9c, J!!... 3c1 + 10 c2
3c2
md (t)
V10m
Sec. 4.6
Damping i n MDOF Systems
This matrix contains damping coefficients that are defined as actions required for unit velocities. That is, any term Cik in an array of viscous damping coefficients is an action of type j equilibrating damping actions associated with a unit velocity of type k. This definition is similar to those for stiffness and mass terms and implies that the damping matrix is also symmetric. To form the damping matrix, we consider first the systems for which this urray is assumed to be linearly related to the mass and stiffness matrices. That is , we take
C (u)
In this case, joint 1 vibrates about the displaced position: (v)
165
= aM + bS
(3)
where a and bare constants. The formula in Eq. (3), attributed to Rayleigh [3], is called proportional damping because the matrix C is proportional to a .linear combination of Sand M. In such a case the equations of motion [Eq. (1)] are uncoupled by the same transformation as that for the undamped system. Thus, in principal coordinates we have (4)
obtained by static analysis.
where (5)
4.6 DAMPING IN MDOF SYSTEMS
Damping in solids and structures is not understood as well as stiffness and mass properties . Often the effects of damping upon the response of a vibratory system cun be ignored, as has been done in Secs. 4.2 through 4.6. For example , the iulluencc of a small amount of damping on the response of a structure during an cxcilution of short duration is not likely to be significant. In addition, damping plays a minor role in the steady-state response of a system to a periodic forcing function when the frequency of the excitation is not near a resonance. However, l'or u periodic function with a frequency at or near a natural frequency, damping is of primary importance and must be taken into account. Because its effects arc usually not known in advance, damping should ordinarily be included in a vibrational analysis until its importance is ascertained. When a discretized solid or structure is assumed to have viscous damping, lhc uction equations of motion may be written as
MD+Ci>+SD = A
(' .
C2, ( ' 11
C,2
(6)
'l'hc diagonal matrix w in this expression contains the characteristic values w? for the undamped case [see Eq. (4.2-17)]. Therefore, the ith equation of motion in normal coordinates is 2
..
DN;
+
' + w;2DN; (a+ bw;2)DNi
c,,,
C22
Cn C21
C211
(' IJ
(' II
( '111
=
(i = l, 2, ... , n)
AN;
C Ni
n;
= 2n; = a + bwr
'}'; = -
111 these relationships the term CN; = 2n; is defined as the modal damping ,·1111sta111 for the ith normal mode, and '}'; represents the corresponding modal rlfl111pi11R ratio. Using the first of these definitions in Eq. (7), we obtain
= 1, 2,
(
11 1
( '11 •
(
"'
(
',,,,
. .. , n)
(9)
l•m·lt ol th1· 11 l'q1111tio11 s 1l' iWl'Sl'lllt'd by lhis l"Xprcss ion is uncoupled from all of 1111' 111111'1 ~ 1'11l'11·lo11•, Wl' l'IIII d 1•ti•1111i111• lli1• ll'M iH>IINI' ol' Jill' /th IIH >(k in llw sum1..• 0
1
(8)
W;
(i
(2)
(7)
To make this expression analogous to that for a SDOF system (see Chapter 2) , we introduce the notations
( I)
whkh upplics only to free nod~! displacements. The damping matrix C pre 111ultiplying lhe velocity vector Din Eq. (l) has the general form
c,,
'l'he symbol CP represents a diagonal array that will be referred to as a principal damping matrix, and it consists of a linear combination of MP and Sp. When the modal matrix is normalized with respect to M, the damping matrix in normal rnordinates becomes
1111\lllll' I II N 111111 1111 II
SI)( II • ~yH tt-111 Wllh Vl ~l 11111, d11111pi11p
166
Normal-Mode Method of Dynamic Analysis
Chap. 4
. From the definitions in Eqs. (8), we may express the modal damping ratio Y1 m terms of the constants a and b, as follows: Yi=
a + bw7 2wi
(10)
This. formula is useful for studying the effects upon the modal damping of varymg the constants a and b in Eq. (3). For example, setting the constant a equal to zero (while bis nonzero) implies that the damping matrix is proportional to the stiffness matrix. This type of damping is sometimes refeJTed to as relative damping because it is associated with relative velocities of displacement coordinates. Thus, under the condition that a = 0, Eq. (10) becomes
bwi 'Yi = 2
(11)
which means that the damping ratio in each principal mode is proportional to the u~1da mped angular frequency of that mode. Therefore, the responses of the hi gher modes of a system will be damped out more quickly than those of the lower modes . O n ~he othe~ h~nd, setting b equal to zero (while a is nonzero) implies that !he cla~pmg matnx 1s proportional to the mass matrix. This type of damping is Homct1mes called absolute damping because it is associated with absolute velocities of displacement coordinates. In this case Eq. (10) simplifies to a 'Yi=-
2wi
(12)
.~o thul the damping ratio in each mode is inversely proportional to the undamped 1111gular frequency. Under this condition the lower modes of a system will be suppressed more strongly than the higher modes. .. .It has been shown by Caughy [6] that the criterion given by Eq. (3) is s11 l llc 1c nt but not necessary for the existence of principal modes in damped 11ystems . .The ess~ntial condition is that the transformation which diagonalizes t hl' dump111g matn x also uncouples the equations of motion . This criterion is less H'Nltictivc tha n that in Eq. (3) and encompasses more possibilities. llowcver, in the most general case, the damping coefficients in matrix (' llll' suc h !hut the damping matrix canno t be diagonalized simultaneously with thl' 1111,ss and stiffness matrices. In this instance, the natural modes that do exist haw phust· re lar io~ships that complica~e the analysis. The eigenvalues for this type o l Nysll'lll arc ei ther real and negative or complex w ith negative real parts. Tlw l'otllplt·x c igcnval.ues ~ccur as conj ugate pairs, and the correspondin g cigcn Vl'l'lms uls.o t·o~1s1st of complex conju gate puirs. In hi ghly damped systems, wh1:1l' lhl· 1nwg 111111 y ll'1111s dm· to di ssipnl iw f'otTL'S nn• sig nifit-11111 , (he 111elhod ol h>HNI / I 11u1y lw llSl'd 'l'hi~ 11ppn111l'11 l11v11lVl's l1 1111s l0111111li1111 of' llw II Sl'eond 111d1•1 l' 11 IIIH'llll)lh•d ilt,1 111d1•1 1•q1111IIOIIH
Sec. 4.6
167
Damping in MDOF Systems
Lightly damped structures need not be treated in such a complicated manner, especially in view of the fact that the nature of damping in physical systems is not well understood. The simplest approach consists of assuming that the equations of motion are uncoupled by use of the modal matrix obtained for the structure without damping. In other words, the matrix cl> is assumed to be orthogonal with respect to not only M and S but also C, as follows:
ct>J C cl>; =
cl>; C cl>j = O
(i -:/= j)
(13)
This expression implies that any off-diagonal terms resulting from the operation CP = cf>T C cl> are small and can be neglected. In addition, it is more convenient to obtain experimentally (or to assume) the damping ratio y; for the natural modes of vibration than to determine the damping coefficients in matrix C directly. We can usually find the damping ratio y 1 for the first mode of vibration by field testing a structure or by previous experience. As mentioned in Sec. 2.4, the range of this constant for metal structures is approximately 0.01 to 0.05, while that for reinforced concrete is about 0.05 to 0.10. With the value of Yi on hand, we can extrapolate to other values of y; using the approximate formula: 'Yi
W·)e1
= 'YI ( W;
(0.5
::5
e1
::5
(14)
0. 7)
This expression suppresses the higher modes in accordance with damping experiments [8], but not as severely as in Eq. (11) . Alternatively, we can simply determine y 1 and then let 'Yi = y 1 for all other modes. Now we rewrite Eq. (9) in terms of 'Yi as (i
= 1, 2, ... ,
n)
(15)
where CN; = 2yiwi. In order that this equation may pertain to a lightly damped structure, let us also specify that O ::5 Yi ::5 0.20 for all modes. The type of damping associated with this set of assumptions is of great practical value, and it will be referred to simply as modal damping. It should be remembered that this concept is based on the normal coordinates for the undamped system and that dumping ratios are specified in those coordinates. When modal damping is assumed in the normal coordinates for a structure, ii may also be of interest to determine the damping matrix C in the original (or physical) coordinates. This array can be found by means of the reverse trnnsformation (16) lustcacl of attempting to invert cl>N, ho wever, we use the relationship 1sec Eq. (4.2-27)) and rewrite Eq. (16) as
•I>/ • iM
(17)
l'hii; f'o nu of' lhl' ll'll11Hlo111111lio11 is t•spl'd11 ll y appropri ate when not all of the 11111d1•, 1111• ind11dl•d 111 till' 1111ul y1. is (111od11 I t111t1l'1ttio11).
11111111 111
168
Normal-Mode Method of Dynamic Ana lysis
Chap. 4
Example 4.10
Sec. 4.7
Damped Response to Periodic Forcin g Functions
169
where
As an example of modal and physical damping, we shall reconsider the 3-DOF plane truss analyzed in Examples 4 .1 and 4.2. For this case, let us assume that the structure is made of steel and that the damping ratio for the first mode of vibration is y 1 = 0.02. Prom Eq. (4.2-f) the three angular frequencies are wi =
0.5197c1
w2
= I.445c1
W3 =
2 .304c1
(a)
1.445
,'2
= 0.02 ( 0.5197
'}'3
2.304 ) = 0.02 ( 0.5197
0 6 ·
+
. 2n;DN;
+
2 W; DN; -
= 0.04887
DNi
a (,.... = Pm; 2 JJi cos ~ Lt
-
O;)
1 (c)
{3; = Y[ l - fl/w;) 2] 2 + (2y;fl/ w;)2
and the phase angle O; is
Using these values in Eq. (16) along with iv 1 from Eq. (4.2-h), we obtain
_
O; - tan
C = ivTCN,v 1
[
1, 2, ... , n)
(3)
(4)
The magnification factor {3; in this expression is
= 2(0.04887)(2.304)c1 = 0.2252c 1
= A YpE
=
W;
= 2(0.02)(0.5197)c1 = 0.02079c 1
0.06154
(i
Pmi cos flt
(b)
CN2 = 2(0.03694)(1.445)c1 = O. I068c1 CN3
(2)
in which p 111; is a constant. This equation has the same form as Eq . (2.4-23), so we can take the damped steady-state response of the ith mode to be
= 0.03694
Then the normal-mode damping constants are CN1
= {P1 , P2, A, ... , Pn}
In Eq. (1) the terms in Pact as scale factors on the function cos flt. Transformation of the action equations of motion to normal coordinates produces the typical modal equation .. DN;
where C1 = (VE/p)/L. Applying Eq. (14) with e1 = 0.6., we find that 0 6 ) ·
P
-0.01270
- 0.01 270 0.02002
-0 .01662
0.01987
- 0.01662] 0.01987
(d)
0.07620
wllich is the symmetrical damping matrix in physical coordinates.
4 .7 DAMPED RESPONSE TO PERIODIC FORCING FUNCTIONS
- I[
2y;fl/ W;
1 _ (fl/ w;)2
J
(5)
(6)
Equations (4) , (5) , and (6) are drawn from Eqs . (2.4-31),.(2.4-32), and (2.4-33) , respectively. The response given by Eq. (4) may then be transformed back to the original coordinates in the usual manner, using Eq. (4.3-5). To determine the response of the mode having its angular frequency w; closest to the impressed angular frequency, we need only use the modal column N; in the transformations to and from normal coordinates . That is, Eq. (4.4-4) is specialized to (7) Pm;= <1>1;P and the back-transformation in Eq. (4.3-5) becomes
As mentioned in Sec. 4.6, damping is of greatest importance when a periodic 1•xcitalion has a frequency that is close to one of the natural frequencies of a MDOf! system . In this section we consider the normal-mode approach for l'llit'uluting steady-state responses of discretized structures to periodic forcing l1111l'tions. Knowing the imposed frequency of such a function and the natural lil·qucncies or the system, we can obtain in a direct manner the steady-state ll'Sponses of the modes having frequencies in the vicinity of the imposed frc '(lll'llt'Y . Both simple harmonic and general periodic forcing functions will bl' di.~l·11 ssed, and modal damping will be assumed, as described in Sec . 4.6. Ir II lightly damped structure is subjected to a set of actions that uru all p1op111tio11ul to tlw siinpk harmonic fun<:tion <.'OS tlw action vector A lllny h1• WI iltl'II HS ,
n,.
A
p
!' ON
01
(Il
D
= N;DN;
(8)
If desired, this process can be repeated for other modes with frequencies in the vicinity of fl. Now we shall consider a lightly damped structure subjected to a set of actions that are all proportional to the general periodic functionf(t). In this case the applied action vector A may be written as A = F(t)
= Pf(t)
(9)
where the vector Pis given by Eq . (2). Proceeding as described in Sec. 2.5 , we i:xpn:ss f(r) in the form of' a Fourier series, as follows: ,.
/'(I)
( 10)
Normal-Mode Method of Dynamic Analysis
1'70
Chap.4
Sec. 4.7
which is drawn from Eq. (2.5-1) with i replaced by j. The coefficients a1, b1 , and in Eq. (10) may be evaluated as indicated by Eqs. (2.5-3). Transformation of the action equations of motion to normal coordinates produces the typical modal equation
(i
=
where p,,.; is again the constant given by Eq. (7). From the solution of Prob. 2..5 5, we talce the damped steady-state response of the ith mode to be DN; =
Pm;{ao + w,
f
{3u[a1 cos (}flt - Ou) + b1 sin (}flt -
j= l
in which the magnification factor
8;J]1
J
- VpiL
The magnification factor for the first mode is obtained from Eq. (5) as
~1=--;========:=~=====(~=:=:=' 0 .25 ) 0.25 7) 4
(12)
( l - 0 .2701
f3u = Y[l - (jfl/w;)2] + (2y;Jfl/w;)2
DNI =
(13)
und lhe phase angle Ou is
= - I[
(d)
11.94
2
+ (0.0 ) 0.2701
From Eq. (4) we find the damped steady-state response of the first mode to be 2
_
~J (c)
2
f3u is 1
eij - tan
vk[0.2803 1.2114 -0. 2995{
l.2114Pi
(11)
1, 2, ... , n)
~
p., - "'"· p
110
171
Damped Response to Periodic Forcing Functions
2y;jfl/ W;
l _ (}fl/ w;)2
]
(14)
Because a multiplicity of terms contribute to the ith mode in Eq. (12), the possibility of resonance (JD, ""' w;) is much greater for a general periodic func1i1111 than for a simple harmonic function . Therefore, it becomes more difficult lo predict in advance which of the natural modes will be strongly affected. l lt1wl~vcr, after the forcing function has been expressed as a Fourier series, each of the JO frequencies can be compared with thew; frequencies for the purpose ol p1'l•dicting large-amplitude forced vibrations.
2114 1. ~~(11.94) sin (flt - 81) 0.2701£ V pAL 2 53 .55P2L p . CA 8)
E\,lpAf,
where 81
=
tan
Sill ut -
(e)
1
l
- 1
(0.04)(0.5) ] (0.5197) = 270 21' _ 0 25 l - 0.2701
(f)
as given by Eq. (6). Transfonning the response of the first mode back to the original coordinates with Eq. (8), we obtain D = NIDNI = {15.01, 64.87, -16.04}~~ sin (flt - (Ji)
I~ nnmk 4.1 t
(g)
Proceeding in a similar manner, we can determine the response contributed by the
I t'1 tlw plane truss in Fig. 4.l(a) be subjected to a simple harmonic forcing function /', Ni ll Ht, applied in the x direction at joint 2 (corresponding to displacement D 2). The 11111111l11r frequency of the forcing function is n = (0.5/L)VE/p. Calculate the steadyHinh· ll'Nponse of the structure, assuming that the modal damping ratios are y 1 = 0.02, Yi 0.0.15, und 'Y1 = 0.05 (see Example 4.10). Till' squurc of the impressed angular frequency is
n2
= 0.25£ L2p
/,,
,,
(h)
D = {-0.09478, 0 .01875, -0.1093}~~ sin (flt - 82) 1111d that due to the third mode is found to be
(i)
D = {- 0.1824, 0.1104, 0 .1114}~~ sin (flt - 83) (a)
The amplitudes in both of these vectors are small compared to those i~ ~q. (g). Further111ore, the innucnce of damping is significant for the first mode but neghg1ble for the other
'l'hlN v11h1l' is t'nirly close to the square of the first angular frequency 0 .270 IH
second mode as
IWO.
(h)
11H 1'iw11111 l'11hlt• I •I 'I h1•H•lo11•, w1• Nho11ld 1•xp1•1•11h1• tl1 NI 11111111· ol 1hl' s1n1vtl11l' 10.lw 1111 1111111111 y lCllllllh111111 hi !Ill' 11·~pn11Nl' IJ~111p Jlq (/) ,WI d1,1r1111h11• lh1• lliN1 IIOlll\111 llllllh lo11d NI 1111 1111 llll In h1
11:,mm1>lt• 4.12 l•li\tlll' 'I .') Mhows 11 1wriodic forci11A 1'1111l'tio11 in lhe sh'.1pc..of' a square wave. lf this llua·tlmi ls 11 pplli•d 111 llw I tlh1•1•1io11111 jnl111 I ol 1111' IIIINN 1111·1~. 4. l(a), find the damped Hh'illl y Hlilh' ll'H Jlllll/'11' 1111 l 'lll hot lh1• IICllllllll 111111h•H 0
172
Normal-Mode Method of Dynamic Analysis
Chap.4
p
0
2'!T
'lT
- P,
3'!T
n
TI
4'lT
n
n
D Ni
t--
Expanding the square wave as a Fourier series (see Prob. 2.5-1), we obtain 4 1 p (sin fit+~3 sin 3fit 7T
+ . . .)
(j)
Trunsformation of the load vector to normal coordinates produces
AN
= J.; A = J,;
F(t)] 0 [
0
=
[cf>N11] Nl2 F(t)
(k)
m 3
111 which the load corresponds to displacement D 1 . In accordance with Eq. (12) the nw mul-modc responses become '
3 m1~11 sin (fit - 811) + /3~ sin (30t - 8 13) + .. -}/ wf l)N = 4P1 m 2{/321 sin (fit - 821) + /323 sin (3fit - 823 ) + . . .} /w~ 7T 'l'N1 3 N13~ 31 sin (fit - 831) + Wlil'tC
13;3 sin (3fit -
833 ) + . . -} /
=
e
-11·/(v I
NOi
cos
Wd; t
+ DNOi +
n;DNOi
~
. sm
WJ;t
(1)
Wdi
which is drawn from Eq. (2.4-6). The angular frequency of damped vibration in l~q. ( 1) is (2) WJ; = V wJ - nT = w; V 1 - YT
Figure 4.9 Periodic forcing function.
F(t) = Pif(t) =
173
Damped Response to Arbitrary Forcing Functions
nre relatively high ( y; > 0.05) , the effects of damping could have some conNc4uence. Therefore, we shall now modify the formulations in Secs. 4.3, 4.4 , 1111d 4.5 to account for the influence of damping on vibrational response in normal coordinates. Action equations and modal damping will be assumed th roughout the discussion. In Sec. 4.3 we formulated the normal-mode responses of a MDOF struc111re to initial conditions of displacement and velocity at time t = 0. In the presence of damping, the free-vibrational response of the ith mode, given by Eq. (•1. 3-4), must be changed to
-
p1
Soc. 4.8
wl
_the magnification factors and phase angles are given by Eqs. (13) and ( 14),
I l'N IWt"t1vc ly,
4 .8 DAMPED RESPONSE TO ARBITRARY FORCING FUNCTIONS
1>11111ping sl_1ou_l~ be included in transient response calculation s whenever ii 111l1,1l~t Ill' s1gn1licnnt. For t•x111nplc, if the durntinn or u forc ing function is ll' l11t1v1·ly 10111• t·o111p1111.·d to illl· 1111turnl pt·1'iods ol II st111 ctun·, dninpiug rn\ild lw 1111po1 l1111t. Also , wlll'11 1111• t11111• of i11t1•1t•sl ii.. i..hrn I h111 till' 111rnlnl d11111pi11g ,·nti wi
111 which w; is the undamped angular frequency. Transformation of the initialvectors D 0 and D0 to normal coordinates remains the same as in Eqs. (•U -2), and back-transformation of the response is still given by Eq. (4 .3-5). Similarly, the calculation of normal-mode responses to applied actions, as d1•scribed in Sec . 4.4, requires only a few modifications associated with modal dumping. Transformation of applied actions to normal coordinates is the same 11N in Eq. (4.4-4), but Duhamel's integral in Eq. (4.4-7) must now be written as
1 lllldition
DN;
e - n;r =WJ;
lt e '
tH'
'
AN;
sm
I
WJ;(t -
t ) dt
I
(3)
O
which is taken from Eq. (2.6-4). Normal-mode responses to support motions, covered in Sec. 4.5 , may also It,• ultered to include the effects of modal damping. For rigid-body ground ,111·l'lcr~tions, there is neither displacement coupling nor velocity coupling be1w1•1•n the masses and the ground in relative coordinates. There exists only 111111tiul coupling with ground , which is the same as that for the structure without il11111ping. To determine nodal responses relative to the ground, we first calculate th1• 1•quivalent nodal actions in the vector At, as given by Eq. (4.5-17). Transli11111111ion of these actions to normal coordinates yields the equivalent modal 11111\ lh (4) 1 lu•u the relative response D~1 in each normal coordinate is obtained from Eq. 1 \ J. wi th Ai 1 replacing ;\ Nf, Finally , these displacements are back-transformed 11· l11p (5)
174
Normal-Mode Method of Dynamic Analysis
Chap.4
Sec. 4.9
175
Step-by-Step Response Calculations
(e)
which_ gives the relative responses in physical coordinates. As before, the absolute d1splace~ents at free nodes may be found with Eq. (4.5- 18). . In ~ertam cases where support displacements are specified, there is velocco_uplmg bet~een fre~ displ~~ement coordinates and support restraints. This s1tu~t1on can ar1~e for e1th~r ng1d-body ground displacements or independent mot1~ns o~ multiple restraints. Methods for handling such circumstances are described m Ref. 1.
4.9 STEP-BY-STEP RESPONSE CALCULATIONS
lt:xnmple 4.13
111 Sec . 2.7 we examined step-by-step solutions for SDOF structures, where the
We s~all now repeat the first part of Example 4.5 , including the effects of modal dt11pmg. Rec~II that ~e three-member truss in Fig. 4.4(a) is subjected to a step force l i, correspondmg to d isplacement D 2 • Symbolically transforming the vector of applied actions to normal coordinates pwduces
lorcing functions are not necessarily analytical expressions. The basic approach 111 that section was to approximate the forcing function (or data points) using 1>iecewise-linear interpolation and then to use the Duhamel integral within small lime steps. We shall now incorporate this technique into the normal-mode 111cthod for calculating transient responses of MDOF structures. As in the pre1·cding sections, modal damping will be assumed throughout. Because of the 1•xtensive calculations required, it is implied that the method of this section is to he program.med for a digital computer. Such a program is described in Sec. 4.10, where numerical examples are also presented. Let us consider again the piecewise-linear type of interpolation illustrated hy Fig. 2.18. Without loss of generality, only one such forcing functionfe(At1) will be handled at a time, and the piecewise-linear action vector Ae, (or M;) may Ill· expressed as
•?
AN=
~U =
(a)
~ { ; ] = [ :: }
I >uc lo the step function, the damped normal-mode responses are
9}/
N21{1 - e-"ir(cos Wd1 t + ;~ sin wd1 1 DN
= P2 N2+ -
e- "2 ' (cos wd2t
N2+ -
e- " '(cos W113t
3
+ ;: sin wd2t)} /
9}/
+ ;: sin wd3
wf w1
where i = 1, 2, 3. The expression in Eq. (e) is a modified version of Eq. (2.7-2c).
(b)
w5
'lhl·sc expressions are drawn from Eq. (2.7-2b). 11:x11m1>tc 4.14 S11.p1iosc l!lat t~e gr:~und in Fig. 4.4(a) accelerates in they direction in accordance with lhl 111111p_fun ct1on D82 = a2t/t2. Formulate the damped responses of the normal modes Nl111t111g from rest. ·' this problem we work in relative coordinates, where the equivalent nodal toad
1:m
(j + 1 = 1, 2, ... , n1)
where At represents a small but finite time step, and n1 is the number of steps. 1 III this form the values of P act as scale factors on the common function f e(At1). 11 more than one such function is applied simultaneously, the responses for each 111 them handled separately can be superimposed. Transformation of the action equations of motion to normal coordinates pt oduces the typical modal equation t' (2) DN,· + 2n,· DN,· + w,2 DN·I = ANI,)· + M N·I ,).At· J
Vl'l'IOI IS
(i=l,2, ... ,n
Mm~:,- mM<:' IIN Jt lVl'n
(<:)
(1)
and
j
+ 1 = I, 2, ... , n1)
wh1•1e t' = t - t . The symbol AN;,J in Eq. (2) represents the ith normal-mode 1 11111d at time t1. Thus, (3)
hy Eq . (4.5- 17). Then the cqu ivulcnt nonnnl mode loads arc //, I
/1111
•I• hll 1(11 lllllljl ltllll llllll, 1(11 d11lll(II
ii
(d)
111 11dditio11, we have the change in the ith modal load during the time step At1, ,11 li111•d 11s follow s: (4)
II
1ltlllli11' llitllh II ~(11111~1 I, (111 lllllt
wl11·11·
lh1·
11y111l111I
A,,, 111 dl•111111•11
ihl' m ltnll
111
111111• 1111 •
176
Normal-Mode Method of Dynamic Analysis
Chap. 4
In a manner similar to that in Sec. 2. 7, we can express the response of the ith mode at the end of the jth time step as the sum of three parts, which are = (DNI
D Ni,j+I
+
DN2
+ DNJ)i,j+ I
(5)
The first part of the response consists of the ith free-vibrational motion caused by the conditions of displacement and velocity at time tj (the beginning of the interval). Therefore , we have (D NI) i,j+ I =
e -n·tJ.t·(D I Ni ,j cos I
A
Wd;utj
+ DNi,j +
n;DNi,j
. sm
A
)
Wd;Utj
Wd;
(6a)
= -ANi,j[ 2- 1 W;
- e - n·tJ.t·( ' 1 cos
wd;Atj
. + -n; sm
wd;Atj)
J
Wd;
whkh is drawn from Eq. (2.7-Sc). We can also write the velocity of the ith mode at the end of the }th tinw Ntl•p in three parts, as follows: (DNI
+
D N2
+
(7)
DNJ)i,j + I
'l'lu:sc lhrec contributions may be formed by extending the notation in Eqs. (J. .7 6) to obtain (/)• NI ) /,/1 I
= e
n•tJ.t·[ 1 1
-
(v
Ni,jWd;
+ n;l)Ni,j +Wdin;DNi,j) sm .
A
Wd;Utj
(811)
+
DN;,J
cos w,11 AtJ (8h)
AA.Nl,/1I '"/ At1
,.
111 11
t 1(cos
w,11 At1 I
111
'",11
DNi,j+I = DNi,j
+
/)Ni,j Atj
+
1(ANi,j
+
1AAN;)(AtY
(9)
•
DNi,j+ 1
' = DNi,j +
(ANi,J
+
A 2I AANi,j ) ut1
(10)
11.quations (9) and (10) pertain to rigid-body motions with no absolute damping.
4.10 PROGRAM NOMO FOR NORMAL-MODE RESPONSE
(6c)
=
pmvide the initial conditions of displacement and velocity at the beginning of Htep j + 1. These expressions may be applied repetitively to obtain the time history of response for each of the normal modes. Then the results for each time 11tution are transformed back to the original coordinates in the usual manner. If the ith mode of a structure is a rigid-body motion, appropriate expres~ions for rigid-body response must be used instead of the recurrence formulas ~iven above. That is , the displacements in Eqs. (5) and (6) are replaced by
(6b)
which is taken from Eq. (2. 7-5b); and that associated with the triangular portion becomes
l)Ni,j+ l
177
Program NOMO for Normal-Mode Response
1111d the velocities in Eqs. (7) and (8) are supplanted by
This fo rmula represents an extension of Eq. (2.7-5a). The other two parts of the response in Eq. (5) are due to the linear forcing function within the time step. The rectangular portion of this impulse yields (D N2 ) i,j+ 1
'foe. 4.10
sin w,11 6.1)
l
1
(He)
l\q1111tlo11s (.,) 1h11111)'h (H) t·rn1 stil11lt• ll'l'lll ll'lll't' 111111111111.~ lon·11kul11ti~11• th1· d111111wd ll'N p1111 ~1· ol 1·11d11111111111l 11111tl1• 1111h1• 1•11d 1111111• /lh lii11t• slt•p 'l'lwy 11bo
l'hc normal-mode method for calculating dynamic responses of structures will 110w be applied in a program named NOMO. This program can be used to 1111nlyze any type of linearly elastic framed structure or discretized continuum. l'he main program for NOMO calls six subprograms, as shown by the double boxes in Flowchart 4 .1. The first subprogram appearing in the flow chart is VIB, which is the program for vibrational analysis described previously in Sec. 3.~. I lerc it is treated as a subprogram that calls the seven other subprograms m l•lowchart 3.1. However, as the last step in Subprogram RESl, the eigenvectors 111 c normalized with respect to the mass matrix. Then the number of loadin_g 111,1·tems NLS is read, the loading number LN is initialized to zero, and LN 1s liK-rcmented by one. Next, Subprogram DYLO reads and writes dynamic load d11(a, and the output includes the loading number LN as well as the n~mber of luutling systems NLS. This is followed by Subprogram TRANOR, which tran~lmms initial conditions and actual or equivalent nodal loads to normal coord11111h.:s, using Eqs. (4.3-2) and (4.4-4). At its beginning , TRA~OR reads and w1itcs the number of modes NMODES for the purpose of usmg modal trun1111 ion. Subprogram TIHIST calculates time histories of normal-mode displace1m111ts and velocities with the step-by-step method described in .sec. 4.9. Be11111se uniform time steps are to be used, the coefficients of DN;,J , DN;,J, ANi,j, and ,\ANt,J in Eqs. (4.9-5) through (4.9-10) becom~ consta~ts t~at need ~e deter11il1ll'd only once at the beginning of the analysis. For v1brat10nal motions, the 11•Nponscs in Eqs . (4.9-5) through (4 .9 -8) can be written in eight parts , as lt1llows: n N1.i1 1 /,NI/II
nNI.,
t (' , 1>N1. 1 t c,A.N1. 1 -1 C4 AAN1 .1
(I )
( \ /)NI,/
I ( ',/>NII I (' 1A.N/,f I ('K/J.A.N/,/
(2)
(',
178
Nurmnl Moctn M111h111l 11l l>y1111111h Annlyala
179
Pro,1rem NOMO for Normal-Mode Response
I h1 l'Ollstunt coefficients C 1 through C8 appearing in Eqs. (1) and (2) have the iii ll111tions Flowchart 4.1 Main program for NOMO•
n~(
C1 = e ' cos
IOI
l. Prngram VIB from Sec. 3.8, plus nom1aliza1ion of eigenvcciors with respcc1 to mass mairix.
C2
A +-~ wd;ut
= - 1- e -n;t!.t sin wd;!:J.t Wd;
• sm
A~ wd;ut
Wd;
C3
1
= -(1 - C1)
Wf
(3)
Read number of loading systems.
Initialize loading number to zero.
Increment loading number by one.
2. Read and write dynamic load data.
3. Transfomi initial conditions and loads to nonnal coordinaics.
4. Calculate lime histories of nonnal-modc responses.
5. Transfomi displacements back 10 physical coordinates.
6. Wrilc ancVor plot rcsulls of response calculations.
Check for las1 loading sys1c111.
Go 10
101
,truclurc l·ND
•Applies lO uny type of lincnrly ch1Ntlc
Nlluctu11,
IS11hp1ow11111~111111 dllkr for cvc1y ty1w ol Hl1m•1111,•
and rrocc,, 111101hcr
1111 Nl' coefficients, as well as simpler constants for rigid-body motions [see Eqs. and (4.9-10)], are coded in Subprogram TIHIST. After the response calculations have been completed, the normal-mode 1ll~pl11cements are transformed back to physical coordinates by Subprogram I 11/\BAC, using Eq. (4.3-5) . The last subprogram, named RES2, optionally '" 111•s and/or plots resulting time histories of nodal displacements and axial l111n·s in members. After writing, the maximum and minimum values of these 11111111tities and the times of occurrence are written as well. At the end of the lh1wchart the test of LN against NLS determines whether to return for another h111dl11g system or another structure. The logic in Program NOMO implies that the time histories of nodal 1IINplacements are stored in a matrix of size NDF x NTS, where NDF is the 1111111bcr of degrees of freedom and NTS is the number of time steps. Although 1111, procedure is conducive to plotting and calculation of internal actions or l11•sses, the use of such a large block of storage is not efficient. We could 111111sfcr blocks of these displacements to auxiliary storage if desired. However, lhl~ upproach would require more data, more intricate logic, and more detailed 1 ,pl11nutions. Program NOMO may be specialized to become NOMOCB for continuous 11111111s, NOMOPT for plane trusses, and so on. The main program for each p1'l' inlization has certain subprograms that are different for each type of struc11111•, us indicated by the second footnote below Flowchart 4.1. That is, the ~11hp1ogram named DYLO becomes DYLOCB for continuous beams, DYLOPT 1111 plune trusses, and so on. As for Program VIB, notation for Program NOMO ,1ppt•11rn in Part 5 of the list of notation near the back of the book. Also, the lh1wch11rt for Program DYNAPT in Appendix C contains detailed steps for the 1,,.,1c in the subprogrums of Program NOMO. Tnhk 4 . I shows prcpurution of d1111m11i<· load data for plane trusses. In the ll1 11t lt1w of' thl' tuhll· 111c thc n11111h1•1 lll 1i111t• st1·ps NTS, the duration of the 111111111111 liit1l' sll'P In: und llw d11111p111p 1ut111 I >AMPR. pcrtaining to nll modes. 1 I 1l lJ)
Normal-Mode Method of Dynamic Ana lysis
180 TABLE 4.1
Program NOMO for Normal-Mode Response
181
Dynamic Load Data for Plane Trusses
Type of Data Dynamic parameters Initial (a) (b) (c)
Sec. 4.10
conditions Condition parameters Displacements Velocities
Applied actions (a) Load parameters (b) Nodal loads (c) Line loads Ground accelerations (a) Acceleration parameter (b) Acceleration factors• Forcing function (a) Function parameter (b) Function ordinates
No. of Lines
Items on Data Lines
1
NTS, DT, DAMPR
1
NNID, NNIV 1, D0(2J-l) , D0(2J) J , V0(2J-l), V0(2J)
NNID NNIV
1
NLN NEL
NLN,NEL J, AS(2J-l ), AS(2J) I, BL!, BL2, BL3, BIA
1 1
IGA GAX,GAY
1
NFO K, T(K), FO(K)
NFO
"Omit when IGA = 0.
Nl'Xt, the initial-condition parameters NNID and NNIV give the number ol nodes with initial displacements and initial velocities, respectively. Each line ol llw data for initial displacements (NNID lines total) contains a node number J, the I component of displacement D0(2J-l), and they-component D0(2J). Sim ilul'ly, the NNIV lines of data for initial velocities indicate terms that are anal opous lo those for initial displacements. Applied actions and ground accelerations carry appropriate dimensions 11ml me to be multiplied by a dimensionless forcing function given at the end ol tlw l11blc. In the data for applied actions, the load parameters are the number ol lo11dL'd nodes NLN and the number of elements with line loads NEL. Data fo1 11od11I loads (NLN lines) consist of a node number J and scale factors for llw 1 rnmponcnl of force AS(2J-1) and they-component AS(2J). Figure 4.10(11) dl•pit•1s lhcse components of the applied force acting at node j of a plane truss. Also, Fig. 4.1 O(b) shows linearly varying line loads (force per unit length) upplied in the x and y directions along the length of a typical plane-truss elemt'lll 1. 1>11111 for this condition of loading appear as part (c) under applied actions, wht•1l' L't1c:h of NEL lines contains an clement number I and scale factors for lh1· load i11tt·11sitics 13LI throuµh BI A shown in Fig . 4. IO(h) . ll can easil y he show11 lhul 1h1• t•q11i v11k·11l 11od11I loud~ i11 slruclunil din•ctiom1 al joints j Hnd k 111'L'
(b)
Figure 4.10 Loads on plane truss: (a) nodal loads; (b) line loads on element.
AS(Jl) = (2BL1 + BL3)L/6 AS(J2) = (2BL2 + BL4)L/6 AS(Kl)
= (BLl +
AS(K2)
= (BL2 + 2BL4)L/6
(4)
2BL3)L/6
where subscripts JI through K2 are obtained from Eqs . (3.5-30). Of course, an infinity of other load sets could be applied to the element and their equivalent uodu I londs dcri vcd .
182
Normal-Mode Method of Dynamic Analysis
Chap.4
Data for ground accelerations consist of two lines, the first of which gives 11 parameter IGA for ground accelerations. If this number is nonzero, ground uccelerations exist; otherwise, a zero indicates nonexistence. On the second line arc scale factors GAX and GAY for ground accelerations in the x and y directions. The possibility of rotational ground acceleration is omitted from this program. For either applied actions or ground accelerations, the load data must include information defining a dimensionless piecewise-linear forcing function. Such a function is given as the last block of data in Table 4.1. On line (a) we have the number of function ordinates NFO. Ordinates of the forcing function 11rc given in NFO lines, each of which contains a subscript K, the time T(K) when the function ordinate occurs, and the value of the function ordinate FO(K) al that time. For simplicity, we restrict the time T(K) to be equal to an even number of time steps DT. If the forcing function has a discontinuity at time T(K), lwo lines are required to define FO(K) on both sides of the discontinuity (for the sume time). Note again that the function ordinates will receive dimensions only when they are multiplied within the program by either applied-action or ground11ecclcration scale factors, thus creating time-varying proportional loads. ln all computer programs for dynamic analysis we write and plot seleclivcly to limit the volume of output. Table 4.2 shows our method for selecting nodal displacements and element stresses for writing and/or plotting in Subpmgrnm RES2. Line (a) of the table contains output parameters that have the lollowing meanings: IWR IPL NNO
= indicator for writing (0 or 1) = indicator for plotting* (0 or 1) = number of nodes for output
Output Selection (11) Output pnrnmctcrs (h) Nodnl displnccmcnts
I
(l') l\lt•lll('III Ntll'SN('N
we calculate and write only the axial force at the k end, which has the M 11me sign as the axial force in an unloaded member. On the other hand, with lwo- and three-dimensional finite elements, the stresses are calculated at sampling points for numerical integration (see Chapters 7 and 8). If desired, ~e ,•ould specialize the variables JNO( ) and IEO( ) to have a second subscnpt cli,noting a particular type of nodal displacement or element stress. l~xnmple 4.15 l'o illustrate using Program NOMOPT, we shall analyze the three-member pla~e truss IINCd previously in Example 3. 4. Dynamic responses of this structure will be obtamed for the following influences: I. Initial displacements of 0.1 in. at all nodal degrees of freedom, with DAMPR = 0.0 2. Piecewise-linear force in x direction at node 2, with DAMPR = 0.0 '- Same as case 1, but with DAMPR = 0.1 4. Same as case 2, but with DAMPR = 0.1
11, 11 11 cases we take the values of E, p, L, and A as given in Example 3.4, where the 11111tcrial is steel and US units apply. Table 4.3 contains a partial listing of the output from Program NOMOPT for the 111111, analyses considered. With the formats coded in the program, this output becomes TABLE 4.3
Computer Output for Example 4.15
STRUCTURAL PARAMETERS NN NE NRN E 3 3 2 3.0000E+04 NODAL COORDINATES X NODE 0.000 1 150.000 2 150.000 3
RHO 7.3500E-07 y
0.000 200.000 0.000
ELEMENT INFORMATION AX J K ELEM. 10.0000 1 2 1 6 . 0000 1 3 2 8 . 0000 2 3 3
Output Selection Data
No. of Lines
11 usses
PROGRAM NOMOPT
lt1 line (b) the node numbers JNO( ) for output of displacements are listed, and t·kmcnt numbers IEO( ) for output of stresses appear in line (c) of the table. J-101 l 1·11111l'd struclures, this output usually consists of generalized (or integrated) Nltt•sses al the j and k ends of members (see Chapter 6). However, for plant•
Type of' Data
Items on Data Lines
NNO
IWR, IPL, NNO, NEO JNO(l), JN0(2) , .. , , JNO(NNO)
NEO
1110 ( I ) , 11 \0(2) , , , , , IEO(NEO)
+111•1·111m• plilllillp 1·11p11hlll1 y v11d1•N 111 1111 I 11•1. II "' . ,,,"
lllllllllll, Pllll'IIIIII II Nl' I " ,
11!1• p1111um•11, 1 11'1.
11111y
ex CY 0.6000 0.8000 1.0000 0.0000 0. 0000 -1. 0000
EL 250.0000 150.0000 200.0000
NODAL RESTRAINTS NODE NRl NR 2 1 3
0 l
l l
NlJMl\lOH 01r PMOH!ilVlS 0 1r J<'Rl'l llJDOM I NUMIIMII OJI NOO/\L lll9!l'rltl\ I N'l'fl I I q1111I 111 11•111
183
Program NOMO for Normal-Mode Response
*** EXAMPLE 4.15: THREE-MEMBER PLANE TRUSS***
NEO = number of elements for output
TABLE 4.2
Soc. 4.10
ht• "''' •1•11r1rN1r.OI
Ml\'1'111 X IJl1t'OMl'O dr.11
NOF • NNH
3 3
184
Normal-Mode Method of Dynamic Analysis
TABLE 4.3
(Continued)
MODE 1 ANGULAR FREQUENCY 4.1995E+02 NODE DJl DJ2 1 2 . 3137E-Ol O.OOOOE+OO 2 l.OOOOE+OO -2 .4 722E-Ol 3 O.OOOOE+OO O.OO OOE+OO MODE 2 ANGULAR FREQUENCY NODE DJl 1 8, 6725E-Ol 2 -l.7149E-Ol 3 O.OOOOE+OO
1.1677E+03 DJ2 O.OOOOE+OO l.OOOOE+OO O.OOOOE+OO
MODE 3 ANGULAR FREQUENCY l.86i8E+03 NODE DJl DJ2 1 l.OOOOE+OO O.OOOOE+OO 2 -6.0504E-Ol -6 .10 68E-Ol 3 O.OOOOE+OO O.OOOOE+OO
*** LOADING NUMBER 1 OF 4 *** DYNAMIC PARAMETERS NTS OT DAMPR 20 l.OOOOE-03 O.OOOOE+OO INITIAL CONDITIONS NNID NNIV 2 0 INITIAL DISPLACEMENTS NODE 001 D02 1 l.OOOOE-01 O. OOOOE+OO 2 l.OOOOE-01 l .OOOOE-01 APPLIED ACTIONS NLN NEL 0 0 GROUND ACCELERATIONS IGA 0 NORMAL MODE SOLUTION NMODES 3 OUTPUT SELECTION IWR IPL NNO NEO 1 1 2 1 NODES:
1
ELEMENTS:
1
2
DISPLACEMENT TIME HISTORY FOR NODE 1 STEP TIMID DJ1 DJ2 0 O.OOOOE+OO l.000010 01 o.oooom•oo 1 1 . OOOOIC 03 6 , 70b lM Oil O.OOOOJlltOO 1. 1,000CJM O'I !, 'J. ., ltil~ () i CJ, 00QCJJ~ I QQ I I, OCJO(IIC !l I 1.uo11m 01 () • (l{l(l()lq I ()(1 4 4, OOOOIC O I I • !\Ii 1111f1l (I ,I o. 11110111q, no
185
Program NOMO for Normal-Mode Response
TABLE 4.3
5 6 7 8 9 10
5.0000E-03 6.0000E-03 7 . 0000E-03 8 . 0000E-03 9 . 0000E-03 1. OOOOE-02 l,lOOOE-02 l.2000E-02 l.3000E-02 l.4000E-02 l.SOOOE-02 l.6000E-02 l.7000E-02 l.SOOOE-02 1. 9000E-02 2.0000E-02
(Continued)
9.7853E-02 4.9458E-02 -7 . 4544E-02 -l.0603E-Ol -5.4252E-02 2.7924E-02 9.2962E-02 4.0849E-02 -8.1017E-02 -6 . 9040E-02 6.8279E-02 l.1979E-Ol 4 . 7138E-02 -3 . 6778E-02 -8.3692E-02 -5 , 2695E-02
O. OOOOE+OO O. OOOOE+OO O.OOOOE+OO O. OOOOE+OO O.OOOOE+OO O. OOOOE+OO O. OOOOE+OO O.OOOOE+OO O. OOOOE+OO O. OOOOE+OO O.OOOOE+OO O. OOOOE+OO O.OOOOE+OO O.OOOOE+OO O.OOOOE+OO O.OOOOE+OO
MAXIMUM TIME OF MAXIMUM MINIMUM TIME OF MI NI MUM
l.1979E-Ol l . 6000E-02 -l.0603E-Ol 8 . 0000E-03
O.OOOOE+OO 2.0000E-02 O.OOOOE+OO 2.0000E-02
DI SPLACEMENT TIME STEP TIME 0 O.OOOOE+OO 1 1.0000E-03 2 2.0000E-03 3 3.0000E-03 4 4 . 0000E-03 5 5.0000E-03 6 6.0000E-03 7 7.0000E-03 8 8 . 0000E-03 9 9.0000E-03 10 l . OOOOE-02 11 l.lOOOE-02 12 l.2000E-02 13 l.JOOOE-02 14 l.4000E-02 15 1. SOOOE-02 16 1. 6000E-02 17 l.7000E-02 18 1.SOOOE-02 19 l.9000E-02 20 2.0000E-02
HISTORY FOR DJl 1. OOOOE-01 8 . 5266E-02 7.2848E-02 6.0750E-02 -5.4078E-03 -8 . 4207E-02 -9 . 8212E-02 -8 . 5678E-02 -9.3358E-02 -8.2739E-02 -5 , 1067E-02 -2.9195E-02 l.9009E-02 9.6665E-02 l.2087E-Ol 8 . 872BE-02 7.6180E-02 7 . 2053E-02 3.4921E-02 -3 . 4530E-03 -3.8910E-02
MAXIMUM TIME OF MAXIMUM MINIMUM TIME OF MINIMUM
l.2087E-Ol l . 0992E-Ol l.4000E-02 l . lOOOE-02 -9.8212E-02 -l,1650E-Ol 6.0000E-03 l.9000E-02
11
12 13 14
15 16 17 18 19 20
MEMBER STEP 0 1 2 3 4
b h
NODE
2 DJ2 1. OOOOE-01 l.6390E-02 -l.0676E-0 1 -1. 0298E-Ol 3 . 5295E-03 1.0144E-Ol l.0809E-Ol 2.3188E-03 -9.5989E-02 -3.8360E-02 9 . 6986E-02 l.0992E-Ol -7,1293E-03 -l.0699E-Ol -l.0634E-Ol -l.2480E-02 8.6587E-02 5.6728E-02 -7 . 7855E-02 -l . 1650E-Ol 2 . 4600E-03
FORCE TIME HISTORY FOR ELEMENT TIME AMl O.OOOOE+OO 9.6000E+Ol 1. OOOOE-03 2 .8848E+Ol 2.0000E-03 -2.6 47 2E+O l 3.0000E- 03 l. 7503E+Ol 4.0000ID 03 1. O<,bbIC101 b. oooorn o 1 I, I'/0410 t OI I, , OOOOltl ll I • 'J. hht1 1 m, no
1
188
No111111J Mocln M11tl1111I 11l l>y1111111 lo An11lyel11
Program NOMO for Normal- Mode Response
TAJJLI~ 4.J (C'or11J11111•dJ 7 8 9 10 11 12 13 14 15 16 17 18 19 20
7 , 0000E - 03 -5 .790 21!l+ OO B.OOOOE-03 -8. 3 0 2 7E+Ol 9,0000E-03 -5 . 7337E +Ol l.OOOOE-02 3.6233E+Ol l . lOOOE-'02 l . 7570E+Ol l.2000E-02 -2.2569E+Ol l . 3000E-02 2.5220E+Ol l.4000E-02 3.4642E+Ol l. 5000E-02 2 . 7426E+OO l.6000E-02 5,l 722E+Ol l.7000E-02 7.2398E+Ol l.BOOOE-02 -2. 3117E+Ol l. 9000E-02 -S.4068E+Ol 2.0000E-02 l . 2287E+Ol
MAXIMUM TIME OF MAXIMUM MINIMUM TIME OF MINIMUM
***
9 , 6000E+Ol O.OOOOE+OO -B.3027E+Ol 8.0000E-03
LOADING NUMBER
2 OF 4 *** DYNAMIC PARAMETERS NTS OT DAMPR 20 l , OOOOE-03 O. OOOOE+OO
INITIAL CONDITIONS NNIO NNIV
0
0
APPLIED ACTIONS NLN NEL l 0 NODAL LOADS NODE AJl 2 2.0000E+Ol
AJ2 0.0000E+OO
GROUND ACCELERATIONS IGA 0
FORCING FUNCTION NFO 7
FUNCTION ORDINATES K TIME 1 0 . 0000E+OO 2 l . OOOOE - 03 3 3.0000E- 0 3 4 4.0000E-03 5 5. 0000E-03 6 7.0000E- 0 3 7 B.OOOOE- 0 3
FACTOR O. OOOOE+OO 1. 5000E-Ol 8 . 5000E- Ol 1. OOOOE+OO 8 , 5000E- Ol l . 5000E- Ol 0.0000E+ OO NORMAL MODE SOLUTION NMODES • 3 OUTPUT SMl ,11CTl ON I WU I Pl, NNO Nl!O I I I I
187
TABLE 4.3 (Continued) NODES :
2
ELEMENTS:
1
DISPLACEMENT TIME STEP TIME 0 O.OOOOE+OO 1 1.0000E-03 2 2.0000E-03 3 3.0000E-03 4 4.0000E-03 5 5 , 0000E-03 6 6.0000E-03 7 7 . 0000E-03 8 8 . 0000E-03 9 9 . 0000E-03 10 1. OOOOE- 02 11 l.lOOOE-02 12 l.2000E-02 13 l . 3000E-02 14 1.4000E-02 15 l.5000E-02 16 l . 6000E-02 17 l.7000E-02 18 1.BOOOE-02 19 1.9000E-02 20 2.0000E-02
HISTORY FOR DJl O.OOOOE+OO 5.3221E-04 4.4538E-03 l,5965E-02 3.7705E-02 6 . 4468E-02 8.7960E-02 9 , 9425E-02 9.4028E-02 7 , 1941E-02 3.7520E-02 -i . 2319E-03 -4 . 4130E-02 -7 . 7419E-02 - 9.6296E-02 -9 . 8483E-02 -8.4354E-02 -5 . 5403E-02 -l.6520E-02 2.4923E-02 6 . 2330E-02
MAXIMUM TIME OF MAXIMUM MINIMUM TIME OF MINIMUM
9 . 9425E-0 2 2 . 4408E-02 7 . 0000E-03 l . 4000E-02 -9 . 8483E-02 -2 . 3223E-02 l . SOOOE-02 7.0000E-03
NODE
2 DJ2 O.OOOOE+OO -2,0013E-05 -5.1199E-04 -3.0147E-03 -8.9069E-03 -l.6887E-02 -2.2416E-02 -2 . 3223E-02 -2 . 1405E-02 -l,8146E-02 -l.1247E-02 -2 . 5762E-05 l . 1875E-02 2.0566E-02 2 . 4408E-02 2 . 3290E-02 l . 9033E-02 1. 3650E_-p2 6 . 0714E-03 -4.9980E-03 -l . 6433E-02
MEMBER FORCE TIME ·HISTORY FOR ELEMENT STEP TIME AMl 0 O. OOOOE+OO O. OOOOE+OO 1 l.OOOOE-03 4.8184E-01 2 2 . 0000E-03 3 . 2844E+OO 3 3.0000E-03 9 . 0341E+OO 4 4 . 0000E-03 l . 647BE+Ol 5 5 . 0000E-03 2.1947E+Ol 6 6 . 0000E-03 2 .7 113E+Ol 7 7 . 0000E-03 3 . 1667E+Ol 8 8 . 0000E-03 3 . 08l4E+Ol 9 9 . 0000E-03 2 . 2442E+Ol 1 0 l . OOOOE-02 l.0842E+Ol 11 l.lOOOE-02 -7.8519E-01 12 l .2 000E-02 - l . 3479E+Ol 1 3 l .3000E-02 -2 . 4655E+Ol 14 l . 4000E-0 2 - 3. 0138E+Ol 1 5 l .5000E - 02 -3 . 0917E+Ol 16 l. 6 000E-0 2 -2. 7 5 91E+Ol 17 l .7 000E-02 - l .77 88 E+O l 1 8 l. BOOOE - 0 2 - 4 . 09 57 E+O O 19 l.9 000 E- 0 2 8 .l366E+OO 2 0 2. ooo om- 02 1. oee1m+o1 MMC IMUM 'I' I MIQ 0 1~ MAX I MUM MINIMUM 'l' IMI~ OI' MINIMIIM
l, l (1t1 7 1QI 01 I, 00 0 01~ 0~ I. II 111 /If, I O I I . hfllllll~ ll ~
1
188
Normal-Mode Method of Dynamic Analysis
Chap.4
self-explanatory. Note that the first page of the listing repeats the vibrational analysis given previously in Table 3 .4. For each load case we use number of time steps NTS = 20 and duration of time steps DT = 1 ms. Partial results for the four load cases are also plotted (by computer) in Figs. 4. 11 and 4. 12. Figure 4.11 shows that the initial displacements excite all three modes of vibration. However, the responses in Fig. 4. l 2(b) due to the applied force in Fig. 4.12(a) demonstrate mostly first-mode contributions.
20
-"
0.
-
w
O
0 D
0.10
z ......
o.os
-,
5
15
10
20
T
(ms)
<: -I 0
C
~ z
10
(\J
DAMPR-o.o -
189
Program NOMO for Normal-Mode Respon se
Sec. 4.10
<:
-20
o.oo
D
,_ -o.os
(a)
<:
, - 0 . 10 0
DAMPR-0-0
(a)
0. 10
DAMP'<=O.O -
0 . 10
.: o.os
C
o.os (\J
~ 0 z
~
o.oo
o.oo
0
z ......
..... -o .os <:
<:
-o.os
-, 0
-0. 10
( b) ( b)
DAMPR-o.o -
0 . 10
Figure 4.12 Three -member plane truss: (a) applied force; (b) responses.
C
- o.os (\J
~
11.,mmplc 4.16
o.oo
z
I
<
N ~
• 0 . 05 0 , 10
·
1
. NOMOPT let us reconsider the symmetnc p ane AN n ~ccond example usmg Programs). For this.structure dynamic responses due to the 1111KN trom Sec . 3.8 (see Example 3. lnlluwing innucnccs arc desired:
0
-
Stcp force or magnitude I 00 kN appI'icd ·m the y direction at node 3, with
,
I >AMPR O.O · · with DAMPR = 0 0 Jlk·t·t·wi1H: li1w11r ~round nccl.llurntlon In the Y direction , . S111111· IIH \' llH~· ) , h11t with ))t\M l'I{ (), I
190
Normal-Mode Method of Dynamic Analysis
Chap.4
For these cases we use the values of E, p, L, and A stated in Example 3.5 , where the material is aluminum and units are SI. To take advantage of symmetry, we must decompose the load for case 1 into symmetric and antisymmetric components and then analyze half the structure twice. Thus, half the force must be applied using symmetric restraints on the plane of symmetry; nnd the other half of the force must be applied with antisymmetric restraints. Of course, the results of these two analyses must be added to find the total solution for the left-hand pa rt of the truss. Figure 4 .13 consists of computer plots showing the total step force of 100 kN in purl (a), symmetric responses to half the force in part (b), and antisymmetric responses
z 100
Program NOMO for Normal-Mode Response
Sec. 4.10
to half the force in part (c). For each load case the number of time steps NTS "'.' 40, and lhc duration of time steps DT = 5 ms. The responses in Fig. 4.13(~) consist of the 1,.translation at node 6 for all symmetric modes as well as truncation to mode. l . Similarly, Fig. 4.13(c) shows the responses from all antisymmetric modes ~nd truncat10n lo mode 2 (for the same nodal translation). We see that the first (symmetric) and second (untisymmetric) modes produce the major con~ributi_ons _to the responses. . . Figure 4.14(a) is a computer plot of the p1ecew1se~h~ear ~round acceleration m the 1, direction. Because this influence induces symmetnc mertial loads, we_ need only ,•nnsider symmetric responses of the structure. Time histories ~f the resultmg y-transliuion at node 10 appear in Fig. 4.14(b), with and without dampmg. Evidently, the first 11 1ode is the primary contributor to both responses.
80 ,.i.,,
w
60
Z
40
0 0
z o.so
0
< 20
I-
<
N
et:
~ 0
so
100
ISO
200
T
(a)
(ms
J
w o.oo .....J w (.) (.)
< 0
ALL MODES
ee
I. 00
'E
I-
-o.so
z
=i
0
4
et: <:!>
-1 . 00
-
I
>-
3
w
(a)
§
2
'<
I
z
N
~ 0 200
T
1
.oo
(ms)
e
..=
0. so
0
::!5
0. 00
0 7
I •
..r - 0 ,50 ('J
-.,
N
n
~ 0
1 00
l1'l111m• 4.11
(b)
l'l1111r IIIINN (II) 11ppllt•d lor n •: (h)
hy1111111•1rl1 111~p1111 \ 1 -
NY lllllll•ltl1
191
IIINjlOII Nl'N; (t' ) 111111
1!'11111111 4, 14
1'111111" IIII NN 111) ~11 1111111 111 l't h•111lio11; (h) ll'NJ1UII Nt'N
192
Normal-Mode Method of Dynamic Analysis
Chap.4
REFERENCES I. Tim?she~ko, S. P. , Young, D. H. , and Weaver, W. , Jr. , Vibration Problems in Engineering, 4th ed., Wiley, New York, 1974 . 2 · Weaver, W., Jr., and Gere, J. M. , Matrix Analysis of Framed Structures 2nd ed ' ., Yan Nostrand Reinhold , New York, 1980.
3. Rayleigh , J. W. S., Theory of Sound, Vol. I, Dover, New York , 1945. 4 · ;ea~er, W., Jr., and Johnston, P. R. , Finite Elements for Structural Analysis rent1ce-Hall , Englewood Cliffs, N.J. , 1984. ' 5. Weaver'. W. , Jr., "Dynamics of Discrete-Parameter Structures," in Developments in Theoretical and Applied Mechanics Vol 2 ed w A Sh p p York, 1965, pp. 629- 651. ' . ' . . . aw, crgamon ress, New 6. Caughy , T· K ·, "Class·1caI N ormaJ M odes m · Damped Linear Systems " J A / ; ech., Yo!. 27, 1960, pp. 269-271; also Vol. 32, 1965, pp. 583-588'. · PP·
?ss, K. A., '_'Coordinates which Uncouple the Equations of Motion of D d Lm~ar Dynamic Systems," J. Appl. Mech., Vol. 25, 1958, pp. 361-364. ampe
7.
8. Loui~,. J. J. ~- , "~amping in Structures-a Review," Engineer thesis, Department of Civil Engmeenng, Stanford University, June 1976.
PROBLEMS 4.J-1.
4.J-2.
~~sumc that the pla~e truss_ i~ H g. 3.12(a). is at rest when joint 2 is sudden! ~t1 uck_s~ ~at 1t acquires an m1tial velocity D 0 1 in the x direction. Determine th~ '.rcc-v1biat1onal ~esponsc of the structure due to this impact. The properties angular frequencies, and mode shapes for this problem are all given in Exam~ pie 3. 1. Calc_ulate the free-vibration~! response of the cantilever beam in Fig. 3.13(a) to t~c sudden release of a static force Po in the y direction at node 2 Exa I 3 2 ~1vcs the properties, angular frequencies , and mode shapes for this pr:e~. · l·o:· t~e pla~c truss in Pro?·. 3.6-2, find the response caused by initial displace, mcnts Doi - D 02 = d at Jomt 1.
4 ..1-4. Obtain the response of the plane truss in Prob 3 6-3 t · ·· l I · · · the y direction at joint I. . . o an m1t1a ve oc1ty Do2 in
4..1-5.
J,'t t~corcc~'.anePc, m_trussthe innegative Pro?. 3.6-~, determine the response to the sudden release o direction at joint I.
4..1-6.
Su1~~'.l~~ that ~ode 2 of the beam in Prob. 3.6-5 has an initial velocity Doi in tlw v d11 cct1on. Find the response of the structure due to this influence.
y
,t
4..1-7. Let the bca'.11 in Prob. 3 .6-6 initially have a small positive rotation < at node 2 and a negative rotation O,o at node 3. Calculate the response of the s(/ucturc th1tl
o
,·esults.
4••1-H. AsN111ne. that
1111
initial posi1iv1; mClmcnt M O ut node I of the beum in Proh ..U, 1 Oh1111n till' l l'hJ)llllNl' o l till' ~II Ul lllll' l'ltll\l'd h y this l'Oll
Ih h
lion
Chap. 4
Problems
193
4.3-9. If joint 1 of the symmetric plane truss in Prob. 3.7-7 has an initial velocity Doi in the x direction, determine the antisymmetric response of the structure. 4.4-1. (a) Calculate the response of the plane truss in Fig. 3.12(a) caused by a step force of magnitude P = Pi applied in the x direction at joint 1. (b) Confirm the reciprocal theorem for dynamic loads by obtaining the response due to the same fo rcing function applied in the y direction at joint 1. Example 3. l gives the properties, angular frequencies, and mode shapes for this truss. 4.4-2. Assume that a hannonically-varying force P = Pi sin Ot acts in the y direction at node 2 of the cantilever beam in Fig. 3. l 3(a). (a) Find the steady-state response at the free end of the beam due to this loading. (b) Confirm the reciprocal theorem for dynamic loads by calculating the response caused by a moment M = M 2 sin Ot in the z sense at node 2, where M 2 is numerically equal to Pi. The properties, angular frequencies and mode shapes for this problem are all given in Example 3.2. 4.4-3. For the plane truss in Prob. 3.6-2, determine the response caused by a ramp force P = Pit/ ti applied in the negative y direction at joint 1. 4.4-4. Obtain the response of the plane truss in Prob. 3.6-3 to a step force of magnitude P = Pi applied in the y direction at the middle of element 1. 4.4-5. Assume that a hannonically-varying force P = P, cos Ot is applied in the x direction at joint l of the plane truss in Prob. 3.6-4. Calculate the steady-state response of the structure due to this influence. 4.4-6. Apply a step moment of magnitude M = M 2 in the z sense at node 2 of the beam in Prob. 3.6-5, and find the response of the structure. 4.4-7. Suppose that a hannonically-varying force P = Pi sin Ot is applied in the y direction at the midpoint of element 2 in Prob. 3.6-6. Determine the steady-state nodal responses of the beam caused by this loading. •1.4-8. Let the beam in Prob. 3 .6-7 be subjected to a ramp moment M = M2t/t1 in the z sense at node 2, and calculate the response of the structure. •t4-9. For the symmetric plane truss in Prob. 3.7-7, obtain the antisymmetric response to a ramp force P = Pit/ t I applied in the negative x direction at joint 1.
4,.!l• 1. For the plane truss in Fig. 4. 7, calculate the relative response of joint 1 to a rigid-body ramp acceleration Dg1 = a 1t/ t1 of ground in the x direction. ,t ~-2. Considering the cantilever beam in Fig. 4.8, let the ground at point 1 have a sudden translation D 82 = din the y direction. Find the response at node 2 due to this influence. ,I. ~.J. Suppose that the ground at point 3 in Prob. 3.6-2 has a small rigid-body step rotation DR6 = 8, . Determine the response of joint 1 in the plane truss caused by thi s motion. ,I,~ 4. For the plane truss in Prob. 3.6-3, evaluate the relative steady-state response at joint I due to a rigid-body ground acceleration Dg2 = a cos Ot in they direction. ,I,~-~. Ass11n1t· th11t the support ut point 3 of the plane truss in Prob. 3.6-4 translates 111dl'Pl'lldl•n1ly in tlw y dircl·tion lll'l'mdin8 to the fun ction D R3 = d sin Ot. Calculnh· thr 1(·~11ltlnµ Ntl•1tdy st11h• H'~p1111s1• 111 joint I.
184
N111111111
Mu 111 M11ll111d "' I>v1111111lc. A1111ly11h1
Chnp
5
4.!i ''· fl lht• ~round in Prob I c , 1 (} in lhc ncgai ivc z ,sc."1s',, ,/;'.~1i':11~1;11i:: "t'1111ll 1l8fd•hody step rotulion " •n •t ~-7. For the continuous b, . ' 111 llu iw1pons0 of the beam ut 11od1• cam 111 l'roh I c, c, f 1 , 1 .. relative steady ~1111 rcspons0 due to a rigid-body , t I' I • i ti 1111111.(lt.l tie node 3. I() U 10 1111 IIC '('t'f11111lio11 l)N(> = {), Sin fit Of grOUlltf Ill
4.~•tt Let lhe support at point 3 in Prob 3 6 7 h OH1 0, si n Ot in the z sense. Calc~lat~ t~e ave a small harmonic rotu11t111 and 2 of the beam caused by thi . fl steady-state responses at nod1·- I s m uence.
Direct Numerical Integration Methods
11.1 INTRODUCTION 111 this chapter we shall discuss various numerical integration methods for calculnling dynamic responses of structures. These techniques are usually more c•I fl cient than the normal-mode method, unless modal truncation is used. Thus, 1•i1hcr approach involves approximations of different types. While the normal11wdc method applies only to linearly elastic structures, direct numerical integralIon can be used for either linear or nonlinear systems. Any method for direct numerical integration of second-order differential l'quations of motion may be visualized as some type of finite-difference formul"tion. Although we use finite elements for discretization in space, we find that the discretization in time is more conveniently handled by finite differences. This upproach is most apparent in the next section, where a second central difference Im rnula is converted to an explicit extrapolator for dynamic response . Although II is not always evident, the other approximation formulas used in subsequent Nl'Clions are also various types of finite-difference expressions. For a SDOF analytical model, the damped equation of motion is
mu + cu + ku = P (t)
(1 )
in which the terms were all defined previously in Chapter 2. On the other hand, the damped equations of motion for a MDOF structure have the matrix form
Mfi +c 6 + SD = A00
ill
11H
196
10/
f>
1111 1111 N11111n11t nl ltllM11fnllo11 M111hod11
1U8
ti111c 1 110111 pH·viow, rcsponscH by 11si11g 11pprox.imulion formu las. 11 1 ,1il1111H111uthmls for doing this tll'i.) tk:sct'ib1.:d in thll fo llowing sections for both
I 111111Hl' 11!
11t SI t11m· 0. Fmn1 II >OFlo system as these qunnlllll II w, 11111v, 11h 11l11tl• lhc luitiul act:clllrnli1111 l11
111 ,1111nd ..
uo
J (
= -;
/{l
kllu
,·,io)
MDOF problems .
I I)
:l • XTRAPOLATION WITH EXPLICIT FORMULAS
is obtained from Eq . (1) at f ime to. Similarly' . (wltil'h l ) yilllds for a MDOF structure ' I q
Do= M-1(Ao - S Do - C Do) ( I1 If till' Vl' i.' lor.muss matrix Mis singular ' the vec tor OO .m Eq. (4) may be taken as u 111111 . i•igure 5. 1 shows a graph of the nu . 1 S l)OF system. This plot is represent d menca solution for the response ol n Ihough it may actually have slight ast. a s.n:iooth curve in the u-t plane, i.'Vl II
d.:
i con mu1t1es. The symbols Uo denote values of u at the time stations t t t ' u1' u2, . . . /11 1, . . . ' and so on . The time interval !:,,.t b . o, 1, 2, . . . 'lJ ,, r,. t11kl.ln to be of uniform duration t:,,.r altho jh etween tim~s !j ~d tr: 1 is usuul lv HI ii y. 'l'hll objective of the '. 1 . ug su~h a specialization is not ne,·1•" numenca integration process is to calculate tlu
"1
i,
"1, UJ' i,
11 1•xplicil 111111• 1 in
extrapolation procedure consists of expressing the displacement at terms of the displacements, velocities, or accelerations at previous 11 1 111111' stotions. Referring to Fig. 5.1 , we can write the second central difference 111 llll' displacement at time ti as the acceleration: 1
Uj = (At)/ui- 1 - 2uj
+
(1) Ui+ 1)
hklt implies a SDOF system with a unifonn time step At. Similarly, the
...
1
l11l'ily at time
tj
may be approximated as .
Uj
=
2
1 !:,,.t (ui+l -
(2) Uj-1)
',11hstituting iij and Uj from Eqs. (1) and (2) into Eq. (5 .1-1) and rearranging the /I
11
~ult produces
",, , P'(t;)
t'
(3)
~ P(I;) - G- !~}, - L(:,)' - i'i.},-,
(4)
l1q11L1tion (3) is known as the central-difference predictor, which can be applied 1rpctitively to obtain ui+ 1 for each time step . Then the acceleration ili and the w locity ui at time ti may also be found using Eqs. (1) and (2) if desired. The central-difference predictor in Eq. (3) is a two-step formula, so it 1 1111not be applied directly in the first time step. In order to derive a starting 111 ocedure, we apply Eqs. (1) and (2) at time t0 and solve for U - i , as follows: U- 1
= uo - uo Lit + ! iio(At) 2
(5)
l'his result can be used in Eq. (4) to evaluate P*(t1) for the first time step. For a MDOF structure, the expressions given above can be generalized in 11 1nulrix format. Thus, Eqs. (1) through (4) become ()
t,
..
t,
11'(11111•1• ~. I
1
D1 ""' (iit) 2 (Di- , - 2Di + Di+1)
S11l111lt111 hy tllil'l'I 111111 111111111 l11h' f''11 1l1111
lllr!I I N111111111t nl lt1h1111ntlon Muthoda
198
I(~r)2 M A*(r1) =
111 th is exu111ple we c 1111 cxp1li1111 tlw t
2 ~( cJ01, 1 • A+(I/)
f
1
l(A
A(tJ - [s - (;t)2 MJ01
)2 1
'
M - 2~rc]Dj,
D- , = Do - Do At + 2 Do(At)
D;,i+ I
=
"'
(i
M; A;' (ti)
=
=2 =~ w,,
= }5_ = w2
k(.!_) 21r
(a)
o, so the initial acceleration is Po
P,
m
m
(b)
iio=-=-
1, 2, ... , n)
- ) with R =Pi.To start the solution, we apply Eq. (5) hh h Is drawn from Eq. (5 . 1 3 o 1o, 11h111in 2 2 2 (c) 1 Pi ( T ) _ Pi (21r) = O 04935~ U - 1 = 2iio(A.t)2 = 2m 20 - 2k T 20 . k
(!_)
. . Eq ( ) ·ves the displacement at time 1111 II the central-difference predictor m . 3 g1
which is simply a multiplication of the effective action Af (ti) by the scul111 (At) 2/ M;. Therefore, high-speed computer core storage need only contain info, mation for one displacement (or more conveniently, one node) at a time. The central-difference predictor is probably the most widely used explkll formula for solving structural dynamics problems. However, all expressions ol this type have a critical time ste[l.., above which the solution becomes numerically unstable and diverges [l]. Nevertheless, among all of the known second-order pied ictors, the central-difference method has the largest stable time step [2]. The value of the critical time step for this technique is
(At)c,
_ Li 0 , we have uo - o -
2
which may be used to start the procedure. If there is no damping and the n111 NN matrix is diagonal , a so-called nodewise solution may be devised. That is, lho solution for a particular displacement D;,i+ I in Eq. (8) becomes
(At)2
. , r ti , sti t'fness k and the period f\l1llls o 1e 2
/I.I 111111• 10 l ..
111111111111 111
I
m
In addition , the generalized form of Eq. (5) is
199
I x 1111pol11tlu11 wllh I xplh It I 111111111111
Chnp
2
u, =
(t:,.: ~1 -
ltxuma,ll• 5.1
as
U -I
(d)
. . Eq (c) A second application of the predictor at time t2 yields "·hkh 1s the same as U- 1 m . .
u2 =
C~
2f, -[k - c: 2Ju, - o}
= [ (; )2(2;)2o
(12)
SuppoNl' thut an undnmpcd SDOF linear system is suhj1.·1.·t1.•<111> 11 step force of mugnit11<1,· /'1, Ht111 ll11g fw111 n•st. l.1·1 us rnkulute the 11pp1oxi 111111l• 11•sponse using lhl' l'1.•111rnl 1lllh·H·m·,· p11•dii101 with 1w,•11ty unifrnm 1i11w ~h·pN 111 d111111in11 t\/ '/'/'20
=
Pi P1 = (0.09870 - 0.04935)k = 0.04935k
0
7T
The symbol w,, in this expression denotes the largest angular frequency in the analytical model, and T,, is the smallest period. Key and Beisinger [3] applied the central-difference predictor to the linem dynamic analysis of thin shells. In addition, Krieg and Key [4] showed that us in~ 11 diagonal mass matrix improves the accuracy of the procedure. Successful use of this method in nonlinear analysis has also been reported by Key [5], who USl'd 111ti!lcial damping to control the inherent instability. Morino et al. [61, searchl•d Im the optimal predictor for systems of second-order differential equations and 1.·onl'l uded that the central-difference formula is best. The main disadvantage ol this method is that for a fine network of e lements a very small time step i11 11.· qu i1c
0- (:)2U-1] (;0)2(2;)2 ~ -
ti
=
_ 0.04935) + c2)(0.04935)]~ (e)
P, 0 . 1925-k
. (3) is a lied in the same manner, and Eqs. (1) and lu i•uch subsequent time step, Eq. p~ I .. 11 I d to find the accelerations and ve oc1t1es as we . t 'J rnuld be emp .oye f b . db this procedure are listed in Table 5.1, The approximate values o u! .o ta1~e y ws lots of the approximate results ,1l11ng with the exact values. In add1.t1on, ~1g. i'2 ~~ 1 PFor the scale used in Fig. 5.2, 1111 both zero damping and a dampmg ratio o r. .. 1111· plotted curves are indistinguishable from exact responses.
I' uauplc 5· 2
. · Exam le 3 1 Recall that the l'luurc 5.3(a) shows the plane truss analyzed prev1ous~\m O 8A ~nd Now we shall ' i 11Ns sectional areas of members l an~ , 2 tare t~!u:o ;e force P (t), applied in the x ,h1fl11ll1i1te the undamped response of this s rue . . F' 5 3(b) Let us 1 1llH'l'llOII Ill jnint I. The time histordy ~; th~.d~n:::~ ~~~~~1~Xrr~=:~~e ~~; di~tor ~ith 20 , 11k11l11tr the displacements I), an i IISlllK , . , t:,. 1 '/'1/ J..O st111 tlllij lrnm rest. 1 1111ltoi 111 111111· Ht0ps ol durnt on . . ' ll'l'H for the free displacements at 1•111111 E x11111pll• \ I llw st1llm•ss 11111 1 11111N~ II 1111 i
A..
200
Direct Numerica l Integration Methods
Chap. 8
oc.
5.2
Extra polation w ith Explicit Formulas
TABLE 5.1 Response for Example 5.1 Using Central-Difference Predictor" y
j
Approx. u
Exact u
j
Approx. u
Exact u
1 2 3 4 5 6 7 8 9 JO
0.04935 0.1925 0.4154 0.6960 1.007 1.317 1.595 1.815 1.955 2.000
0.04894 0.1910 0.4 122 0.69 10 1.000 1.309 1.588 1.809 J.951 2.000
11 12 13 14 15 16 17 18 19 20
1.947 1.800 1.574 1.292 0.9804 0.6712 0. 3944 0.1774 0.04 157 0.00034
1.951 1.809 1.588 1.309 1.000 0.6910 0.4 122 0. 1910 0.04894 0
P(t)
"Tabulated values to be multiplied by P1/ k.
u
i------
-
- -0.6L (a)
p
P,
I' I k
1.----------,
Ol__---- - -- -+:-----~--Tt.t2:o-- t,o / ' P, (b)
()
I
Fi~urc 5 .3 I
2
(a) Plane trnss; (b) applied force.
201
202
Direct Numerical Integration Methods
Chap. 5
Sec. 5.3
TABLE 5.2 Responses for Example 5.2 Using Central-Difference Predictor•
joint 1 are
S=
s[ - 0.36 0.48
M = m[ol 01]
-0.48] 1.64
(f)
in which s = EA/Land m = 3.28 pAL/6. In addition, we have
wf
=
0.2 .:_ m
w~ = 1.8 .:.
(g)
m
Using the first expression in Eqs. (g), we can relate the mass constant m to the stiffness ' constant s and the fundamental period 7; , as follows: 2
m = 0.2s ( 7i71")
(h)
2
~.l tim~ to = 0 the initial displace~ents an~ :~locities are (D 1) 0 = (D2) 0 = (D 1) 0 = ( >2)0 0. Therefore, Eq. (5.1-4) gives the m1t1al accelerations as
Do=
M - 'Ao = ;[~
UJ ~
~][~] =
(i)
Before beginning the step-by-step procedure, we apply Eq. (10) to find D- 1 =
=
I2 Do(At)
2
=
I[ ~ (1i)
2
1 ] 2 0 m 20
~ [~] /;s (2;)2(~~)2 =
[ o.2;67]
D, = (~ =
[~
~]{[~]
-
(:;)2 [~
{[o.~34] _ [ o.2;67]}
9
=
(j) ti
produces
~Jn-J
2
[ o.2;67]
9
(kl
2
[1 O]{[P'] O 1 0
_ s[-0.48 0.36
.. (2!.20.) _!!_ ( 271") [1 - 0.08883] 0 .2s Ti O + 0.1184 2
2
J
o] }
-0.48] ~ [1 1.64 D, + (Llt) 2 o I D,
+ 2[0 .2467] ~ o
l
0.4496 -I [0.4935]} ~ = ro.9431 p 1 , 0.05845 o s Io.os845 -; {'
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
Approx. D1
0.2467 0.943 1 1.979 3.227 4.574 5.928 7.188 8.223 8.898 9.107 7.839 5.330 1.952 - 1.929 -6.024 -10.06 -13.70 - 16.50 - 18.07 -18. 15
Exact D ,
Approx. D2
0.2432 0.9322 1.963 3.210 4.556 5.901 7.148 8.179 8.868 9.111 8.382 6.315 3.221 - 0.51 87 - 4.556 - 8.592 -12.33 - 15.43 -17.49 -18.22
0 0.05845 0.2930 0.7593 1.375 1.962 2.365 2.557 2.627 2.679 2.720 2.416 1.420 -0.2636 - 2.191 - 3.772 -4.683 -5.049 - 5.238 -5.467
Exact D2
0.0047 1 0.06831 0.2932 0.7350 1.333 1.932 2.374 2.598 2.662 2.667 2.653 2.462 1.787 0.4618 -1.333 - 3.128 -4.454 -5 .128 -5.319 - 5.333
exact values are also listed. Figure 5.4 depicts plots of these responses for both the 11pproximate and the exact analyses. Here we see that the curves for the approximate responses deviate slightly from those for the exact responses.
6.3 ITERATION WITH IMPLICIT FORMULAS
The technique to be explained in this section is called the predictor-corrector method [6, 7]. In each time step an explicit formula (a predictor) is first used to estimate the response at the end of the step. This is followed by one or more applications of an implicit formula (a corrector) to improve the results. Although such an iterative procedure is not required for linear analysis, it is included here l'o r use in nonlinear problems, where physical properties can change in each
which is lhc same as D- , in Eq. (j). Applying Eq. (8) again at time t2 yields
(Llt) m
j
• Tabulated values to be multiplied by P, / s.
9
lJsl ng this result in the matrix form of the predictor [Eq. (8)] at time
02_
203
Iteration w ith Implicit Formulas
s
({')
In .suhsl·~ru.e nt ~·n lculutions Wl' upply Eq . (8) rcpctili vllly 10 find rhc response at cnch ol ihl tl111111111111g 11ml' st111lo11s. Noll' ih nl lhc 11111ss n1111il x 111 thi s l'x11111pll• is diugonnl so 11q (K) dm•s not 1t•q11i1 t• sol vl11i.: sl 11111l11111l·t111 s l'q1 111th111s •' , · 1<1·~1111~ 1111111 lhl~ 11pp1 wd111111t• 111111lys is l111 /l 111111I /l , 111r 11 rv,•11 111 'f'nhk' 'i l, wlwtl'
eycle of iteratio n. By an approach that we shall refer to as the average-acceleration method, lhc veloc ity u11 1 for a SDOF system at time tj+ t is approximated as (1)
in whic h 1i1 is the ve lt)cily Ht the preceding time station 11 (see Fig. 5 .1). This lrn nnlla , know n lo nu,rn·ri rnl 111111lysts as till' t1'l'zoidul rttfr, implies that the
205 204
Direct Numerical Integration Methods
Chap.
6
Sec. 5.3
Iteration with Implicit Formulas
acceleration in the stepJ s taken to be the average of ii1 and ii1+,. Similarly , the displacement u + at the end of the step is approximated by the trapezoidal rule
D,
1
10 ~s
1
with the expression
Approx. Exact
(2)
where the velocity in the step is taken to be the average of u1 and u1+1· Substitution of Eq. (1) into Eq. (2) yields
5~ s
(3)
When applying this method , we do not use Eq. (3) directly; but Eqs. (1) and (2) are used in succession. Because the value of ii1+ 1 is not known in advance, the approximation is said to be implicit, so the solution must be iterative within each step. The following recurrence equations represent the ith iteration of the Jth step: - 5~ s
10 ~s
(i > 1)
(4)
= R1 + Hu1+i); At1
(i 2: 1)
(5)
l = -(~+1
(i 2: 1)
(6)
(u1+ 1);
= Q1 +
(u1+1); (ii1+1);
m
Hu1+1);- 1At1
- ku1+1 - cu1+1);
where
(7)
15 ~s
and R1 = 20 "1 s (a)
Uj
+ ! Uj
(8)
At1
This iterative procedure is not self-starting because it requires a supplementary formula for determining the first estimate of u1+ 1 in each time step. After evaluating the initial acceleration from Eq. (5.1 -3) , we may start the iteration for the first step by approximating 1 with Euler's extrapolation formula, as follows:
u
(j
=
0; i
= 1)
(9)
Then the first approximations for u 1 and ii 1 are obtained from Eqs. (5) and (6). All subsequent iterations for the first time step involve the repetitive use of Eqs. (4), (5), and (6). To start the iteration in the Jth time step, we may again apply Euler's formula to determine a first estimate of
u1+ 1 as (i = 1)
fl'l1411n1~.•I (h) /l ,
(10)
Hoth Eqs. (9) and ( I 0) imply constant values of the accelerations within the steps . To improve the accuracy of the rcsu lts fo r the first iteration of the jth step, we l:HI\ 11se till' slil'htl y IIHll'l' cl,,borntc fo1 mul11 thul is valid only for uniform time
206
Direct Numerical Integration Methods
111
6.3
207
Iteration with Implicit I ormul tJII
I q1111tion (16) is the same as Eq_. ( l) of the aver.age-acceleration method, but Eq.
steps:
1 l /) is slightly different_from its clount~erpart
(i = 1) This expression spans two equal time steps from tj- I to tj+i (see Fig. 5.1) and utilizes the midpoint acceleration at time tj. Equations ( 10) and ( 11) are called explicit predictors because they provido estimates of Uj+I in terms of previous values of u and u. On the other hand, Eq (1) is referred to as an implicit corrector that yields an improved value of u11 1 after an estimation of uj+I has been obtained. The method described here in volves one application of a predictor, followed by repetitive applications of thl' corrector. An iterative type of solution requires some criterion for stopping or chani,t ing the step size, such as a limit on the number of iterations. A convenient method for measuring the rate of convergence is to control the number of significant figures in uj+ 1, as follows:
I(uj+1); -
(uj+1);-1 I <
(12)
where € 11 is some small number selected by the analyst. For example, an accuracy of approximately four digits may be specified by talcing €,, = 0.0001. That lewl of accuracy is used in the numerical examples of this section. For a MDOI• structure, we use the length of a translational or rotational vector in Eq. (12), which is equal to the square root of the sum of the squares of its components . Another implicit approach for approximating responses is known as tht• fl11ear-acceleration method. As indicated by its name, this technique has thl' ussumption that the acceleration varies linearly within each time step. Thus , nn expression for a during the step f:.tj may be written a (t ')
=
aj
+
(iij+ 1 -
uj) f:.t.
. + Ujt .. + (.,Uj+ I
= U;
I
2
(u1+1);
= Rj + 'i
R
*=
J
Uj
+ ("Uj+ 1
1i1 I
1Ui1
(20) •
. the recurrence equatio.ns for F MDOF structures, we can genera1ize . II• rntio~rinto matrix formats. Considering first the average-acceleration method, 1w replace Eqs. (4) through (8) by (21) (i > 1) (Dj+1); = Q1 + Hi>1+1);- 1 D.tj (22) (i ~ 1) (D1+1); = Rj + HDj+l)i D.tj . (23) (i ~ 1) (D1+1); = M - l(Aj+I - s Dj+l - C D1+1); Wlll'l'C
(24)
I\IHo, Eqs. (9), (10), and (11) are supplanted by the matrix expressions (D1)1 = Do + Do f:.to • o" A (D;+1)1 = D1 + i utj
J
-
") (t')J f:.tj
Uj
6
( 15) •
I il1,,)At1
+
o··
2
j
(j = O; i = 1) (i = 1)
(26)
(i = l)
(28)
A
t
(27)
' 'lli 11 IIurl y' for the linear-acceleration method' Eqs. ( 19) and (20) generarize to * 1 (D. ) 1::,. (i > 1) (29) (D;+1)1 = R1 + !! 1+1 ; t1 -
( 1(1)
Rj
1111d
( I '/l
(25)
= Dj + !Dj D.tj
Rj
( 14)
At the end of' the step the velocity and displacement become ,;, ,,
(19)
,
1111 t the iteration in each step.
.
(t')2 = Uj +•I+" Ujl Uj 2
as:
A + 32 Uj• ut; + oI Uj.. ( uA(j )2
(D·+ 1)1 = Dj-1
')
U1+1 ;
u1+1
" lll'l'C
und U (t
,
,1 1111
.. ) (t f:.t· ')
Uj
-gu1
l lt11s, we form the recurrence equation for the ith iteration of * I ( • ) D,.t (i ~ 1)
2
-
t1
CIJ >
where t' is measured from the beginning of the step (see Fig. 5.1). If till' uccelcration varies linearly, the corresponding velocity and displacement wil I vary quadratically and cubically with time. Therefore,
•( ')
= u1 + H2u1 + Uj+I
I
J
U t
U1+1
I Ill' formulas given earlier [see Eqs. (9) and (10) or (11)] may be used agam to
E,,I (uj+1)d
f
:~o~f~
~3!anner analogous to that We will apply the lmear-acce era ion me . E (l) . h Because Eq ( 16) 1s the same as q · , li11 the average-acceleration appro~c : . . . is the same as that in Eq. 1111 recurrence expression for the ith iteration of u;+ i . I for ii· in 1 Il To obtain a direct relationship between U1+ J an~ Uj+ '.' we so ve J+, I q ( 16) and substitute the result into Eq. (17), which yields . . ) /::,. + I .. (f:.t )2 ( 18)
I II
(' IHll'Ht',
D1 I
~ I>,
A.11 I
~ i>1 (Atj}2
(30)
. ( I) (l\) ( tl) ('JC,) (J'f), uml (2K) apply to both methods. I•.qH. 2 t • t 2 t t
208
1>1111 I N1111111rl1 11l l111n111n1lu 11 Mulhod1
1111111111111 wi th lmpllc It
II is we ll known thar !ht· 11111111 11111 I« 111111111 111t•lhod is somewhur 1111 uccurure than the average-accch.-111111111111r 1I111d I / l llowever, it has been shu 181that the former technique is onl y , ,111,/11/1111,tl/\ .1·tahle. Therefore, as in th central-difference procedure, the solut ion diwqit·11 if the time step is too 111111 On the other hand, the average-accelcnition nicthod is unconditionally .1·tuli/ although less accurate. As with the central-difference predictor, a nodt·wiM iterati ve solution of a MDOF problem is feasible if the mass matrix is diago1111I
1:q. (•I ):
201
I 1111111,I H
/ '1
I
•
Ill
,., '/'
( Ii iJ I
(I
(111 ),1
= 0 .04816 ~ 0 I 0 .!MHXO T k k ( ' '120/' ' ) ( 21r)
() I/ ~ I H)
I()
O Oi! HHO
Ill
1
2
Eq . (5): J',q.
2
(11 1)4 = ~ m ( I - 0 .048 16) = 0.95 18 ~
(6)·.
Example 5.3 We shall now repeat Example 5. 1 using the iteration methods described in this sect,1111 Recall that m = k(T / 211')2, as given by Eq . (5.2-a), and that the in itial acceleration 111 ti me to = 0 becomes u0 = P1/ m, in accordance with Eq . (5 .2-b) . To apply the average-acceleration method, we start the first iteration in the fir I time step us ing Eq . (9) to estimate the velocity at time t 1 = At= T/ 20, as followh (u1)1
= uo + f1o tlt = 0 + (~)
(:0) =
nse has converged to within fo ur significa nt 111111 lh iterat ion we see that the re~po . T ble 5 3 along with the number of II~ II l(csults for 20 time steps are given m a . ' II i,1!11111~ required in each step .
It 1111
"1
TABLE 5.3
Average-Acceleration Method
0 .05 P~T
Then the displacement at time t 1 is fou nd from Eq. (5) to be (u1) 1 = Ro + I (u. 1)1 At= 0
2
0.05 + 2k
2 )
(Pi T (2T1r) 20
2
k
= 0. 04935 P1
•I
wh1d(6) 1 isasthe same as u1 in Eq . (5 .2-d). Next, we obtain the acceleration at time t1 from 11q,
(ui)1
1 = -(P1 -
ku1)1
m
Pi = -(I m
- 0 .04935)
= 0 .9507 -Pi m
Por the second iteration in the first time step, Eqs. (4), (5), and (6) yield
'> 11
'
H 1/ 10
II
1:2 Eq. (4): Eq. (5): Eq. (6):
(ii1h
=
(I
P1T + 0.9507)P1T - = 0 .04877-
40m
( ) =O+ Ui 2
Pi . (u1)i = -(I m
0 .04877 (Pi T 2k 20
m
2 )
( 21r) T
2
= O.O48 l 3 ~k
Pi 0 .95 19 -
m
Third iteration :
Eq. (4): Eq. (5): Eq . (6) :
l•o1111h lll' rn lio n:
11
1,1 l:i JI, )/
IH
- 0.04813) =
Respons e for Example S 3 Using Iteration Methodsa
-
11) )0
Linear-Acceleration Method
II;
Approx. u
n;
Approx . u
4 4 4 4 4 4 4 3 3 3 3 3 3 4 4 4 4 4 4 5
0.04816 0. 1880 0.4061 0.6813 0.9873 1.294 1.573 1.797 1.944 2.000 1.959 1.827 1.6 14 1.343 1.038 0.7300 0.4478 0.2 187 0.06497 0.00 126
4 4 4 4 3 3 3 3 3 3 3 3 3 3 3 4 4 4 4 5
0.04855 0.1895 0.409 1 0.6861 0. 9936 1.302 1.581 1. 803 1.948 2.000 1.955 l.8 18 1.60 1 1.326 1.0 19 0.7 105 0.4299 0.2047 0. 05665 0.00025
• l'uhulatcd values to be multiplied by P1/k.
(ti1 h
P1T P1T = ( 1 + 0.95 19) 40 m = 0 .04880-;-
(u1 )1 = 0
+
0.04880(P1T2)(21r)2 = 0.048 16~ 2k 20 T k 0 .048 16)
0,9 5 18 Pi Ill
. . ~thod using Eqs. (19) and (20) in place Nex t, we apply the lmear-accelerat~on: res o~ses are calculated somewhat more 111 1 q~. (5) and (8). In this c~se th~ appr~:1:~s fo/this second analysis also appear in ,111 1111111.: ly and with few er iterations. 1,,hlt• '> . L
I• \llllljlh.• 5.4 truss in Example 5 .2 which was shown 1, 1 11~ 111111ly1.c hy_ itcra~ive methods thct ancthe relationships m =· 0 .2s(Ti / 21r )2 and 111 l'lp '> . 1(a ). h ,r this prohl':m we av~ 5 2-·) 1111 ! l , 0}/ '1/ 111 , as giw n hy l:qs. (5 .2-h) and (. t .
210
Direct Numerical Integration Methods
nc. 5.4
Direct Linear Extrapolation
t b For t_he a~erage-acceleration method, we begin the first iteration in the first timo s ep Y estimatmg the velocities at time t 1 with Eq. (26), as follows:
(D,),
=Do +
D0 At= o +
[~](:) (~) =
TABLE 5.4 Responses for Example 5.4 Using Iteration Methods•
[ois] P;
=Ro+ i(D,)
1
At=
j
;;_ic(~;;
O+ [0.05] = [0.2467] [l 0 0.PiT..2s (T..)(27T) 40 T.. O s (cl
i::: same as D1 in Eq. (5.2-k). Then the acceleration at time t, is calculated from
(f)
For the second iteration in the first time step, Eqs. (21) , (22), and (23) produt·t•
=
ico
At={[~]+ [o.9;12]} ~ (~t) [1.911 2] !j_ (Ti) = [0.04778 JP, I;
(D,h = Q0 +
1) 1
0 . 11 84 m 40
Eq. (22):
Eq . (23):
(D,)2
0.002960
m
=Ro+ ~(D,h At= 0 +[0.04778 J P,Ti (Ti)(21r)2 2 0.002960 0.2s 40 Ti = [0.2358 J!j_ 0.01461 s
(D,h =
M- 1(A1 -
= _!_
[1
D;
Approx. D,
Approx. D2
6 6 5 4 4 4 4 3 3 3 4 4 4 6 4 4 4 4 3 3
0.2369 0.9121 l.932 3.173 4.515 5.851 7.081 8. 102 8.807 9.101 8.691 7.150 4.451 0.9388 -3.005 -7.028 -10.82 -14.09 -16.53 - 17.87
0.01168 0.08378 0.2961 0.7005 1.263 1.864 2.357 2.648 2.734 2.693 2.596 2.409 1.950 1.014 -0.4398 -2.193 -3.841 -5.009 -5.537 -5.533
5 5 4 4 3 3 3 3 3 3 4 3 3 5 3 4 4 3 3 3
0.2400 0.9220 1.948 3.192 4.535 5.875 7.112 8.139 8.838 9.110 8.742 7.146 4.394 0.8498 -3.106 -7 .142 - 10.97 -14.27 - 16.73 - 18.02
0.00834 0.0761 8 0.2941 0. 7165 1.298 1.902 2.373 2.629 2.697 2.670 2.618 2.482 2.031 1.024 - 0.5508 -2.386 - 4.002 -5.034 - 5.425 -5.395
"Tabulated values to be multiplied by
P, / s.
t \ti)
O]{[P'] _ s[ -0.48 0.36 0
m O 1
0.9221 JPi -_ [ 0.08922 ;;;
- 0.48] [0.2358 1.64 0.01461
J!j_}s
11 ,4 DIRECT LINEAR EXTRAPOLATION
Such ,7ul~ul~tions arc re~eated u_ntil convergence of displacements is obtained to within lour s1gn1f1cant figures (m the sixth iteration). At that stage the values of the displ .. 111t·11ts cnlculatcd from Eq. (22) are · au
'l'11hil• "i.il t:011t11i11s tilt• displ11t•1•111<'11ts /) 1 1111111lw1 ol 1tt•111t101111 11, 101 1•111 h ~ti•p
Approx. D2
As a second analysis, we apply the linear-acceleration method, using Eqs. (29) and in place of Eqs. (22) and (24) . The results for this case are also listed in Table 5.4. c 11111parison of these responses with the exact results in Table 5.2 shows that they are 1111111• llCCurate than those for the average-acceleration method.
s D,h
(D i)t> _
Approx. D 1
2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
- 0.48] [0.2467] !j_} 1.64 0 s
Eq. (21):
D;
I 2
.
Linear-Acceleration Method
Average-Acceleration Method
Next, we find the displacements at time t 1 from Eq. (22) to be
(D,),
211
f
0.2369
0.01 168
1111<11) 1
l
P1
,\'
Im· lwi•uty linw stt•ps , 11s wl'II Its
(g) fill'
11 lht· equations of motion for a MDOF structure are linear, it is possible to avoid lti 11t1ion of implicit formulas for numerical solutions. Instead, direct linear , l111polation procedures may be devised for both the average- and linear11111•lt·r11tion methods. We can formulate either total-response algorithms for l1t1r111 systems or incremental-response methods for nonlinear systems. Because 1h, 11111nhcr of nrithmetic operations is about the same for either approach, we 1 h111111l' lo dl•vl'lop lhl' i1 wn.•111l•nt11l tt•chniqm• , which upplies to both linear and 111111l1rn•111 p1ohlt·1n~ F111 lht·1111111t•, 11 ,., ft•11s1hlt• to st·l up simultuncous equations
213
Direct Numerical Integration Methods
212
!loc. 5.4
for incremental accelerations, velocities, or displacements. Here we will use incremental displacements as unknowns and solve a pseudostatic problem for each time step. To save space, only MDOF structures will be considered; and any SDOF system becomes merely a special case. At time ti (see Fig. 5.1), the damped equations of motion for a MDOI• linearly elastic structure are
M Di + C Di + S Di = Ai
.. !::,.Di
Subtraction of Eq. (1) from Eq. (2) produces the incremental equations of motion as
t::,.D Q i- i
(11)
2
-
(12)
4
1
Now substitute Eqs. (11) and (12) into Eq. (3) to obtain
_ 4_ 6.D- _ M [ (!::,,.tj)2 J
Similarly, at time ti+ 1
(2 l
= (Cl.ti )2
6.Dj = 6.t 6.Dj - Rj
( I)
= ti + D.tj, the equations of motion become M (Dj + D.Dj) + C (i>j + D.Dj) + S (Dj + 6.Dj) = Aj + D.Aj
Direct Linear Extrapolation
t 'ollecting
QljJ + C [~ !::,,.D- l:::,,.tj
R} + S 6.Di
1
;J
= Cl.Ai
terms, we rewrite this equation in the form
S 6.Di = 6.Ai 111 which
(14)
(3) These equations will be used for both the average- and linear-acceleration algorithms developed in the following discussion. For the average-acceleration method, the incremental velocities obtained by the trapezoidal rule at the end of time step D.ti are •
6.Di
1
.•
••
••
1
••
= 2 (Di + Di+1) D.ti = Di D.ti + 2 6.Di D.ti
Similarly, the incremental displacements at the end of the step become _
1
D.Di -
•
2 (Di
•
_
.
I
'
+ Di+ 1) !::,.ti - Di D.ti + 2 6.Di !::,.ti
(5)
Substitution of Eq. (4) into Eq. (5) yields
D.Di
•
1 ..
2
I
..
= Di b.ti + 2 Di(b.ti) + 4 b.Di (!::,.ti)
2
(6)
4
= (!::,,.ti )2 !::,.Di -
(7)
Substituting this expression into Eq. (4) produces
. 2 . !::,,.D = - tJ.D- - 2D1 Lltj J J
(8)
Now we define vectors Qi and R1 that contain only combinations of i>i and
~ R1
4 -
A
u.11
D1+ 2D1
(15)
thus, the pseudostatic equations represented by Eq. (1~) are to be solved fo~ the lnl'l'emental displacements 6.Di in each step. Then the mcrement~l accelerations .\i) and velociti~s ai>i mar. be found using Eqs. (11) and (12) . Finally, the total 1 vulues of Di+1, Dj+l, and Dj+l are
D1+1 = Di + 6.Di
(16)
i>j+1 = i>j + !::,,.l)j
(17)
= i>j + anj
(18)
Dj+l
i>1•
Ulrcct Linear Extrapolation by the Average-Acceleration Method Determine Qi and Ri from Eqs. (9) ·and (10) . i 1:ind Sand !::,.Ai using Eqs. (14) and (15) . Solve Eq. (13) for t::,.D1. Ca lculate 6.Di and t::,.D1 with Eqs. (11) and (12). . \ Add the incremental displacements , velocities, and accelerations to pren·ding values using Eqs. (16) , (17), and (18).
('))
2f>1
11~1111-1 llw~t· 1k•f1111tio11~. Wl' ll'wiitl' Eqi. (7) 1111d (K)
6.Ai = 6.Ai + M Qi + C R1
11 ucture consists of the following calculations m each time step:
4 . .. !::,.ti Di - 2Di
as follows:
1111d
In summary , the procedure for obtaining ~ynamic ~esponses of a MDOF
Solving for the incremental accelerations in Eq. (6) gives
.. !::,.Di
(13)
( IOI il !j
'l'lll lling now to the linear-acceleration ~ethod , we may again derive the 1111 1ri 111• 111ol vc lod ty vector froni the trnpc1.01dal rule to be 6.1>, 1>1 6.11 I ; 6.1>1 6.11 (4)
215 110. 5.4
Direct Numerical Integration Methods
214
llowever, the incremental displacements in the step must be found f
(5 .3- 17), as follows:
E rom
ADj
+ H2Dj + Dj+1)(Atj)2 i>j Atj + !Dj(AtY + !ADj(AtJ2
Dj
=
=
ll
Atj
( 19)
Solving for the incremental accelerations in Eq. (19) gives
..
6
ADj
= (Atj) 2
Atj
(20)
Dj - 3Dj
Now substitute Eq. (20) into Eq . (4) to obtain
AD. · = -3 1.1AD · 1 1
1.1
Then define
* = A6
Qj
.
1.1tj
R j*
--
30·
j
..
Dj + 3Dj
(22)
2Atj Dj..
(23)
+
Qo
4.
(b)
+ -4m- = k +
= (Atj )2 ADj - Q/ J
Eq. (15):
Mo
Eq. (13):
Auo =
Eq. (11):
( 0.04816.. = - 4 - Auo - -Qo = 4 (20) P ) - 2A Auo (At)2 T k m
(At)2
Mo T =
S* AD1 = AA1*
(26)
6 - M+-C 3 S* = S + (Atj)2 Atj
(27)
(d)
2A
A
A) -
tt
= ( =
1
A
= 4 -20) -k ( 0.04816( 27T m k
Auo =
Pi = - 0.048152m m
(f)
2
ti..uo - Ro = 2( ~) (o.04816~) - 0
Pi) (2T )
r
40) (, \0.04816;;;
7T
2
(g)
AT = 0.04880--;;
A
Eq. (16):
U1
Eq. (17):
u, = uo + Auo = 0 + 0.04880m
Uo
(e)
_ k = 0.04816,; 41 53 2
Nl'Xl, we substitute Eqs. (24) and (25) into Eq. (3) and collect terms to find
= 41.53k
= Mo + mQo = 0 + 2P, = 2Pi
(25)
J
I
(c)
Eq. (14):
(24)
AD = l_ AD - R* At·
2
4k( - T ) (20) 21r T
2
6
(a)
m
-k = k
Eq. (12): J
Pi
P, m
+ 2- = 2-
= 0
= 2uo = o
1111d rewrite Eqs. (20) and (21) as
.. ADj
..
= -At Uo + 2uo
2
(2 1I
2
Atj
Eq. (9):
Eq. (10): Ro
.. 3D· 1 - -At1 D· 1
-
11 ~nmple 5.5 o\fLuin , we shall repeat Example 5.1, using the direct linear extrapolation methods 1h1tlved in this section. For that purpose, we need the relationships m = k(T/211-)2 and rl11 P1/m from Eqs. (5.2-a) and (5.2-b). Starting with the average-acceleration method, we apply Eqs. (9) through (18) in 1111 first time step. When doing so, we use notation for the undamped SDOF system, as l11llows:
6 ADj -
Direct Linear Extrapolation
A
+ Auo = 0 + 0.04816,; = 0.04816,;
(h)
i11 wllkh
1111d
AA/
= AAi +
M
Q/
Eq. (18):
+ CR/
(28)
l \quutll)IIS (~6), (27), and (28) are analogous to Eqs. ( 13) (14) and 15 dt•11wd previou sly for the average-acceleration method Thus 'th , I ( ·1·'· Ii •. I' . . . . . · , , c proceu,rc tu , th • , • 1 Ht 1 111~ 111 cxtrnpolat 1on hy the linear accclcralion method foll H(t•p H HS h •I'll' , , I • OWS C SH llll , t t c, tXl'cpt t 1111 hqs . (9) through (I~) nrc rc1 , lnccd by Fqs ("'JJ lh1011µ,lt ( )8) .
' '.
~
.
.
.
PiT
AT m
= 0 .04880-
p p p ii., = ii. +Mo=_!. - 0.04815_!. = 0.9518_!. 0
m
m
(i)
(j)
m
t~11h• lhllt the values in Eqs. (h), (i), and (j) are the same as those obtained in Example
~ I hy the method of iteration. This procedure is repeated for each of 20 time steps.
V11lucs of the response from direct linear cx.trapolation by the average-acceleration 11111hod 111l' listl·d in Table 5.5. Also given in the tublc arc slightly more accurate values , 11111p11tl•d hy tlw 1111~111 lll'l'l' krntion ,m,thml.
1111 •1 I N1111111h ul l11t11urntlo11 Motl1111l1
lln
•I l•.q , ( 11 ),
-
t.\1>11
-
j
3 4
5 6 7
8 9 10
0.048 16 0. 1880 0.4061 0.68 13 0.9873 1.295 J.573 1.797 1.944 2.000
0 .04855 0 .1895 0.4091
11
l ,IJ;W
12 13
0.6861 0 .9936
14 15
1.827 1.614 1.343 1.038
1.302 1.581
16 17
0.7300 0.4478
1.803 1.947 2.000
18 19 20
0.2188 0.06500
1.019 0 .7105 0.4299 0.2047 0.05673
0.00130
0.00033
Av11 At·wl.
I
1.955 1.81 8
l '/' I
~· = :/'· +
Eq. (10):
Ro = 2D0 = O
13q. (14):
S=
Eq . (12):
= s[ 8.466 - 0.48
J!2 _[ 2J!l
0.01167 s
O m (p)
· 2 (40)(0 .2369 JP, ilDo = -Lit ilDo - Ro= -Ti 0.01167 -s - 0 2
L
=
I 0.04801 JP, Ti L0.002365 m
(q)
0.2369 ) P, Eq. ( 16): D, =Do+ L\Do = 0 + [ 0.01167 -s =
2
[1
~mo
-0.48] + (4)(20) 1.64
Eq. (17):
n
Eq. (18):
. .
)Ar..
(s)
m
. [l]A
D1 = Do + ilDo =
0 -m +
( -0.07976JA 0.09459 -m
}!l
= f 0.9202 0.09459 m
L
)2
(t)
1li•ll' the values in Eqs. (r), (s) , and (t) are the same as those found previously by iteration
0 ~Jo.2s({~
- 0.48 J 9.746
s
= [0.04801 0.002365
(kl
(r)
. . . (0.04801 J PiTi D, = Do + ilDo = 0 + 0.002365 - m
~]
2[1
ro.2369 }!l L0.01161
(/)
S + _ 4_ M = S + (4)(20) 0.36 - 0.48
S
= (40)f0.2369 J0.2Pi(!l) T, 0.01167 m 27T
20, = 0+ 2[:J; = [~];
CM2
J'-'
0.09459 m
1.601 1.326
For the average-acceleration method in mat . f shown. Thus, for the first time step we have nx orm, we use Eqs. (9) through ( l Kl
Eq . (9):
() ,() 11 67
= [ - 0.07976J!l
11:xnmplc 5.6 Now we will calculate the responses of the Ian . t•xtrnpolution methods. From Eqs (5 2 h) pd (; ~ss m Example 5.2 by direct lim·111 nnd l>o {l, O}Pi/m. . . . an .2-1) we know that m = 0.2s(7; /211 )
Eq . ( 15):
(,\)( l())J\ (),) \<,I)
()11
(27T)2m
"Tabulated values to be multiphed by P,/k.
s[
t.\1>11 1
Lin . At·t·t•I
=
J
== (4)(20) (0.2s) l0.2369
Li n.-Acccl.
Avg.-Acccl.
=
I
2
llN
(t.\/)
(Ill)
LlAo = ilAo + M Qo
1111\,rnmple 5.4. Following this time step, 19 other sets of calculations are performed in 1111 ~111nc manner, and Table 5 .6 gives the results. ror the linear-acceleration method, we apply Eqs. (22) through (28) in place of I q~. (9) through ( 15). Responses for this second analysis are also listed in Table 5. 6. N1111• lhat all of the values in the table are practically the same as those obtained by 111111tion in Table 5.4.
= 0 + m[~ ~] [~]~ = [~JP,
l!q . ( 13):
L\Do =
S , L\Ao =
_ l_[9.746
82.28 0.48
y,
,0 .2'.\69 ().() 1167 ,\'
(11)
•
0.48 J [ 2JA 8.466 0 ~ (o)
NEWMARK'S GENERALIZED ACCELERATION METHOD
111 this section we study a family of methods that were described by Nathan M. NL•wmark in a renowned ASCE paper. Two other variants, derived by Wilson , I 111 . tmd Hilber et al., will also be discussed.
219
2.11
N11w111n1k'1 Un11n1nll1111I /1,111111111111011 Mnlhoil
lllr1 I N111111111t •I l111nu,nllol1 Mnthocl•
rhl' p11111111t'll'I /I 1111'11 I 'l 1111111111, 1hr vc1111111011 of acceleration within the h·p For this 1cu~o11 !Ill' ll'l h11h1111• IM,l'lccced to as Newmark's generalized , , lr1111 ion method (or Ncw111111 k' s {J 111cthod). For example, if we take f3 = 0 ,
1111
·-
Avcragc-Accclc111111111 Method
j
Approx. D, I
2
0.2369 0.9121
3 4
5 6 7 8 9 10 11 12 13
16 17 18 19 20
1.932 3.173
0.00835 0.0762 1 0.2941 0.7 164
4.515 5.851
1.264 1.865
4.536 5.875
1.297 1.900
7.080 8.101 8.806
2.359 2.649
7. 113 8.139
2.370 2.627
2.737 2.697 2.600 2.409 1.948
8.838 9.108 8.741 7.145 4.397
2.699 2.675 2.623 2.482 2.022
10.89 14.09 16.53 17.86
I •I
Approx. D2
0 .2400 0.9220 1.948 3.192
0.9392 - 3.005 -7.028 -
ApjllOX. D,
0.01167 0.08374 0.2961 0.7006
9.099 8.689 7.149 4.452
14 15
Approx . /Ji
I 1111•111 Al'CCICnllion Mr lhod
1.013 -0.4404 - 2.193 - 3.843
0.8543 -3.104 -7.143 -10.97
-5.011 -5.541
- 14.28 - 16.73
1.011 -0.5576 -2.379 -3.988 - 5.024 -5.428
- 5.544
- 18.01
5.412
u111
(3)
f3 = L
Eq. (2) yields
(4)
lhlMa prcssion is the same as that in Eq. (5.3-3) for the average-acceleration m, r/111rl . When we take f3 = i, Eq. (2) produces 2 Uj+I = Uj + Uj Di.tj + H2iij + U1+1)(Di.tj) (5) 1111111s case the formula is identical to Eq. (5.3-17) for the linear-acceleration '"' 1!,11d . let us consider a MDOF structure and cast the Newark-/3 method into Now 111,,tii x l'ormat. In each time step we shall solve a pseudostatic problem for
11111,•111cntal displacements by direct linear extrapolation , as in Sec. 5.4. For this p111pnsc, Eq. (1) is restated in incremental matrix form, as follows: t:i.i>j = [(l - y)Dj + yDj+ tJ t:i.tj
= oj t:i.tj +
(6)
'Y t:i.Dj t:i.tj
111 11ddition, Eq. (2) is restated in incremental matrix fonn as • 1 " " 2 Di.Di = Di !::i.tj + [( i - {3)Dj + /3Dj+1](Atj) (7)
• 1 " 2 .. 2 = Dj Di.tj + 2 D/!::i.tj) + f3 t:i.Dj(t:i.tj)
In hi s 1959 paper Newmark [8] . intcgrut ion methods that h' ad b . generalized certain direct numcrirnl een m use up to that t' H 111· npproximaling the velocity and d' I ime. e presented equatio1111 I fo llows· IIN isp acement of a SDOF system at time tJI I• "
+ [(l - y)uj + 'Yii1+ 1] !::.tj = u1 + u1 t:i.tj + [0 - f3)u · + 13 ... ](A )2 Uj
(Il
u, + t ut1 ( 'l . (or algor 'th . ) d 11 ll· p11ru111etcr 'Yin Eq . (I) produ ces numerical :t· t1111c step !::.11, If 'Y is taken to be less than , . l .mic ar~1pi11g withi n u 11ult1-1. On the other hand ·r . . i , an arttfic1al negative dampin): . , 1 'Y is g1eater than ! such th · · .. uvoll 1 n111m·ncul damping ·1llogethcr ti I '. . 'mpmg ,s pos1t1vc. To (I) l>t•t·o111t·s lhc trnpczoid~d rule. , 1c vu uc ol 'Y must he equal to j; und Hq
'l'I
becomes
II pl11l'cmcnt. II we let
Newmerk-/J Method
=
1)
I hi l111111ula is known as the constant-acceleration method, because the accel1,1111111 fl at the beginning of the time step Di.ti is taken to be constant within the I h I' 11.quation (3) also corresponds to a truncated Taylor series that results from 11 1111' Huler's formula [see Eq. (5.3-10)] for velocity and the trapezoidal rule for
"Tabulated values to be multiplied by P,/ s.
Ujt I
I
'
'-i11lving for Di.Di in Eq. (7) produces ..
ADj
1
= f3 (Ati )2 t:i.Dj - .f3
',111l11tilution of Eq. (8) into Eq. (6) yields
t,:01 = ~ ~
11
6.01 -
7i 01 -
1 . 1 .. Ati Dj - 2/3 Di
u~ -
(8)
1) t.t f>
1 1
(9)
hn rnnvcnicncc, we define the vectors
(10)
221
Direct Numerical Integration Methods
220
Rj =
Sec. 5.5
j 1\ + ( 2~ - 1) Ati f>i
Newmark's Generalized Acceleration Method ii
Now rewrite Eqs. (8) and (9) in the forms •• ADi =
/3 (Ati )2 ADi -
1
' ADi =
/3
A
(1 2)
Qi
'Y Ati ADi - Ri A
(13)
Then substitute Eqs. (12) and (13) into the incremental equations of motion, given previously as Eq. (5.4-3); and collect terms to obtain
S ADi =
AA.i
(14)
in which 1
A
S
= S + /3 (At)2 M + /3
LM,_JI;+, r·
0
'Y
(15)
Ati C
~M,~
and (16)
We solve the pseudostatic problem in Eq. (14) for the incremental displacements ADi and substitute them into Eqs. p2) and (13) to find the incremental acceler ations and velocities ADi and ADi. Then the total values of displacements, velocities , and accelerations at time ti+ I are determined using Eqs. (5.4-16), (5.4-17), and (5.4-18).
Figure 5.5
We solve Eq. (18) for AD 8 to obtain
.. 6 6 . - .. . AD = - A D e - - Di 3D1 e (At8 )2 Ate .
ADe = -
Wilson et al. [9] , extended the linear-acceleration method in a manner that makes it numerically stable. The basic assumption of the Wilson-0 approach is that the acceleration ii varies linearly over an extended time step Ate = 0 Atj, as depicted in Fig. 5.5. During that time step the incremental acceleration is Aue = 0 Aui. It has been shown [ 10] that the optimum value of the parameter 0 is 1.420815, which can be rounded to 1.42. As with the Newmark-,B method, we shall construct the Wilson-0 variant in an incremental matrix format applicable to a MDOF structure. From Eq. (5.4-4) for the linear-acceleration technique, we have the incremental velocities at the end of the extended time step At0 , as follows: ,
..
I
..
AD o = Di Ato + ~ AD o Ato
( 17)
where the symbol AD 0 denotes a vector of incremental accelerations. Simi larly, Eq . (5.4- 19) gives the incrl•mentul displucellll'llls us •
I ..
D1 At11 I 21>1(A111)
)
I
.
I ,.Al>11(A111l
l
( 18)
(19)
. . · t Eq (17) which results in l'hen substitute this expression m o . '
Wilson-0 Method
Al>11
Linear-acceleration method extended to Wilson-6 method .
3
.
Ate
Now define
Q* e
ADe - 3Di -
= ~ i).1 + 3:0j
(20)
(21)
b.te
. · Rt= 3Di 11 11d
Ate .. Di
2
+
b.te .. Di
2
(22)
restate Eqs. (19) and (20) as 6 AD .. Q* ADe = (b.te)2 ee
3 AOo - R! b.to . • ( c; 4 1) nnd collect terms to find 'i11hHt11utl' l~qs. (23) 11ml (24) into 1~q. · st M>,1 AAt
ADo -
(23) (24)
(25)
'l'J'J. 'l'hl•
ll!t It N11111n1 h ,11 l1111111111tlo11 Mnthod11
st11111.! d ll!1111s in this
1 It ll
l''lll1tfhtt1 1111
st
~ ,"l
I
N11111111 le 111
223
t.tnhlllty 111111 /\1, 111 111 y
n NUMERICAL STABILITY ANO ACCURACY (1 A
(ul,i)
)
I 11 ~tucly the stability and accuracy of various one-step direct numerical
M
11111 p,111tio n procedures, we may cast them into
und
U1+1 = A U1 + L (J. I)
where
AAo =
e AA1
(.'HI
The problem in · so1ved ior .: the mcremental . AD pseudostatic Tl . Eq . (25) . is disp1acenw111 II· 1en we ca~ determme the mcremental accelerations AD 8 from Eq (2 II · HiltI reduce them lmear1y by the formula
.. ADJ
1
I 'I (I) represents a column vector containing the three response quantities u1, ,,, 1111d
a1 at
the time station t1. That is,
U1 = {uj, 111111
Next, the incremental velocities AD1. and displacements AD are obt · d f Pei · (5 4 4) d (5 L.l J ll.me ro11 1 · s. · - an .4-19) for the time step At. As before Eqs (5 4-1() ( '\ .4 17), and (5 .4-18) give total values of the dis I ' · ··· l ' lll'l'l'lcrntions at t' t· p acements, veloc1t1es, and 1me 1+ 1 .
u1, u1}
(2)
the vector U1+ 1 is similarly defined at time tj+ t to be
U1+1 = (2'))
(1)
Pj+I
1ltlNl'xpression pertains to a SOOP system that might equally well be considered , 1111c natural mode of vibration for a MDOF structure. The symbol U1 in
..
= -AD e o
operator form [12], as follows:
{u1+1, u1+ 1,
(3)
ii1+1}
1111' t·oefficient matrix A in Eq. (1) is a 3 x 3 array called the amplification 1t1tt11·/, that we shall examine to answer questions about stability and accuracy.
I l1111lly, the symbol L denotes a column vector called the load operator, which I 11111ltiplied by the load lj+ 1 at time t1+ i· If there is no loading, Eq. (1) simpli111 H to
(4) Hllber-a Method
lt11 free-vibrational response.
. To imp~·ove control of numerical damping, Hilber et al. [11 ] introduced p111 amctcr a into the equations of motion at time t· .: IIows.. J+ l , as 10
MD1+1 + C D,+1 . + (1 + a)S D1+1 -
aS D1
= A,+ i
11
(30)
~11btracting similar equations of motion at time t from Eq. (30) prod ti . tru.:rnmcntal equations , uces 1(
M AD1 + C AD1 + (1 + a)S AD1 - aS AD,- ,= AA1
(3 1)
Ncl>lw substitute Eqs. (12) and (13) from Newmark's method into Eq (31) and cct terms to find ·
l'O
To investigate the stability of a numerical algorithm, we apply spectral ,l,•,·0111position [13] to the amplification matrix A, as follows: (5)
ltt this equation A is the spectral matrix of A, containing eigenvalues A1, A2 , and \ In diagonal positions; and is the 3 x 3 modal matrix of A, with 1 1 l~l·nvcctors <1> 1, <1>2 , and <1> 3 listed columnwise. If we start at time t0 == 0 and l ll~ l' 11 time steps using Eq. (4) , we have 1 Vn = AnjUo (6) 1
wlwrc the vector U0 contains initial conditions, and vector Un gives the response )
Sa ADJ= AA.a}
(32)
111 which
v11 l11cs at time t,.J' Raising the decomposed form of matrix A in Eq. (5) to the powe r n1 yields (7)
(33) N11w let us define the spectral radius of matrix A as
1111d
,.,
,.,
AA<\i = AA1 + aS AD1
1
(J4 )
Expressi<~ns for S and A~ were derived previously us Eqs. ( 15) and ( 16) F . lhl• llrst time step, we luke A Di 1 _ in Eq . ( ,,I). · or
o
(r)A
= max IA,\
(i
= 1, 2, 3)
(8)
I hl'II Eq . (7) shows that we must hnvc (1 )A
(9)
224
1 11c.
Direct Numerical Integration Methods
in o~d~r to keep the numerical solution from row· . , c~nd1_hon is known as the stability criterion for a !ven:!t:~h°;t boun~. Th~H ~p~ym~ ~111 ~ntenon to the constant acceleration method (see Sec 5 5) . time step for this conditionally stable approach to be. . ' we n t e cntical
J
(At)c,
= _!!T.1T = 0.318Tn
(J 0)
This value is the same as that given b E (S 2 . . Y q. · -12) for the central-difference predictor For th r 1 . e mear-acce erat1on method, the critical time step is
(At)c,
= V3 '7T [;_ = 0.551T.,,
(JI l
On the other hand the spectral d' f; h always unity. The;efore it has ra IU~f o~ t_e average-ac~eler~tion method is dilionally stable. ' no en ica time step and is said to be uncon The matter of accuracy of a n · 1· . re lated to that of stability Figure 5 6 u:enc\ mteration procedure is closely system to an initial dis la~e . s ows t e un amped response of a SDOJ• Ihose labeled 2 3 ~ 4 ment uo. The_ curve labeled 1 is the exact result, and ' ' an represent various approximations Curve 2 d Nlrat<.:s an amplitude increase (Al) that · r · . emon shows no amplitude change and curve ie~ an unstabl~ algonthm. Curve .l Ut•cuuse of the stabTt -' . . epicts an amplitude decrease (AD) 1 1 · Y en 1enon m Eq (9) only curves of t 3 d · ndlllissiblc approximations Curve 3 · b ' . ypes an 4 arc uvcrugc-acceleration meth;d h. h hmay e cons1der~d to be _the result for the siblc algorithms are represen~e: b1yc cu~/;pech~ahl ~adml_s of umty. Other_admis ' w IC imp 1es a spectral radms Jess
i:~
LJ
--.. PE
'
Al
AD
1111(1111•
~.<,
S 111h1l11 y 1111d
un111m y 111 11111111,111·111 lr11t•p1111i1111 llll'lhrnlN
5.7
Program DYNA for Dynamic Response
225
lhnn unity. Thus, one important type of error to be considered is amplitude 11tppression, as exhibited by curve 4. All of the approximate responses in Fig. 5.6 also show period elongation l PE), which is a second type of error introduced by any numerical algorithm. lloth the amplitude suppression and the period elongation may be made negli1,11ble by using sufficiently small time steps. Newmark [8] recommended a time M l<.:p of duration equal to! or of Tn, which is the smallest period of a MDOF Nltucture. However, a more commonly used time step is lit= Tn/10. At first glance, the average-acceleration method appears to be the best 1 hoice among the implicit approaches, because it has no amplitude suppression 1111d the least period elongation [12]. However, a small amount of amplitude ~uppression is desirable to reduce or eliminate unwanted responses of higher 111odes in an undamped MDOF structure. But if the spectral radius of an 11111plification matrix is too small, the response of the structure will be unduly ~uppressed, as in the Houbolt method [14]. Although the optimized Wilson-8 lmmulation produces a reasonable level of amplitude suppression, it also exlllhits an undesirable tendency to overshoot the true response in the first few time ~1,·ps [15]. Thus, we conclude that the best choice of algorithm is the New11111rk-/3 method with the Hilber-a modification, as described in Sec. 5 .5 . Probahl y, the optimum selection of parameters for this approach is to let a = -0.1, /I 0.3025, and 'Y = 0.6.
i
,7 PROGRAM DYNA FOR DYNAMIC RESPONSE
Now we shall describe a general-purpose program named DYNA that calculates dy11umic responses of linearly elastic structures. This program includes vi111111 ional analysis from Chapter 3, normal-mode responses from Chapter 4, and dl1l'Ct numerical integration from the present chapter. By virtue of a branch in 1111• main program, responses may be obtained using either the normal-mode 1111•1hod or direct numerical integration. For the latter approach, we select the Nt•wmark-/3 method with the Hilber-a variant discussed in Secs. 5.5 and 5.6. Flowchart 5.1 shows the main progrnm for DYNA, which calls the five , 111,programs indicated in double boxes. As in Flowchart 4.1, Subprogram VJB 11111s the seven other subprograms given in Flowchart 3.1, including nor111111izution of the eigenvectors with respect to the mass matrix. Then the program 1, 11tls the number of loading systems NLS and checks it against zero to determine wlwlh<.:r to process a loading system or another structure. Next, the loading 1111111hcr LN is initialized to zero, and then it is increased by one. 'l'h<.: s<.:cond subprogram to be called is DYLO, which is almost the same ,, 111 Flowe ha, I 4 . I. I lowevcr, th<.: first dynamic parameter now becomes the l111ll111lor ISOLVE for type of sol11lio11 . The two choices of solution are the 11111111111 111e111l' 1tll'lhod (ISOLVE 0) 1111d dirl'L'l numcricnl integration
M11w1'11111 I 'I, I
1\,111111 1111111111111 t111 U\' NA•
227
~. I
IOI V IIJ1
J
1'11111111111
VIII
11 11111
S1'1' I H,
JIiii , lllllllllllilllllllll Oi Cl)IC ll Vl'('l(ll N II 111 1 1,•,1w1'I Iii 11111,s IIIHIII X.
R1·ud 111111,hcr of loading syslcms.
If NLS = O, go to IOI and process
another structure.
Initialize loading number to zero.
Increment loading number by one.
2. Read and write dynamic load data, including indicator !SOLVE for type of solulion.
lf!SOLVE = l ,goto 103.
3 · Calculate dynamic responses
by nom1al-mode method.
Go to 104.
4 · Calculate dynamic responses by direct numerical integration.
5 · Write and/or plo1 results of response calculations.
Check lo r las, loading sysle m.
Go to IO I nncl process a11o 1hc, s 1ruc 111rc.
l !ND '~lll'lll'S 111 1111y IYl'l' ol l1111•11oly 1'111M 11 ,,11111'lll ll' IS11 hp111~111111~ 111111 dilll•1 1111 l'Vl'IY ' Yl'l' 111 , 1111l lilll'
I) . SubNt'qlll'lltl y. 1lll' p111g1 11111 l'tllllpt1r0s !SOLVE with unity to I , hh• whk h ly pl! of solulion prnn•d111t• is to bi! used for dynamic response
,h 11l111ions. II' ISOLYE I- I , Subprngrum NORMOD for the normal-mode 111, 1h,,d in turn calls Subprograms TRANOR , TIHIST, and TRABAC, which are 1111111hl'1\·d 3, 4, and 5 in Flowchart 4. I . Alternatively , if ISOLVE = 1, Sub1111111111111 NUMINT computes dynamic responses by direct numerical integra111111 For this purpose , it reads and writes the integration parameters ALPHA, Ill I A, und GAMMA required in the Hilber-a formulas. Within Subprogram NUMINT it is also necessary to generate the damping 111,111 Ix. C to be used in the response calculations. Toward this end, we substitute lh1 d11mping ratio DAMPR and the results from Subprogram VIB into Eq. 1I,, 17) to create the damping matrix in physical coordinates. The final subprogram called by the main program is RES2, which writes 111111/rn· plots selectively the results of response calculations as before. Testing LN 1111111111! NLS at the end of the flowchart determines whether to return for another li 1111ll11g system or another structure. Program DYNA may be specialized to become DYNACB for continuous 111 1111111, DYNAPT for plane trusses, and so on. As before, the main program for , ,11h specialization has subprograms that are different for each type of applicallt 1111 us indicated by the second footnote below Flowchart 5.1. Note that Subp111grnms NORMOD and NUMINT are the same for all types of structures. As 1111 l'mgram NOMO, notation for Program DYNA is included in Part 5 of the 11-1 of' notation near the back of the book. Details of the logic in Subprogram NI IMINT appear in the flowchart for Program DYNAPT, given in Appendix C. Thus, in several stages we have devised a program that will handle not only ~ll11111ional analysis but also two types of dynamic response calculations. Vari1111111 of the program apply to all linearly elastic framed structures and discretized , 1111llnua discussed in this book. I•,111111,tc 5.7 I 11 Nhow how Program DYNAPT is used, we shall repeat Example 4.15 (the three1111111hcr plane truss), using the same number and size of time steps. In this application ~, 1,i11ploy direct numerical integration instead of the normal-mode method by setting the p111 11111l1tcr ISOLVE equal to unity. Also, the values of the integration parameters read hy S11hprogram NUMINT are taken to be ALPHA= - 0.1, BETA= 0.3025, and t 1/\MMA
= 0.6 .
I ,ct us examine again the responses of the structure to the dynamic influences given 11111,.,nmplc 4. J 5. Table 5.7 lists part of the output from Program DYNAPT for these four ,11111lys1•s. figures 5.7 and 5.8 (on pages 233 and 234) also show computer plots of the 11 ~1,IIN ohJained . The responses in Fig. 5.7 due to initial conditions follow the same 11111r111I lt'l!nds us those in Fig. 4. 11 , but they differ because the time step is too large. 1111 1hr nlhcr hand, !he smoother responses in Fig. 5 .8(b) caused by the applied force in 111• 't .K(u) m·c prnctically indistinguishuhlc from those in Fig. 4.12(b). This good 111111•l111lon INdm• to thl' facl that the ti11w s11·p Is short enough to model the slowly varying II ~plllllll 'N
lll'l'llnlll'ly,
228 Direct Numerical Integration Methods
Program DYNA for Dynamic Response
TABLE S.7 Computer Output for Example S.7
229
TABLE S.7 (Continued)
PROGRAM DYNAPT *** EXAMPLE 5.7: ~HREE-MEMBER PLANE TRUSS*** STRUCTURAL PARAMETERS NN NE NRN E 3 3 2 3.0000E+04 NODAL COORDINATES NODE X l 0.000 2 150.000 3 150.000
0
0
GROUND ACCELERATIONS IGA EL 250.0000 150.0000 200.0000
ex
CY 0.6000 0.8000 1.0000 0.0000 0. 0000 -1. 0000
0
DIRECT NUMERICAL INTEGRATION ALPHA= -0 . 1000 BETA 0.3025
GAMMA
OUTPUT SELECTION IWR IPL NNO NEO 1 1 2 1
NUMBER OF DEGREES OF FREEDOM • NUMBER OF NODAL RESTRAINTS: •
NDF NNR
STIFFNESS MATRIX DECOMPOSED MODE 1 ANGULAR FREQUENCY 4.1995E+02 NODE DJl DJ2 l 2.3137E-Ol O.OOOOE +OO 2 1. OOOOE+OO -2.4722E-Ol 3 0.0000E+OO O.OOOOE+OO l.1677E+03 DJ2 O. OOOOE+OO l .OOO OE+OO 0.0000E+OO
MODE 3 ANGULAR FREQUENCY l.8 618E+03 NODE DJl DJ 2 l l.OOOOE+OO O.OOOOE+OO 2 -6 .0504E-Ol -6.1068E-Ol 3 O.OOOOE+OO O.OOOOE+OO LOADING NUMBER
0
APPLIED ACTIONS NLN NEL
y
0.000 200 .0 00 0.000
NODAL RESTRAINTS NODE NRl NR2 l O l 3 l l
MODE 2 ANGULAR FREQUENCY NODE DJ l 1 8.6725E- Ol 2 -L 7149E-Ol 3 0.0000 E+O O
2
INITIAL DISPLACEMENTS NODE DOl D02 1 l.OOOOE-01 O.OOOOE+OO 2 l.OOOOE-01 l . OOOOE-01
RHO 7.3500E-07
ELEMENT I NFORMATI ON ELEM. J K AX l l 2 10.0000 2 l 3 6.0000 3 2 3 8.0000
INITIAL CONDITIONS NNID NNIV
l OF
DYNAMIC PARAMETERS JSOLVE NTS DT l 20 l.OOOOE - 03
4 ***
DAMPR O.OOOOE+OO
3 3
NODES:
1
ELEMENTS:
1
2
DISPLACEMENT TIME STEP TIME 0 O.OOOOE+OO 1 l.OOOOE-03 2 2.0000E-03 3 3.0000E-03 4 4.0000E-03 5 5.0000E-03 6 6.0000E-03 7 7.0000E-03 8 8 . 0000E-03 9 9 .000 0E-03 10 l . OOOOE-02 11 l . lOOOE-02 12 l .2000E-02 13 l .3000E-02 14 l.4000E-02 15 l.5000E-02 16 l.6000E-02 17 l.7000E-02 18 l.8000E-02 19 l.9000E-02 20 2.0000E-02
HISTORY FOR NODE 1 DJl DJ2 l.OOOOE-01 O.O OOOE+OO 6.9151E-02 O.OOOOE+OO -9. 7574E-03 O. OOOOE+OO -8. 0921E- 02 O.OOOOE+OO -6 . 9989E-02 O. OOOOE+OO 2.1718E-02 O. OOOOE+OO 8.8292E-02 O.OOOOE+OO 3.9844E-02 O. OOOOE+OO -7.5170E-02 O.OOOOE+OO -l.2500E-Ol O.OOOOE+OO -5.7660E-02 O.OOOOE+OO 4.7566E-02 O.OOOOE+OO 8 . 9567E-02 O.OOOOE+OO 5.1576E-02 O.OOOOE+OO -l . 2272E-02 O.OOOOE+OO -4 . 6102E-Oi O.OOOOE+OO -2.5622E-02 O.OOOOE+OO 3.6195E-02 O.OOOOE+OO 8.5913E-02 O.OOOOE+OO 6 . 2893E-02 O.OOOOE+OO -2.9158E-02 O.OOOOE+OO
MAXIMUM TIME OF MAXIMUM MINIMUM TIME OF MINIMUM
l.OOOOE-01 O.OOOOE+OO -l.2500E-Ol 9.0000E-03
O. OOOOE+OO 2.0000E-02 O.OOOOE+OO 2.0000E-02
0.6000
230
Direct Numerical Integration Methods
TABLE 5.7 (Continued)
TABLE 5.7 (Continued)
DISPLACEMENT TIME STEP TIME 0 O.OOOOE+OO 1 l.OOOOE-03 2 2.0000E-03 3 3.0000E-03 4 4.0000E-03 5 5.0000E-03 6 6.0000E-03 7 7 ,0 000E-03 8 8 . 0000E-03 9 9.0000E-03 10 l.OOOOE-02 11 l . lOOOE-02 12 l.2000E-02 13 l . 3000E-02 14 l.4000E-02 15 l,5000E-02 16 l . 6000E-02 17 l.7000E-02 18 l.BOOOE-02 19 l.9000E-02 20 2.0000E-02
HISTORY FOR DJl l.OOOOE-01 8.8727E-02 6.8467E-02 4,9368E-02 l.2394E-02 -5,2366E-02 -l,1040E-01 -l,2015E-01 -8,9808E-02 -6,2312E-02 -5.2629E-02 -3,4703E-02 9.6841E-03 6,2708E-02 9 .8915E-02 l.1326E-01 l.0852E-01 8.0000E-02 3.1105E-02 -l,48 6BE-02 -3 ,9867E-02
MAXIMUM TIME OF MAXIMUM MINIMUM TIME OF MINIMUM
l . 1326E-01 l . 1863E-Ol l.5000E-02 6.0000E-03 -l.2015E-01 -l,3161E-Ol 7.0000E-03 l.5000E-02
MEMBER STEP 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
HISTORY FOR ELEMENT AMl 9 . 6000E+Ol 4.6852E+Ol -2.5370E+Ol -2,4355E+Ol 1. 6683E+Ol 1. 2893E+Ol -2,9174E+Ol -4.2452E+Ol -2,3751E+Ol -2.4867E+Ol -4 .330 1E+Ol -2. 3119E+Ol 3.4926E+Ol 5.8727E+Ol 2.1256E+Ol -l.159BE+Ol l.3378E+Ol 5.3656E+Ol 4.3667E+Ol -4.5712E+OO -3. 2732E+Ol
FORCE TIME TIME O.OOOOE+OO l.OOOOE-03 2.0000E-03 3.0000E-03 4,0000E-03 5.0000E-03 6.0000E-03 7.0000E-03 8.0000E-03 9.0000E-03 l.OOOOE-02 l. lOOOE-02 l.2000E-02 l . 3000E-02 l,4000E-02 l . 5000E-02 1. 6000E-02 l,7000E-02 l.BOOOE-02 l.9000E- 02 2.0000E-02
MAXIMUM TIME OF MAXIMUM MINIMUM TIME OF MINIMUM
9.6000E+Ol O.OOOOE+OO - 4.3301E+Ol l.OOOOE - 02
231
Program DYNA for Dynamic Response
***
NODE
2 DJ2 l.OOOOE-01 3.4123E-02 -B.5095E-02 -1. 2309E-Ol -4.4408E-02 6.8993E-02 l.1863E-01 7.5778E-02 -l,3763E-02 -7 ,2918E-02 -4,8878E-02 3.7619E-02 9.6294E-02 5.2826E-02 -6.1248E-02 -l.3161E-01 -B.6674E-02 2.3038E-02 8.6592E-02 5.3559E-02 -2.6064E-02
LOADING NUMBER
2 OF
DYNAMIC PARAMETERS !SOLVENTS DT 1 20 l.OOOOE-03
4
***
DAMPR O.OOOOE+OO
INITIAL CONDITIONS NNID NNIV
0
0
APPLIED ACTIONS NLN NEL 1 0 NODAL LOADS NODE AJl 2 2.0000E+Ol
AJ2 O.OOOOE+OO
GROUND ACCELERATIONS !GA 0
FORCING FUNCTION NFO 7
1
FUNCTION ORDINATES TIME K 1 O.OOOOE+OO 2 1. OOOOE-03 3 3.0000E-03 4 4,0000E-03 5 5.0000E-03 6 7.0000E-03 7 8.0000E-03
FACTOR O.OOOOE+OO 1. 5000E-01 8.5000E-01 1. OOOOE+OO B,5000E-01 l,5000E-01 O,OOOOE+OO
DIRECT NUMERICAL INTEGRATION ALPHA= -0 .1000 BETA 0,3025
GAMMA
0.6000
OUTPUT SELECTION !WR IPL NNO NEO 1 1 1 1 NODES:
2
ELEMENTS:
1
2 DISPLACEMENT TIME HISTORY FOR NODE DJl DJ2 STEP TIME 0 O.OOOOE+OO O,OOOOE+OO O.OOOOE+OO 1 1. OOOOE-03 B,4579E-04 -l.OOOBE-04 2 2,00 0 0E-03 5.5055E-03 -8.5533E-04 3 3.0000E-03 l,7942E-02 -3,4897E-03 4 4.00 0 0E-03 3.8969E-02 -8,9831E-03 5 5.0000E-03 6,3694E-02 -l,6261E-02 6 6.0000E-03 8.4309E- 02 -2,1959E-02 7 7.0000E-03 9.4437E-02 -2 ,3226E-02 8 O.OOOOE- 03 9.0 37 4E- 02 -2. 0476E-02 9 q,OOOOF. 03 7.0970E 02 l.6052E-02 1,07/ bE 01 1. 0521E-02 10 l.0000~ 02
232
Direct Numerical Integration Methods TABLE 5.7 (Continued)
11 12 13 14 15 16 17 18 19 20
l.lOOOE-02 l. 2000E-02 l . 3000E-02 l.4000E-02 l.SOOOE-02 l . 6000E-02 l.7000E-02 l. SOOOE-02 l.9000E-02 2 . 0000E-02
9.4437E-02 2 . 3972E-02 7 . 0000E-03 l.4000E-0 2 -9 . 568BE-0 2 -2 . 3226E-02 l . SOOOE- 02 7 . 0000E- 03
MEMBER STEP 0 1 2 3 4 5 6 7 B 9 10 11 12 13 14 15 16 17 18 19 20
HISTORY FOR ELEMENT AMl O. OOOOE+OO 6 . 1168E-Ol 3 . 4981E+OO 9 . 63 72E+OO l. 7047E+Ol 2 . 2466E+O l 2 . 5649E+Ol 2 . 8511E+Ol 2 . 9596E+ Ol 2.4238E+Ol l.1821E+Ol -1. 9340E+OO -l.2285E+Ol -2 . 0281E+Ol -2 . 7599E+Ol -3 . 1270E+Ol -2 .7 564E+Ol -1. 7935E+O l -6.9027E+OO 3 . 9111E+OO l.5351E+Ol
FORCE TIME TIME O. OOOOE+OO l. OOO OE-03 2 . 0000E- 03 3.0000E-03 4.0000E-03 5 . 0000E-03 6 . 0000E-03 7 . 000 0E-03 B. OOO OE-03 9 . 0000E-03 l . OOOOE-02 l.lOOOE-02 l . 2000E-02 l . 3000E-02 l . 4000E-02 l . SOOOE-02 l . 6000E-02 l . 7000E-02 l. BOOOE-02 l.9000E-02 2 . 0000E-02
MAXIMUM TIME OF MAXIMUM MINIMUM TIME OF MINIMUM
DA'1PR=O.O
l . 3673E-04 - 2.3257E-03 -3 . 7755E-02 8 . 5393E-03 -6 . 9338E-02 l . 8614E-02 -9 . 0054E- 02 2 . 3972E-02 -9.5688E-02 2. 3788E-02 -8 . 4382E-02 l . 9802E-02 -5 . 8635E-02 l . 354 2E-02 -2.3925E-0 2 5 . 5924E-03 l.4150E-02 - 3. 2004E-03 5 . 0421E-02 -l.1543E-02
MAXIMUM TIME OF MAXIMUM MINIMUM TIME OF MINIMUM
233
Program DYNA for Dynamic Response
Soc. 5.7
-
0. 10
- o.os ~ o.oo 0
z
-o . 05
1-
<
, -o. 0
( a)
l
2.9596E+Ol 8 . 0000E-03 -3.1270E+Ol l .SOOOE- 02
-
0.10 C
- o.os N
~ o.oo 0
z
I-
<
-o .os
, -o . 0
(b)
DAMPR=O . O -
0 . 10 C
- o.os N
~ o.oo
l r.101111l)IC
5.8
N1•x1, we repeat Example 4. 16 (the symmetric pl ane truss), with the same number 1111d time steps. llowcvcr , the first load case in that example will be treated diffcrcnl ly l11Nll•11d ol' using modal truncarion , we shall compare results from the Ncwmark-/3 method 111111h1sl those obtained by the l lilbcr-a 1cchniquc. In the lirsl instance, we h11vc Oil' i111q~rntion p11r11 1111i tcrs ALP! IA 0, BETA 0.25, und OA MM A 0.5; und in !hr Nt1t'1>11tl wt• t11k1• ALPHA 0 . 1, BETA 0. 102'>, n11d ClAMMA 0.6. Ni11• of
0
z
,_ -o.os <
N
ci " 0, 10
-
~.7 'l'l!i,·t•
1111, 111 tw1 pl 1111l' 11 u NN: H'Nl)lHl NCN 10
inilinl displacements.
234
z 100
20
80
"'C. -
235
Program DYNA for Dynamic Response
Sec. 5.7
Direct Numerical Integration Methods
10
w 60 D 0
z
(\/
s
0 0
10
15
T
z -;
20
<(
20 (ms
J
I--
<(
40
I--
w O
(\/
~ 0
so
0
I 00
l
so
200
(a )
T
(ms)
T
(ms
-, <(
-20
NEIIMARK ~ 4
(a)
E
DAMPR =O.O Q. J 0
'::;: I (\/
-, 0
D
_': 0 OS
JOO
( b)
200
J
(\/
~ o. oo 0 z E 4
I--
E
< - o.os
..,
lD
0
NE I/MARK 3
w
8z 2
- 0. I 0
~l (\/
6
( b)
0
Figure 5.8 Three-member plane truss: (a) applied force; (b) responses.
Figures 5.9(b) and (c) show computer plots of the symmetric and anti-symmcl r rt l'lllllributions to they translation at node 6 for the Newmark and Hilber methods. Thcs1• H'Nponscs should be the same as those for all modes given previously in Figs. 4. 1J(hl 1111d (l') . They all mulch fairly well , even though the lime step is rather long . Nole th111 lhl• 11pprniti111011· rl·sponscs in Fig. 5.9(c) huvc II signilirnnl f)Ni()(/ 1'lo111,1ar/011 rclutiw lo lhut i11 Fjg, 4. 1 l(l') , Also, llw llilhl•r plol in Jllg , 'i .')(c) hus II rmlkl·11hk (//ll/1llr11tl1• 1·111i1111·.1·.1'io11 rd111iv,· tu llw N1•w111111k plot , 11s it should
Figure 5.9 Plane truss: (a) applied force; (b) symmetric responses; (c) antisymmetric responses.
Responses to the piecewise-linear ground acceleration in Fig. 5.lO~a) .are/~ot!~d I11 Fi 5 ·, O(b.) i-or the two cases of DAM PR = 0.0 and 0.1. They were o tame y e !:I· •n · tll)JWOttch using the same .intcgrn11on · I lilhcr parameters as before .hThese plots l are d · ' , J" 4 14(1 ) hich were found by t e norma -mo e li11IINti11guiNht1hlc from those 111 ·tg. , · 1 , w 11111(hotl ,
23 1 2311
1111 ,11 1 N1111111lt11l li1ln11111tlm1 M11thod11
...
I , y, S W , uml IkiNiiW'' , Z I\ , "' I 11111NH'111 I >y1111111ic Analysis of Thin Shells by 1111 1111111~· 1ih;n w11t Method ," l'11w . ltd C'CII(/ Mal. Methods Struct. Mech., AFIT, Wlifr,hl Pulh.irson Al
I 00
('l
"
-' l
z
0
o. so
....
n::
w o.oo
...J
w
(.) (.)
-o.so
z
::i 0
n::
'f'
- 1 . 00
-
>-
(a )
I . 00
6
~
o.so
[
0
~
o.oo
0
7
....
< -o.so
.., N
,,
Cl
- 1 . 00
-
( b)
PROBLEMS*
Figure 5.10 Pla ne t russ : ( a) ground acceleration; (b) responses.
REFERENCES I. Dnhlqulst, G. G., "A Special Stabilit Pr Nord. 'l'lrl.1·kr. !11.f. Bl'hw1d/i11g Vol 3 YI 9;1blcm ~2or ). K i , ' · • · , pp. 7 · rill,\ , I,. I)., " Unconditional Stnbilit · N .I. Ap11!, M1•c ·lt Vo l 110 N11 2 I 117; 111 111111,111•111 '' · · · , '/ ., , pp. •I I I ,111
· L111car Multi step Methods," 43 ,: ·
1111w ln11;11n1tlon MulhodN" '
I.• I. Confirm the approximate results of Example 5 .1 in Table 5 .1.
1•2. Repeat Prob. 2.7-2 using the central-difference predictor. 1.•.\. Repeat Prob. 2.7-3 using the central-difference predictor. Repeal Prob. 2. 7-4 using the central-difference predictor. i-~. Ri;peat Prob. 2.7-5 using the central-difference predictor. l (1, Repeat Pmb . 2.7-6 using the central-difference predictor. + Solutions for pmhlc111s in this ch11p1c1 til l' , 111hcr tedious and should be handled using a jt• 1-1111111 Vlllllplltl'I,
238
Direct Numerical Integration Methods
5.2-7. 5.2-8. 5.2-9. 5.2-10.
Repeat Prob. Repeat Prob. Repeat Prob. Repeat Prob.
l
2.7-7 using the central-difference predictor. 2.7-8 using the central-difference predictor. 2.7-9 using the central-difference predictor. 2.7-10 using the central-difference predictor.
l111p. 5
~.4-5.
5.2-13. Re~at Prob. 4.4-4 (plane truss) for 20 time steps, using the central-d·rn . 1 erenlr predictor with !::.t = Ti./20.
5.3-1. Confi_nn the iterativ~ results of Example 5.3 in Table 5.3 for both the avera 'l'
~.4-6. ~.4-7.
~.4-8.
~.4-9.
g
and hnear-accelerat1on methods.
5.3-2. Repeat Prob. 2.7-2 using iteration by the average- and linear-acceleratiori methods.
~.4-10.
5.3-3. Repeat Prob. 2 ·7· 3 us1·n g I·terat·10n by the average- and linear-acceleration
.4, 11.
5.3-4. Repeat Prob. 2 .7-4 us1·ng 1·terat1on · by the average- and 1· J methods. mear-aece eration
~.4-12.
S.3-5. Repeat Prob. 2.7-5 using iteration by the average- and linear-acceleration
.4-13.
5 ..1-6. Repeat Prob. 2. 7-6 using iteration by the average- and linear-acceleration
,4, 14.
methods.
methods.
methods. 5.3-7. Repeat Prob. 2.7-7 using iteration by the average- and linear-acceleration methods. 5.J-8. Repeat Prob. 2.7-8 using iteration by the averageand linear-acceleration methods. 5.3-9. Repeat Prob. 2.7-9 using iteration by the averageand linear-accelerat ion methods.
5.3-10. Repeat Prob. 2. 7-10 using iteration by the average- and linear-accelerat ion methods.
,•l-15.
8,4- 16.
4 17.
5.3-11. Confirm the iterative results for the plane truss of Example 5 4 · T bl 5 4
.· bh h . .mae . , using ot t e average- and Imear-acceleration methods. !i.J. J2. Repeat Prob. _4.4-3 (plane truss) for 20 time steps, using iteration b thl average- and ltnear-aceeleration methods with !::.t = Ti./20 = ti. y
4 Iti.
!i.J-1.l Re~eat Prob. _4.4-4 (plane truss) for 20 time steps , using iteration by !ht· average- and linear-acceleration methods with !::.t = Ti./20.
·I 19.
5.4- 1. Confirm the a_pproximate results of Example 5.5 in Table 5.5 ~
b h !ht· or ot
•I lO.
!i.4-2. Rlipcat P.rob. 2.7-2 using direct linear extrapolation by the avcrnge- and li tll'llt Ut'l't•lerut 1011 met hods.
5.4-J ,
Rl' Pl' HI Jlmh . '2.7 l usinR cli1vt·t lint•11r cxtrnpol111ion hy tlw uwrugc- und lint•ru 1111'1 hrnJi,
lll'l'l' ll'l llt IOII
239
t 4-4. Repeat Prob. 2.7-4 using direct linear extrapolation by the average- and linear-
5.2-11. Confinn the approximate results for the plane truss of Example 5.2 in Table 5.2 2 5, -12. Rep~at Pro~. 4.4-3 (plane truss) for 20 time steps, using the central-differenl·c predictor with !::.t = Ti /20 = r1 •
average- and hnear-aeceleration methods.
Problems
,,t,l l .
acceleration methods. Repeat Prob. 2.7-5 using direct linear extrapolation by the average- and linearacceleration methods. Repeat Prob . 2.7-6 using direct linear extrapolation by the average- and linearacceleration methods. Repeat Prob . 2.7-7 using direct linear extrapolation by the average- and linearacceleration methods. Repeat Prob. 2.7-8 using direct linear extrapolation by the average- and linearacceleration methods. Repeat Prob. 2.7-9 using direct linear extrapolation by the average- and linearacceleration methods. Repeat Prob. 2. 7-10 using direct linear extrapolation by the average- and linearacceleration methods. Confirm the approximate results for the plane truss of Example 5. 6 in Table 5. 6, using both the average- and linear-acceleration methods. Repeat Prob. 4.4-3 (plane truss) for 20 time steps, using direct linear extrapolation by the average- and linear-acceleration methods with !::.t = Ti / 20 = t1 • Repeat Prob. 4.4-4 (plane truss) for 20 time steps, using direct linear extrapolation by the average- and linear-acceleration methods with !::.t = Ti. / 20. The equations of motion for the average-acceleration method may be converted to the form: S Di+ t = Ai+•· Derive e.:pressions for the matrices Sand Ai+t to be used in this approach. The equations of motion for the average-acceleration method may be converted to the form: C Di+ 1 = Ai+1 . Derive expressions for the matrices C and Ai+ 1 to be used in this approach. The equations of motion for the average-acceleration method may be converted to the form: M Di+1 = A1+ 1 . Derive expressions for the matrices M and Ai+ 1 to be used in this approach. The incremental equations of motion for the average-acceleration method may be converted to the form: C !::.Di = !::.Ai. Derive expressions for the matrices C and !::.Ai to be used in this approach . The incremental equations of motion for the average-acceleration method may be converted to the form: M !::.D1 = !::.Ai · Derive expressions for the matrices M and !::.Ai to be used in this approach. The equations of motion for the linear-acceleration method may be converted to the form: S * D1+ 1 = A *1+1 . Derive expressions for the matrices S* and AJ+1 to be used in this approach. T he equations of motion for the linear-acceleration method may be converted to the form: C * D 1 , 1 = Aj, •· Derive expressions for the matrices C* and Aj+ 1 to he used in this approach . 'f'hc cqu utinns of motion for the linear-acceleration method may be converted to lhl' f'w 111: M"' i)1 A 1, 1 , Dt• iiw expressions for the matrices M* and AJ+ 1 to ht• IISl'd 111 lhl~ 11ppm11~· h
240
6
Direct Numerical Integration Methods
5.4-22. The incremental equations of_ motion for the linear-acceleration method may h . expressions . converted J . Denve for the matrices ( •• d AA* to the form:. c * .AD·J = AA* an
I
to be used m this approach.
5.4-23. The incremental equations of ~otion for the linear-acceleration method may h D enve · expressions · converted to the form· M* d AA* b . M* . AD . J -- AA* J • for the matri<.'l'"' an
I
to e used m th1s approach.
Framed Structures
5.5-1. ~he ~rations_of, motion ~or the Ne':mark-/3 method may be converted to 1hr . orm: D1+ 1 - A 1+ 1. Denve expressions for the matrices S and A- to be " I 1+1 usu m th1s approach.
5.5-2. iThe equations . CD. _of' motion for . the Newmark-/3 method may be converted to 1hr
. 0 rm: J+ 1 - A1+ 1. Denve expressions for the matrices Cand A. t b ,, I m this approach. 1+1 o e use,
5.5-3. :0~:~1ttons : 1;1otion fo~ the New~ark-/3 method may be, conver,ted to thl' .
. 1+ 1
A 1+ ,.
used m this approach.
Denve express10ns for the matrices M and A. 1+ 1
t I o ll
5.5-4. The incremental equ,atio~s of motion for the Newmark-/3 method b . may econ verted to the form· C AD · - AA D . AA' b d . . . J J· enve express10ns for the matrices C u111I 1
to e use m this approach.
5.5-5. The incremental equ,atiol)..S of motion for the Newmark-/3 method b may e con vcrtcd to the form· MAD - AA D . AA' t b . . . J J· enve expressions for the matrices M u111I J
o e used m this approach.
0.1 INTRODUCTION I\ framed structure consists of members that are relatively long (or slender) to their cross-sectional dimensions. Points where members intersect, li ce ends of members, and points of support are called joints of the structure. In ll11ile-element terminology, we refer to the members as linear (or perhaps curvilinear) elements, and we call the joints nodes. Thus, the matrix analysis of II nmed structures (1] becomes a subset of the more general theory of finite 1•1l'ments [2] for discretizing and analyzing continua. However, a framed struc1111 c inherently is divided into elements, unless it becomes necessary to subdivide 11wmbers into smaller elements. T here are six distinct types of framed structures that designers use to resist , 01nmensurate sets of loading systems. Six commonly occurring types of force 11wtems are: (a) parallel-coplanar, (b) concurrent-coplanar, (c) general-coplanar, (ii) parallel in space, (e) concurren.t in space, and (f) general in space. Each of 1h1• rramed structures illustrated in Fig. 6.1 is specifically designed to resist one 111 lhesc load sets at its joints. When the structures are loaded, their joints 1111d<.·rgo lranslational displacements (corresponding to forces) and rotational 1tl11placcmenls (corresponding to moments) . In addition, certain types of internal 111111,us (or stress rcsullants) arise in the members, depending on the type of · (111l'111rc . We will bric lly describe the characteristics of each type of framed t1lll'IIIIL' 11ppcnri11g in Fi!' , () . I , l'VCll lhough beams and plane trusses already h11w hl'l'll 11~l·d 101 1•x111Hpll·i, 1111d pmhll'IIIN iu pn•vio11s (' haptcrs .
1 rnnpared
242 Framed Structures
I
y
/ z
1
.t
/ /-x
.
I
2 /
z
/
(a)
i (b)
y
-I
-
/
--x
z
/
i
(c)
z
t2
/
/
3
(d)
·-
/
. t2 I e -... ..... /,.-; (e) (I)
l<'l1t11n• 6.1 pl11111•
Typl·~ ul 1111 11 , 1
11111111•; (cl) wlcl
•
6
,
(1•) :,1:,:./',',',',lH·~ur,1,· ~, (11) rn,111i11111111- 111·11111, Ch) pl11111· hllNN, (l') "•
" P1111·
1111111•
1
,t
oc. 6.1
Introduction
243
Figure 6. l(a) shows a straight continuous beam, subjected to a parallel1•oplanar set of forces in thex-y plane, which is a principal plane of bending . Any 11pplied moments must act in the z sense, as implied by the double-headed arrow 111 the right end of the beam . This type of loading system produces at a free (unrestrained) joint a translation in the y direction and a rotation in the z sen se. l'hese displacements are indicated in the figure by the arrows numbered 1 and ) at a typical joint j. From flexural theory, internal member actions (generalized Ntrcsses) within the beam are a shearing force in they direction and a bending moment in the z sense. If a force applied to a beam has a component in the x direction , the resulting internal axial stress must be combined with the flexural ~tresses due to bending, even though the analyses are uncoupled . Similarly , Fig . 6. l(b) depicts a plane truss in the x-y plane with forces 11pplied to its members as well as its hinged joints. As with a beam, if a moment IN applied to a member, its direction must be parallel to the z axis. Reactions at lhe pinned ends of members caused by loads become forces at the joints when lheir signs are reversed . Thus , the joints resist systems of concurrent-coplanar lorces, which produce axial forces in the members. Due to the accompanying 11xial strains, joint displacements become translations in the x and y directions , nN indicated by the numbered arrows in the figure. For a member with loads 11pplied directly to it, the axial stress from the truss analysis must be combined wi! h the flexural stresses arising from local bending , although the analyses are uncoupled. T he rigidly connected plane frame in Fig. 6. l(c) carries a general-coplanar •r( of forces in the x-y plane, while applied moments act in the z sense. Resulting di splacements at a free joint are translations in the x and y directions and a 111lution in the z sense. Internal actions occurring in the members consist of an 11xiul force, a shearing force , and a bending moment. Figure 6.l(d) illustrates another type of planar structure, called a grid, which lies in the x-y plane. This structure usually has rigid joints and is designed 11, resist forces in space that are all normal to its plane (or parallel to the z axis). 11 lollows that any applied moments have their vectors in the x-y plane. Displace1111•11ts a t a typical free joint consist of rotations in the x and y senses and a 111111slation in the z direction. Internal member actions are a shearing force, a hl'nding moment, and a torsional moment, or torque. The space truss shown in Fig. 6. l(e) is similar to a plane truss, except that th1• llle mbers may have any directions in space. This type of structure carries l111 1·1~s at its hinged (or universal) joints that are concurrent in space. The forces ell t111g on members and joints may be in arbitrary directions, but any moment uppl k·d to a member must have its vector normal to the axis of that member. The 1r11No11 f'or this restricti on is that a truss member is incapable of resisting torque. I llNpl111·1·mcnts at u free j o int ure three components o f trans lation in the x, y, and d111·l'lio11s. Dlll: to npplicd louds, u 111c1ulwr of II space tru ss may have local lln1111· 111 two pr111c1p11I plum·:, of lwmlinA "" Wl'II "" 1111 uxiul for1·c from the truss 11111tly1i1N
245 244
Framed Structures
Chap.
8
Figure 6.l(f) shows a space frame, which is the most versatile and complicated type of framed structure. Locations of joints and directions of members are completely arbitrary, and the structure is designed to carry forces that are general in space. At a free joint there are three components of translation (as in a space truss) and three components of rotation as well. Internal member actions consist of an axial force, two shearing forces in principal planes of bending, two bending moments in the same principal planes, and a torsional moment. A typical prismatic member in each type of framed structure has stiffnesses, masses, and equivalent nodal loads drawn from those of the axial, torsional and flexural elements described in Sec. 3.4. Such member properties are first expressed in local directions and then transformed to global directions by the rotation-of-axes technique explained in Sec. 3.5. Next, we can assemble member contributions to form stiffness , mass, and load matrices in the equation!! of motion for the whole structure [see Eqs. (3.5-14), (3.5-15), and (3.5-16)]. These equations are solved for joint displacements in structural directions , and the displacements pertaining to individual members are rotated to local clirections. Using such local displacements, we can find internal actions at any point for axial , torsional, and flexural deformations with Eqs. (3.4-48), (3.4-50), and (3.4-52). Of course, internal actions due to static influences at time t = 0 must be added to those associated with dynamic response to obtain the total values. Support reactions can also be calculated from member end-actions if desired . In this chapter we deal only with linearly elastic framed structures having small relative displacements and small absolute rotations due to dynamic loads. We also assume that there are no interactions among axial, torsional, and flexural deformations. Because continuous beams and plane trusses have been discussed before, no separate sections are devoted to them. However, Secs. 6.2 through 6.5 give further information about plane frames , grids, space trusses, and space frames. Then dynamic analysis programs are discussed in Sec. 6.6 for these types of framed structures, as well as for continuous beams. Next, Secs. 6. 7 and 6. 8 cover methods for reducing the number of degrees of freedom for beams, grids , plane frames, and space frames. Finally, Sec. 6.9 describc11 specialized computer programs that use these reduction methods in the dynamic analyses of plane and space frames.
!Jue. 6.2
Plane Frames
K'
=
[~t !t]
r, O 12 I Sym. 2 O 6L 4L l -------------~----,, O O l r, o -12 -6L i O 12 2 0 6L 2L2 I O - 6L 4L
y'
Z,
z'
y k2
x,,
- --- -
--- - -
7 k3
t
kl
k
I IY,. I
I i3/
6.2 PLANE FRAMES
Figure 6.2(a) shows a typical member i of a plane frame with local (primed) and global (unprimed) axes. The three numbered di spl acements in local directions ul each end consist of a translation in the x' direction , a translation in the 1•' direction, and a rotation in the z ' (or z) Sl' IISl' . Ass11111i11g the memht~r is prh. matic, we cun write its<, X 6 stiff1wss 11111l11 x Im locul dlll'l't11u1s us
j1
/ lb)
l•'IMUl'I'
(i,l
Phuw I111111l'
w, (1t) lm·nl
111r111I
dill'l 11 nnN, (h) ~lohnl dh cc1io11s.
(1)
246
Framed Structures
uc. 6.2
which is partitioned in accordance with joints j and k. In matrix K' the dimen sionless ratio r1 is AL2 / lz, where / , is the second moment of area of the crosH section with respect to the z axis. We form the stiffness matrix for a plane frame member by adding contributions from an axial element and a flexural element , which were derived in Sec. 3.4 [see Eqs. (3.4-4) and (3.4-24)). Note that the terms from these two types of elements are uncoupled for local directions. Similarly, we can form the 6 x 6 consistent-mass matrix in local direc tions, as follows: 140
0 pAL 420
I
156
Sym.
I
I
I
2
0
22L
4L
70
0
0
I
140
0 0
54 - 13L
13L
I
O O
----- - --------, - -----3L
I
2
I
f4
where
f
= L
lllr
,
=
[bx'] by•
_
, _
L dg
df3
,
_
d/5
L dg
, _
(4 )
(611)
'l'ht !'unctions in this submatrix for the j end are
df6 = M, df6
/, () 0 I I /, I,, ()
(7b)
L dg
(8a)
+ 1)
_ g)
L
=M
(3g2 _ 2g)
= -pM2
(8b) (8c) (8d)
z
111 which the subscript M replaces the subscript b. Note that no integrations are
1111
follows:
(9) 111 which Cx
= [ -;Y
~l 6 ~J
Cy
(10)
. h f th' 3 x 3 rotation matrix R consist of direction cosines I 1·1111s m t e rows o is · . S 35 ply matrix 1 .. ' andz' withrespecttox,y, andz. Asm ec. · , weap I I lllWS X ' y '
u 11 11d
its transpose to ob:i:
(6h) 1 ' // , .
(3g2 _ 4g '
L dg
PM6 - M, dx'
where
io whkh the dimensionless coordinate is ~
_ g)
= M, dfs = _ 6M,(g2
PMs - M, dx'
R
J
(g3 - g2)L
L
= M, df3 = M
(5)
[/1O !20 hOJ
/6 =
3g2
11•quired for evaluating the~e formulas. d l d atrices to global (or strufln order to convert stiffness, mass, an oa m . . R 1111111) directions, we compose a 6 x 6 rotation-of-axes transformation matnx '
which contains forces (per unit length) in the x' and y' directions, as indicatt•d in Fig. 6.2(a). The 2 x 6 matrix of displacement shape functions in Eq. (3) ha~ the partitioned form
f. =
+
= M, dfi = 6M, (g2
dfi
PM2 - M, dx'
PM3 - M, dx'
(3)
-2g3
~=~:a~:~::t:~y also be subjected to ~ime-varyF~ng ~o~c(e)n]tr~~~ P. 11 moment M at any pomt [see ig. . a . d lnrces Px· an y' as we as a . :Uanner similar to that for distributed , uncentrated forces may be han~led m .a . d On the other hand, for lurces b , and b . expect that no mtegrat10ns are reqmre . ' nd x Ille moment vecyt,or M , we need first derivatives with respect to x off2,hJs, a /ri to determine the following equivalent nodal loads:
(2)
In this expression, the vector of time-varying body forces b'(t) is
b '(t)
Is =
ax:~~::e
156 -22L 4L2
fTb'(t) dt
=g
. . ·n Eqs . (6b) and (7b) are drawn from Eqs. (3.4-1) and (3.4-17) l•1111ct1ons given 1
Again, this matrix is obtained by adding terms from those in Sec. 3.4 for an axial clement and a flexural element [see Eqs. (3.4-5) and (3.4-26)). 1n addition, equivalent nodal loads in local directions due to distributed forces on a plane frame member may be calculated from
Pl (t)
247
Plane Frames
RTK'R = [Kii
Kik]
(ll)
Kk1 Kkk
Also , for the k end we huvr
(711,
M
(12)
Framed Structures
Chap,
8
and
Pb(t) = RTpb(t) = [Pb/t)J
Pbk(t)
(Bl
Perfonning the multiplications indicated in Eq submatnces · (11) yields the stiffness
= L3z
Kkk
= -L3
EI, [r1c; (r I
-
+
12c; 12)CxCy
6Lcy
10c.
6.3
Grids
II' such forces happen to be in structural directions instead, a conversion to 111ember directions must precede use of the formulas in Eqs. (16). After stiffnesses, masses , and equivalent nodal loads for individual members have been transformed to structural directions , we can assemble them in the 111anner given by Eqs. (3.5-14) and (3.5-15). For this purpose we must calculate the global indexes j l through k3 shown in Fig. 6.2(b) , as follows :
Sym.J
r1c; + 12c; 6Lcx
kl
(14a)
4L2
El [-r1c;- 12c; -(r1 -12)cxcy 6Lcy ] -(r 1 - 12)cxcy -r1C Y2 - 12c2x -6LCx -6Lcy 6Lcx 2L2
Kkj
1
jl
EI [r1c; + 12c; Kii = L3z (r1 - 12)cxCy -6Lcy
(14b)
Sym.J
-r1Cy2 + 12c; -6Lcx
= pAL [140c_~ + 156c;
M 11
(14c) 4L2
Mkj
420
=
kk
22Lcx
pAL [70c; + 54c; 420 l6CxCy I3Lcy
AL [140c; + = L_
M
420
.
+ 156c;
140c;
16cxcy 70c;
+
54c;
- 13Lcx
(l 5 a)
= 3j
k2
=
- 1
j3
3k - 1
k3
= 3j = 3k
(17)
ltxample 6.1 (ligure 6.2(b) includes a unifonnly distributed body force by sin Ot, acting in the y 1lircction. Let us calculate the equivalent nodal loads in structural directions due to this lnnuence. First, we transform the body force to member directions, using a 2 X 2 rotation
b' (t)=Rb(t)=[-::
: J[~]bysinOt=[: } ysinOt
(a)
Then we apply Eq. (3) to obtain
4L2 - I3Lcy] 13Lcx - 3L2
(b)
(15b) which are forces and moments in local directions. As the last step, Eqs. (16) give
156cY 140c~
+
Sym.J 156c;
-22Lcx
[wi~~;;~ ~t;,Jrs
Pb (t) = {O, 6, Lex, 0, 6, -Lex}~~ sin Ot
(15c)
4L2
OI course, K;k = K t and Mk = MT bee . Fl11ully' Eq. ( l 3) results in ~he foni ' . aus; matnces Kan? Mare symmetric .
=
2
j2
2
-16cxcy 22Lcy
Pbi(t)
2
111atrix, as follows:
Sym.J
-16cxCy -22Lcy
= 3j = 3k -
These indexes serve as subscripts for placing terms into the stiffness, mass, and load matrices for the whole structure.
Also, multiplications given in Eq . (1 2) produce the consistent-mass . submatrices
.
l
249
248
of equtvalent nodal 1;·:::
(c)
lly inspection, we can see that these results are correct.
0 .3 GRIDS
1
p 1,3
p/,~(\,l
p /14C',
P1ik(t) -
r
p/,1<\· 1 1,;,,,-.
( 16h)
1
/1 t,r,
11~·11· Wl' i111ply 111111 hud y lon'l'S lll'l' >,:iv1•11 111 tlw d l11•1 lh111Nol lm·11l 1111·111lwr ;llWS
'l'o make the analysis of a grid similar to that of a plane frame, we place the iltructure in thex-y plane, as illustrated by the typical grid member in Fig. 6.3(a). 'l 'hc three numbered displacements in local directions at each end of the member 111 l· H rotation in 1he x' sense, a rotation in they' sense, and a translation in the 1 (rn z) direction. If the memhcr is prismatic, its 6 X 6 stiffness matrix for local 1ll1l'l'tio11 s
lwro1111•1-1
250 Framed Structures
110. 6.3
Grids
251
I
I
0
K' =
[
Ki KJk] K;j K;k
=
4L2
g_ __Q_ ___:-.§f___l]_ L3
-r2L2
0
0 0
2L2 6L
0
111 this matrix the dimensionless ratio r 2 is Glx/Ely, where Ix is the torsion , ,mstant of the cross section and l y is its second moment of area about the y ' axis . I krc we combine stiffnesses from a torsional element and a flexural element, as 11ivcn by Eqs. (3.4-12) and (3.4-24). In this case the torsional and flexural terms ,ut· uncoupled for local directions. We can also write the 6 x 6 consistent-mass matrix in local directions as
Sym.
/
I
Tr~L-2 __ _ I
-6L I 0
4L 2
-12 / O
6L
12
140r~
y
y'
M'
=
x,.
[M» MJk ] = pAL Mkj Mkk 420
-----
4L2 -22L
0 0
I
Sym.
156 1
---------- ---~------0 - 3L2 - l3L
?Or:
0 0
0 J 140ri 13L I O
4L2
54
22L
J
O
(2) 156
ri
= J / A is the radius of gyration squared. Terms in matrix M' come where I10 111 consistent mass matrices for torsional and flexural elements [see Eqs. ( 1.4- 15) and (3 .4-26)]. Furthermore, equivalent nodal loads in local directions for a grid member 111ny be obtained using the previously stated expression (6.2-3) z, z' (a)
However, in this case the vector of time-varying body forces is
y
b' (t)
x,,
------ --
tk2.
k1
= [ ::-]
(3)
1'11is vector contains a distributed axial torque mx· (per unit length) in the x' dli cclion and a force b, (per unit length) in the z direction, as indicated in Fig. <1 \(a). As before, the 2 x 6 matrix of displacement shape functions in Eq. 111 2· 3) has the partitioned form
f = [f;
fd
(6.2-5)
f; are switched, and the sign of reversed. Similarly, functionsf5 and/;; in submatrix fk are interchanged, and tlw sign of f 6 is reversed. Thus , the submatrix tj becomes 11111 now the functionsf2 and f 3 in the submatrix
i3/ i
z
11 - - - - - -- - - - L - - - - -x
/1 ls
/
0
(b)
Odd llll'1t1hc r: (11 ) lm·ul dh n 111111N, Ch) 11luh11l dlt1•l'l in11~
~] (4)
,11111 !he s 11h111n1rix 1·4 is
21>2
253
I rumod Struoturoe
.I,
log
0.11
Speoo Trusses
(JI ./11 h
0
I {)
=
1111,
(8a)
O
- (f'
g )L 2
O ] - 2g3 + 3g2
l
~or a grid member, we may also have cone
(8b)
.
P, apphed at any point, as shown in Fig. 6 3( ) ;;trated act10ns Mx', Myi, 111111 P, may be treated in the same mann . a . e moment Mx' and the Im , . er as m , and b but w·th t · . 1 x " ou mtegrat11111 ut the functions in Eqs. (4) and (5 ) H negatives of first derivatives with re~:~~~~ ;?r the mome~t My'. we need lh rows of Eqs. (4) and ( ) b P. of the functions m the scl'oiul 5
l40r;c; + 4L2 c; 22Lcx
. ' ecause positive rotation . th , negative slopes. Thus, we have s m e Y sense are equnl 111
-M, dfi __ My' d/2 T dg y dx' PMS=
P M6
-M ,d(-/5) Y dx'
6M , --tce - g)
(C,h)
= My,
= - M , dfs = Y dx'
=
df5 2 L dg - My,(3g - 2g)
My, dfs - 6My, ( 2 L dg - T g - g) = -p;,,,3
K
- E1y (- r2c; + 2c;)L2 -(r2+2)L2CxCy [ -6Lcy
kj-u -
Kkk -
he;+ 4c;)L
ElY
U
(r2 - 4)L2cxCy
[
111cmber.
For this example we can integrate in accordance with Eq. (6.2-3), as follows:
Jr[~J # 1
pi,(t) = L
0
cos Ot
= {0, -L, 6 , 0 , L, 6}
Lb,
l2 cos Ot
(a)
(-r2c;+2c;)L2 6Lcx
J
6LCy -6Lcx -12
S ym.
'l'hcn Eqs. (6.2-16) produce
pb(t) = {Ley, -Lex, 6, -Ley, Lex, 6}
~~ cos Ot
(b)
(7h)
J
Ciel - 6Lcy 6L<\ 12 S11111 larly, th1.; consistent-mass subrnal . • . f lions lll'l' • • rices or 11 1•, Id riwmhcr in slructun1l din•c ,. .
A uniformly distributed body force b, cos Ot acts in the z direction on the grid member 111 Fig. 6.3(b). Find the equivalent nodal loads in structural directions at the ends of the
(711)
12 X
(r2c; + 4c;)L2
156
Subvectors of equivalent nodal loads in structural directions for a grid member 111 c the same as those in Eqs. (6.2-16) for a plane frame member. However, the lll'tions in each subvector become a moment in the x direction, a moment in the y direction, and a force in the z direction. Assemblage of stiffnesses, masses, and equivalent nodal loads follows the pultem described in Sec. 6. 2 for plane frames. Global indexes j 1 through k 3 used as subscripts for this purpose appear in Fig. 6.3(b), and we may calculate their values with Eqs. (6.2-17).
Sym.J
-Cri + 2)L2c cy
2
(8c)
ICxample 6.2 (C,d)
/\gain, these terms ~o ~ot require integrations. Because the gnd hes in the x 1 the sequence x-y-z, rotation-of-axes-~rfn:~i:d. the nodal displacements art• 111 plane frame. The resulting terms in stiff ~hons_ are the same as those fo1 n become ness su matnces for structural directio1111
2 Ely[(r2c; + 4c;)L 2 Kg = U (r2 - 4)L cxcy (r2 c; + 4c;)L2 6Lcy -6Lcx
Sym.]
It is easy to see that these values are correct.
6 .4 SPACE TRUSSES
A typical member of a space truss appears in Fig. 6.4(a), having universal hinges 111 ioints j and k. Due to this idealized type of connection, rotations at the ends
111 the 111c111lwr llt'l' rn1111idcrcd to be immutcrinl for the analysis of the truss . Local
264 f r nmnd Struoturn 1
II 4
1-1111
8p11co I r11u
266
n1
1
11 and z ', in conjum;tion with axis x ' o f the member , de fine two principal
1ln111•H of bending. At each end o f the member we see three numbered arrows I 11 lnrnl directions, representing translations in the x ' , y ' , and z directions. If 1111 111l· rnber is prismatic, its 6 x 6 stiffness matrix for local directions is '
1
K' [K» Kkj
1(
0 0 Sym. I 1 0 0 0 K]k =EA - --- --- --+- ----- L - 1 0 0 11 K kk I 0 0 O I' O 0 0 0 olo 0 0
J
(1)
N111lll:ro terms in this matrix, pertaining to translations 1 and 4 in the x 1 ill11•l·tion, are drawn from Eq. (3.4-4) for an axial element. However, most of 1111' lt•ims in matrix K' are zero, because a truss member has no joint stiffnesses 111 dtn:ctio ns perpendicular to its axis . In a similar manner, we can write the 6 x 6 consistent-mass matrix in 1111111 directions for the member, as follows:
2
z (a)
0
2
I I
Sym .
O O 2 -_ 6pAL 1--0-01_ 2______ 1
M'
y
(2)
I
0
k2
1
o o
p
-7'
k3 /
-
k1
0 10
1
2
Io o
2
I Ir I l' we use terms for the x ' direction that were derived previously for the axial i1,~11u.:nl in Eq. (3.4-5). Also, due to accelerations in they ' and z' directions, the 11lht1r consistent-mass terms in Eq. (2) are the same as for the x ' direction. Equivalent nodal loads in local directions caused by distributed body l 111 l·l\S on a space truss member are calculated as
k
I I Y;, I
(6.2- 3)
X
h11 this type of member, we can accommodate three time-varying components 111 lt11e loads, z
Figure 6.4
(b)
b'(t)
S
pace truss member: (a) local directions; (b) "lobal direct· c,
IOl!S .
=
{bx• , by' , b,,}
(3)
I r 1111s in th is column vector are forces (per unit length) in the x', y 1 , and z 1 ,lhc•cti ons, as shown in F ig. 6 .4(a). The matrix of displacement shape functions 111 11.q (6 .2-3) may again be stated as
r
Iii
f1 I
(6.2-5)
257
256
Framed Structures
Chap. 11
Space Trusses
In these expressions the symbol e represents a unit vector in the direction
However, we now have two 3 x 3 submatrices,
f; - I,!, - [
Sec. 6.4
i !~}
Indicated by its subscripts. In particular, the unit vector ex· is (8a)
(4) where
und
C
Y
fk
t
= l)/4 =
1 01 0OJ f4
0
[0
0
(5)
L
t,
PM2 = M,· df, dx'
= M, , df, = - M,,
PM3 = -My• df1 dx'
L dg
=-
(6a)
L
My• df1 - My, L dg -y
PM6
= M,, di4 = M,, dx'
=
-M , df4 Y dx '
df4 L dg
= M,, -
I
df4 _ _ My' = L dg L - PM3
I
t•,· )( ,..
(8b)
(8c)
R, we have (9)
In which the direction cosines of axes x', y', and z' are listed in rows 1, 2, and 3. For the purpose of transforming stiffnesses and equivalent nodal loads to structural directions , we form a 6 x 6 rotation-of-axes operator R as in Eq. (6.2-9). Using matrix R as indicated in Eq. (6.2-11) , we find the member stiffness matrix in structural coordinates to be 2 Cy
Cxe, eyCz -e; - cxcy 2 -exey -ey -cxe, -eye,
I
2
e, -exc, -eye, _ 2 e,
I I I I
Sym. (10)
- - - - - - - - - - - - - ---,-- - - - - -
(6d)
c~ I i CxCy Cy2 I I exC, Cye,
1
c2
z
Notice that only the direction cosines of axis x' (the axis of the member) Influence these terms. Applying matrix R in accordance with Eq. (6.2-13) pmduces the equivalent nodal loads -
A~
'()
-
Pb t - R Pb t ()
[RTPbj(t)J RTPbk(t)
(11)
('/a)
which are now in structural directions . Of course, the detailed forms of the terms 111 1>1i (t) depend on the nature of the loads o n the member and the location of the 1hlrd point p. Finally, the consistent-mass matrix for a space truss member IN l11vniin11t with rotation of axes, jl1Sl HS it is for ll plane truss member (see
('/I))
S1•1-. I , 'i ),
J
11 11d
(',,•
ZJ = ..J._ L
A similar description may be given for the unit vector ejp, using the coordinates of points j and p. Collecting the three required unit vectors into a rotation matrix
(6c)
As w '.' )1 c?~centrated. forces' these terms need no integrations. I o (01111 a rotat10n matrix for a space t point p (in add ition to j and k) to d fi 1:1ss. member, we can use a third <>A(n) shows such a point lying in\hne a,Pr~nctal plane of bending. FigurtWht•ncvcr possible, this point would be\~ -y pane and. ~ot ?n the x' axis . lor whid, the coordinates in spa kn en as another JOmt m. the structure , n1111ri x may be found using proper~:t~; th ownt. T(hen the terms m the rotation e vec or or cross) product, as follow~ : _ Ce X C,,, Cr• - ,, c,, X ('1,,
z
(6b)
L - -p M2
= _ My•
C
L
= Y xJk + yJk + z}k
CxCy PM5
Y·k _L
und
1
where Ji = 1 andf4 = as before. A space truss member may al b b· d . forces and moments at any point ;~ e su ~~ctef to time-varying concentrated lhc moments My• and M,, included ~: ;fns1st o the forces P,,, Py,, ~nd P,,. and hccause a truss me b · · g. 6 .4(a). (Note that Mx' is omitted m er 1s mcapable of re · t' · 1 !ruled forces may be handled in th s1s;ng ~n axia moment.) The concen no integrations are required For t:;anner t:nbed for distributed forces, bul and ./, with respect to x,, b~t for M , ::men z' we ~eed fir~t derivatives of Ji In Sec1 6 3 Th Y must use their negatives, as explained us, . . .
=
259 258
Sec. 6.5
Framed Structures
As with plane frames in Sec. 6.2 and grids in Sec. 6.3, we can assemble stiffnesses, masses, and equivalent nodal loads by assessing these properties from individual members. Figure 6.4(b) shows the global indexesjl through k3 used as subscripts in the assembly process, and their values are again calculated using Eqs. (6.2-17). At the end of Sec. 3.5, we mentioned that the analytical model for a truss, as described here, is not really suitable for a structure having only a few members. For better accuracy in the dynamic analysis of such trusses, we recommend the component-mode method discussed in Sec. 10.6.
P'(t)
=
Px·] [cos if!] [-0.6889] Z,: = sinO if! P (t) = 0.7;49 P(t) [
(h)
Then the equivalent nodal loads in local directions become
pplt)
= fJP'(t) = PPk(t) = fIP' (t) (i)
= {-0.3445, 0.3625, O}P(t)
l'or which the subscript p replaces the subscript b. Also, their counterparts in global directions are
Example 6.3 In Fig. 6.4(b) a time-varying force vector P(t) acts at the midpoint m of member i. This force is directed toward point p, which is located on they axis. As in Fig. 6.4(a), the third point p, along with points j and k, defines the x '-y' principal plane of bending. Dimensions appearing in the figure are Xj k
=
Xk -
Xj
= 4
Yik = Yk - Yi = 3
Zjk
=
Zk -
Zj
= 2
YiP = YP - Yi
L
X·k
= \!xJk + Y% + zJk = V29 = 5.385 cY
= 0.7428
(b)
Z ·k
= Yik L = 0 .5571
= {-0.4365, 0.1091, - 0.2182}P(t)
c, = { = 0.3714
I OJ
(c)
.
frame which is similar to . . ' . ~ truss member- but now rotations are included at Joints J and k. As befor~, II space ' . h , , lane and the x, -z, plane. Six the principal planes of bendmg are t e x -y p . t f tr s-
a~~
nu~bered displa~em~nttsh, in~icyat~d e:~~i::~t~nt:\~i~:~;ri~::!:~ ;em~:r, lat1ons and rotations m ex , , f · 6 x 6 the 12 X 12 stiffness matrix for local axes is composed of the o11owmg 1mbmatrices:
r1I, E
0
0 0
(c)
Sym.
12/,
0 0
12/y
0 0 0
0
r2 L2ly
-6Lly
6Lfz
0
0 0
KJ; = L3 0
Applying Eq. (7a), we obtain _ ) 0 8944
· f
1:i ure 6.5(a) depicts a typical member z o a space
(d)
e,, = [ -0.3714 0 0.7428) = [ _ 0 4472 V(0.3714) 2 + (0.7428)2 ·
(j)
. b E (11) Because of the central location of P(t), the equivalent nodal loads ns given y q. · nrc the same at both ends of the member.
For this example the unit vector eiP is eip = [O
= RTPPi(t) = PPk(t) = RTPPk(t)
6.5 SPACE FRAMES
and the direction cosines for its axis are = {
PPi(t)
(a)
=2
Find the equivalent nodal loads for both local and global directions due to the concentrated force. From the given dimensions, the length of the member is
Cx
Space Frames
(la)
4L2 ly 4L2 I, 0
and Eq . (7b) gives e y•
= [-0.4983
1111d
0.8305
- 0.2491)
(f)
-r1lz
Altogether, the rotation matrix becomes
[ R
=
0.7428
0.5571
- 0.4983
0.8305
0.4472
0
E
0.37141 - 0.2491
K'kJ - LJ
(g)
0 .8944
which contnins thl.! uni! v1.·1.·1ors c,,, c,,., and c , lis11.·d 10w wlst•. llsin11 1111.· 1tt·omt·II y of II innglc jmp, w1.• rnn llncl th,• nnitll· 1/1 lo lw 2 , 11 111d 'l'h1.•1l'11111•, 1111' 1111111H1111•11t N111 tlw 1111\' l' wrtm P(I) In 1111111cll11•1.•11111111 1111.•
0 0 0 0 {)
111111
0 - 12/, 0 0 0
6U
0 0 0 6Lfz 0 0 0 0 6Lly 0 - 12/y 2 0 0 - r2L ly 0 2 0 2L ly 0 6Uy 2L2 l , 0 {) 0 0
(lb)
261 260
Framed Structures
Kkk =
r 1 I, 0 0 E LJ 0 0 0
12/, 0
Chap. 6
Sym.
12/y 0 0 r2L2f y 0 6Lly 0 4L2I y -6LI, 0 0 0
( le)
Sec. 6.5
Space Frames
Nonzero terms in these submatrices come from Eqs. (3.4-4) , (3.4-12) , and (3 .4-24) for axial , torsional, and (two) flexural elements. All of the structural parameters in Eqs. (1) have been defined previously. Similarly, the 12 x 12 consistent-mass matrix M' for local directions contains the 6 x 6 submatrices
4L2/, ,
=
Mji
140 0 pAL 0 420 0 0 0
156 0 0 0 22L
Sym. 156 l40r~ 0 4L2 0 -22L 0 0 0
(2a)
4L2
und
M' kj
z
=
(a)
M'
y
kk
p
k1 / , k3 k6
. t i2
k4
I I
I Y;k I
=
70 0 pAL 0 420 0 0 0
140 0 pAL 0 420 0 0 0
0 54 0 0 0 - 13L
0 0 0 0 0 54 70r 2g 0 0 l3L 0 0
156 156 0 0 0 22L 0 0 - 22L
I
x,. ----- ~ /
/
11'1111111• ,,. ~
/ z,k
(2b)
Sym. (2c)
140r} 0 0
4L2 0
4L2
(6.2-3)
I
I
0
0 l3L 0 0 0 -3L2
Nonzero entries in these submatrices are taken from Eqs . (3.4-5), (3.4-15), and ( L4-26), and all of the structural parameters were given earlier. Distributed body forces applied in local directions to a space frame memhcr cause the equivalent nodal loads
I /J3 /1 /4 ______ :_I_ /__,_ _ _ x
,1e
0 0 -l3L 0 -3L2
lhnt urc also in local directions. As for the space truss member, the three
1•11111puncnls of force (per unit length) arc (6.4-3)
Iii) l.]' 11111 I
I1111111
1111"111 Illl l (II) I111111
illt1·1 ll1tt1II,
(It)
11l11h1tl 1ll11•1•IIIIIIN
whk h 11pp1•1 11 nl 11 1-f,l'111~1k p11l 11I on 1111• 1111•111hl' I' 111 Fl1 . (1 , 'i(n) . In Eq . (6 .2-3) the
262
Framed Structures
Chap. 8
Sec. 6.5
263
Space Frames
matrix of displacement shape functions is once more
= [~
f
fd
(6.2-5)
t, - [~
But now the two 3 x 6 submatrices have the forms
~-G and
ck = [fio~
~J 1 ~ ~ ~ ~J 0
!2
0 0
0
!2
o
ls
0 0 0
( 4)
= {f,·, P y', P,,}
(6)
w.hich. require no integrations. Furthermore, three moment components in local
<11rcct1ons are
M'(t) = {Mx·, My', M,,}
(7)
To cal~ulate e9uivalent nodal loads caused by these moments, we set up the following matnx-vector multiplication:
= f:,.,M'(t)
(8)
~hi~~ ~ecds no in~egrations . However, Eq. (8) does require the matrix f , rnntammg appropriate first derivatives of displacement shape functions wi;,; respect to x '. The first submatrix in f ,x' is
-1.x' 0 0
0
/6,x'
Ax·
0
0 0
0
}J
(10a)
.,
2 I: .,
(10b)
To convert stiffnesses , masses, and equivalent nodal loads from local to global directions, we form a 12 x 12 rotation-of-axes transformation matrix R, us follows:
R- [[
f ; ~]
(11)
The 3 x 3 rotation matrix R , appearing four times in matrix R, is identical to lhat in Eq. (6.4-9) for the space truss member. Again, the third point p is ussumed to lie in thex '-y ' principal plane of bending, as indicated in Fig. 6.5(a). lJsing matrix R, we apply Eqs. (6.2-11), (6.2-12), and (6.2-13) to obtain the matrices K, M, and pb(t) for structural directions. As for the other types of framed structures in previous sections, the stiffnesses, masses, and equivalent nodal loads are assembled from member conldbutions. To transfer terms from the member arrays to the structural matrices, we need the global indexesj 1 through k6 shown in Fig. 6.5(b). Their values are l'alculated by the formulas
(9u)
0
wlwrc
dx'
M,,.
0
h.x·
= df6 = 3 J:2 -
Functions / 1 and / 4 appearing in Eqs. (9a) and (1 Oa) are not differentiated with respect to x ', because they are to be simply multiplied by the torque Mx·. On the other hand, nonzero terms in the second row of each submatrix are the negatives of first derivatives that multiply the moment My'· Finally, nonzero terms in the third row consist of positive values of first derivatives that multiply the moment
(5)
pf,(t) = fTP'(t)
0 fi.x•
- Js,x·
I' ' J 6,x
Due to these forces, the equivalent nodal loads become
~..,· = [~ ~
0
(3)
o -A o
Ii
f4
-/3
. Figur~ 6.5(a) also .shows the possibility of six time-varying concentrated ~ctlons ap?he~ at any pomt on the space frame member. Three force components m local d!fecttons are given in the column vector ·
pl,(t)
0
where
0 0
Functions Ji throughA in these arrays were defined previously in Eqs (6 2-6b) and (6.2-7b). · ·
P'(t)
0 0
j l = 6j - 5
j2 = 6j - 4
j 3 = 6j - 3
j4 = 6j - 2
j5 = 6j - 1
j6 = 6j
kl
= 6k -
k2 = 6k - 4
k3 = 6k - 3
k4
= 6k
5
- 2
k5
= 6k
- 1
(12)
k6 = 6k
which arc used as subscripts in the assembly process.
h ,,'
d/i
"I
I
111 rnld1lto11 , 1111· ~1'1'111111 , 111111111111 11 IM
.h ,·
l "I
I
.,A/: ., I I
ll:xum1>lc 6.4 .
(%)
l •l~ll ll'
6 'i( h) shows
II
ti11 ll' v111ying unifrn 111ly distributl'd force of intensity b1 (t) , acting
111 tl w 1• di1 t·1·t 1t111 1111 tht• N(ltll 1' I 1111111• 1111•111h1•1 I .111•111 io11Nol pnints .I, k, nnd p arc the same 111, 1111 thr ~plll'l' IIII NII 1111· 111(!1'1 111 I ·.ll.llllll'k Ii \ ( hllN, lhl' 1ot11t1rn1 1111 111 ix
264 Framed Structures
0.7428 0.5571 R =
[
- 0.4983 0.8305 -0.4472
Chap
0.3714] -0.2491
O
li ons a~sing the rotation matrix R , we can find the components of by(t) in local dirl'l
-+:,J ~ [~::~:},(,)
lul
Then apply Eq. (6.2-3) to obtain
p;,ir)
= L
f
fJ b'(t) d(
= {0.2786, 0.4153, 0, 0, 0, 0.0692Il}lby{t)
(h)
Hild
pJ,k(t) = l
I >YNAPF for plane frames, DYNAGR for grids , DYNAST for space trusses, DYNASF for space frames . They represent the dynamic counterparts of the 1111ic analysis programs described in Ref. 1. First, let us consider the task of generating Program DYNACB for con1l11uous beams , using Program DYNAPT for plane trusses as a guide. Both types ,ii structures have two displacements per node, but those for a beam consist of H translation in they direction and a rotation in the z sense, as indicated in Fig. II I(a). Modifications of the structural data for plane trusses (see Table 3.2) are minimal. The y coordinates of nodes can be left blank, and we need only add lhe moment of inertia ZI(I) of the cross section to each of the lines for element l11formation . In addition, the meanings of terms in the nodal restraint list be1111ne: NRL(2J- 1) = restraint against y translation and NRL(2J) = restraint IIHllinst z rotation. Dynamic load data for continuous beams may also be specified as ftlOdifications of those for plane trusses (see Table 4.1). However, the two nodal 11t'lions for a beam are a force in the y direction and a moment in the z sense, u~ shown in Fig. 6.6(a). Because forces applied to a flexural element act only 111 the y direction, the linearly varying line loading depicted in Fig. 6.6(b) tt•quires only two parameters (BLl and BL2) for its definition. The equivalent
rlb'(t) d( y
= {0.2786, 0.4153, 0, O, O, -0.0692Il}lby{t)
,,
;
f
265
Programs for Framed Structures
,111d
0.8944 still pertains. We shall detennine equivalent n d 1 1 directions caused by the distributed force. o a oads for both local and gloh11I
b'(t) - R b(t)
nc. 6.6
(l
I
(A, )2, ,
.h ~~x~~~~r:~~~~i.n (~:~~~~~e: nfio:dal loads in member-oriented directions at joints} uml
t / i
OJ
RT p,,J(t) = [ 0 R T p;,j(t)
= {O, 0.5000, 0,
-
-
-
- x
(A, )2,
-0.03095L, 0, 0.06190l}Lby(t)
(d)
'
(a)
lllld
p,,k(I)
= [ ROT
OJ
R T Pbk(l)
,
= {O, 0.5000, 0, 0 .03095l, 0, - 0.06 190L}Lb,.(t) (i•) I 1•1111N in these vectors consist of equivalent n0 d I I . . . . tu l1111f1·d dlrct:tions. a oads at Joints J and km struchUl'
y
8.8 PROGRAMS FOR FRAMED STRUCTURES
In a 1111111111•1· r11wlopo11s lo lhut fi>r Proorar11 l>YNt\l'T ( • • d1•v •I • " sn I op w1111n111•r ()1C1frn 1111, for llrl' dyr1111111r u1111ly<11 ~ ol rlw lr111111•d ~r,11r r1111·~ 'llr1·N1· profr11111N1111• n1111wd l>\'Nt\< 'II I
~· • . ., 7 ., . ) we c1111 li v1• oflll'r • I , I • . YPt'\ c,
•"'t t ,
or 11u1t111110111, lit·a rttN,
I
/
J_
b,,
ii 11111111111 l l 11 I IO _ x 0 k
Jill I 111111• 1111 1·11111 l11111111N 111111111 (II ) 111111111 l11111IN ,
(h) 11111• 11111d
1111 1•1!111111111
266
Framed Structures
Chap. 9
f
od;l loads at points j and k may be taken as the solution for Prob 3 4 _10 AI ;~Y.rograthm Df YNACB the only scale_ factor needed for ground ·ac~eler~tio:<:~ ' so e actor GAX must be omitted. Example 6.5
'loc. 6.6
l'hc tenn x;(t) in this formula is the distance that the load has traveled along element i 111 time t, as illustrated in Fig. 6.7(b) . Because the functions in matrix fare cubic, the 1•quivalent nodal loads given by Eq. (a) will also be at least cubic in time, depending on how x;(t) varies. For a moving load with constant acceleration, the fonnula for x;(t) is
Figure 6. 7(a) shows a simply sup rt d · . . po e pnsmattc beam, divided into four flexural elements of equal lengths A from left to right so that .. t mo~~ng load _P of constant magnitude traverses the span translational resp~nses oft~: at various speeds [3].
~~=~ :~ ::2e 1;
1
;t:~nc~~on _of)time. We will calculate the llll pomt caused by the load moving
Equivalent nodal loads for this prob) b i . . ment shape functions for a flexural eleme~~r:?Eq~ ~~~~l~t:i~t:,1!tfo~~:;~placcP&;(t) = Pr[x;(t)]
(a)
y
1
2
4
• ©
5
~-/----x
x;(t) = Vopt
~x(t)----1., i + - - - X;(t) ----..i~.
t t- - -__________ p
1-/ - - 0
___,;,
Pb3
/ /'
(I ))
l•'l111m,
,,.7
2-
(b)
(i - 1)£
f = 0), and a0p is the value of its constant acceleration. Substitution of Eq. (b) and the !unctions from Eq. (3.4- 17) into Eq. (a) produces the desired equivalent nodal loadspb1 through p1,4 indicated in Fig. 6. 7(b). Of course, we need to extend Program DYNACB to accommodate a moving load. The load parameters NLN and NEL (see Table 4.1) are augmented with a third paramtitcr, IML, which is an indicator for a moving load. If IML = 0, there is no moving load; hut if IML = 1, a moving load is present. We also need the additional load data: (d) Moving load (P, VOP, AOP) to be input for the case when IML = 1. Subprogram I >YLOCB reads and writes this data and calculates nonproportional equivalent nodal loads in accordance with Eqs. (a), (b), and (3.4-17). We must also modify the loads processed by either Subprogram NORMOD or Subprogram NUMINT by adding to them the equivalent nodal loads due to the moving load. If desired, we could further extend the code to handle more than one moving load simultaneously. Now let us assume that the beam in Fig. 6.7(a) is reinforced concrete and has the physical properties:
E = 3.6
X
103 k/in. 2
7
2
4
L = 4£ = 240 in.
4
P = 10 k
p = 2.25 X 10- k-s / in. 2
/,
= 3375 in.
where the units are US. Figure 6.8 shows computer plots of Di, at node 3, obtained by the modified version of Program DYNACB and using Subprogram NUMINT for 3 lhe response calculations. The case of constant velocity (VOP = 3.585 x 10 in. /s) produces a maximum translation of 0.4036 in. at node 3, and that for constant accelerntlon (AOP = 1.071 x 105 in./s2 , with zero initial velocity) gives 0.3625 in. Their 111tios to the static deflection of PL 3/ 48£/, = 0.2370 in. (due to the load applied grad1111lly at node 3) are 1.703 and 1.530, respectively. The values ofVOP and AOP selected In thi s example both produce travel times equal to the fundamental period of the beam, which is 66.95 ms.
y
p"'
t2
1unning
(a)
z
+ aop
In this expression Vop is the velocity of the load when it first contacts the beam (at time
A= 12 x 15 = 180 in.
P., t /-.
267
Programs for Framed Structures
(11) Sl111ph• lw11111 wifh n111vl11n lune! , Ch) rq11lv111«,n1 11ucl11J lrnulN
)(
Second, we shall describe Program DYNAPF for plane frames, using Programs DYNAPT and DYNACB to guide us. As shown in Fig. 6.l(c), a typical node j in a plane frame has three displacements, which are translations 111 the x and y directions and a rotation in the z sense. Table 6.1 lists the form of structural data required for plane frames. As with continuous beams, each line t•ontuining clement information must include the moment of inertia ZI(I) of the noss section. Also , a line of nodal restraints has three types, which denote 1t•strnints against x trunslution, y trnnslntion , and z rotation, respectively. 'l'nhk' c, .) shows dyn11111ic lond dntu Im pln,w r,·11111l'S. In this cusc, the lines lor 11ii1l11I dlNplm•t•111t·11t., 1111d w locitlt·~ rn11t11i11 tllll't' q1111111i1it•f'I i11slt•nd or two .
268
Sec. 6.6
Framed Structures
TABLE 6.2 Dynamic Load Data for Plane Frames
VOP CONSr .
0.40 AOP CONST.
No. of Lines
Items on Data Lines
l
!SOLVE, NTS, OT, DAMPR
Initial conditions (a) Condition parameters (b) Displacements (c) Velocities
l NNID NNIV
NNID, NNIV J , D0(3J-2), D0(3J-1 ), 00(3J) J, V0(3J-2), V0(3J- l ), V0(3J)
Applied actions (a) Load parameters (b) Nodal loads (c) Line loads
I NLN NEL
NLN,NEL J, AS(3J-2), AS(3J-l), AS(3J) I, BLl, BL2, BL3, BI.A
Type of Data Dynamic parameters
0.30
C:
n
~ o. 20 0
z
Ground accelerations (a) Acceleration parameter (b) Acceleration factors•
}
a 0.10
1 I
Forcing function (a) Function parameter (b) Function ordinates •omit when IGA
o.oo
IGA GAX,GAY
1 NFO
NFO K , T(K), FO(K)
= 0.
flexural elements), we find equivalent nodal loads in member directions. Then the equivalent nodal loads in structural directions can be computed from Eqs. (6.2- 16).
(ms)
Figure 6.8
269
Programs for Framed Structures
Translational responses at center of beam.
Example 6.6
The plane frame in Fig . 6. 10 consists of three prismatic members and has an initial load Po applied by a cable connected to node 1. If the cable suddenly breaks, the frame responds to the initial displacements caused by the load Po applied statically. Such a response is the sum of free vibrations of the natural modes excited by the initial displacements; and in the presence of damping, they will decay with time. We shall assume that all elements in the frame are steel W 12 x 85 sections with the following properties: E
No. of Lines
Problem identification Structurul pnrurnctcrs l'h111t· lrn1m• <111111
-
(11) Nodul rn111cli1111tr N
Ch) 1' 11 11111•111 1111111111111111 11 l N111l11I 11·~11,11111~
(1
~
I I NN
NI! Nl(N
106 kN/ m2
A = 1.61 x 10- m
2
p = ·7.85 Mg/m3
l , = 3.01
X
4
10- m
l=2m 4
for which the units are SI. From static analysis , initial displacements due to the load P0 = 10 kN are: (1) at node 1, (Dot , = 0.5697 mm, (Dot2 = -0.2923 mm, and (D0 )1 3 = - 4.739 x 10- 5 ; (2) at node 2, (Dot , = 0.3642 mm , (Dot2 = 0.1149 mm, nnd (D0)13 = 6.598 x 10- 5 . We used these values as input data for Program DYNAPF, with DAMPR = 0.02 and solution by Subprogram NUMINT. Figure 6.1 l(a) shows computer plots of translational responses Di I and Di2 at node l due to the load release; nnd in part (b) we have the bending moments AM3 and AM6 for the j and k ends (nodes I 1111d 2) of clement I . For all of these responses, the maximum (or minimum) values m 'l ' III' at time I O und diminish thcrcuftcr bccnusc of damping.
Items on Data Lines
-
X
2
TABLE 6.1 Structural Data for Plane Frames Type of Data
= 207
Descriptive title NN,NE,NRN, E, RHO
I, :\CI), Y( I) I, INCi), KN(I), AXC I). './,1(1) ' I, NIU ( II JI, NIU ( 11 I), NIU .( II) ·-
2/0 I 11111111«1 i,1111111111111
C l111p
I 11011ntl 111111111 111
l'roi,110111 1111
y
271
(I
I
y
/
1 •
--,,
2
i3
L
z
2L
__L_,
(a)
/ ~l- L-L /
I
Figure 6.10
Plane frame with load release.
lrn sion constant Xl(I) and the moment of inertia Yl(I) replace the moment of 1t1t•r tia ZI(I) on each line containing element information. Furthermore, the
in the nodal restraint list become restraints against rotations in the x and senses and translation in the z direction. Similarly, the dynamic load data for grids are symbolically the same as 1111 plane frames in Table 6.2, but the meanings are different. When the structure 111 11 grid, the first two initial displacements and velocities at a joint are rotational (in x and y directions); and the third is translational (in the z direction). Nodal l1111ds consist of moments in the x and y senses and a force in the z direction, as Nlmwn in Fig. 6.12(a). In addition, the linearly varying line load illustrated in l•1g. 6.12(b) acts in the z direction and is defined by only two parameters (BL1 11ml BL2). Equivalent nodal moments at joints j and k due to the line load are 1•111,ily converted to structural directions using Eqs. (6.2-16) withpb 1 = pb4 = 0. I •inally, the scale factor GAZ for ground acceleration in the z direction replaces lht· factors GAX and GAY. lr 11ns
y
1•
i
bl1
/ - - - -- x
(b)
lt'lgurc 6.9
Load I s on p ane frame: (a) nodal loads; (b) line loads on element.
_Third, let us brie fly examine Pro ram DYN
.
of ~11ds. Because this type of structu! ha h A~R for the dynamic analysis 111oiirn111 will be very similar to Program D~~l~/tplacements per node, tlw flw lh1'l'L' types of displacements at a t . I I .. or plane frames. However , 111 1 11111! \' senses und ~ lranslatio~ in th : P'c,~ . ~~~ l' J ~ gr!d are rotations in tlw
T 0 I 1111 c z ' l l t l 11011 H~ indicated in F'1 6 I ( I l ' 1 Al' llw plu1ll' f'rume structural d11111 1 ', ·· • ?' · '). wt· n1us1 11
' o
II ~111111111111 p11ra11tL'lt·rs . Also', lht•
11,)(umplc 6. 7
I ill· grid illustrated in Fig. 6.1 3 is made of brass and has rather small dimensions and l111uls. The cross section of each prismatic clement is a solid square with side 0 .25 in. , ,11ul lhc applied actions consist of step nodal und line loads. Physical parameters in this I
lllllltpJC UI~'
I•'
1.5 X 10 1 k/111
1
(i
()
'i (1 "
101 k/111.'
f'
R.10
X
10- 7 k-s 2/in .4
212
Framed Structures
'loc. 6.6
0 , 60
~
e e
273
Programs for Framed Structures z
0,40 w
25
0.20
~
o.oo
z
(/J
6
-o . 20 X
I-
< _, -o. 40
(/J
z <
a:: -0,60
I-
(a)
(a)
z
E z
AN3 10
-"
I-
z w I: w
s b ,,
~o
X
z (/J
z
0
-s
I(.)
< 0
I
-10
z
w
( b) 1
1 1gure 6.11
k
Responses of plane frame: (a) displacements; (b) bending moments. (bl
I,
4 in.
,.
0.02 k
A = 6.25 X 10- 2 in. 2 b1..
Ix= 2!y = 6.510
X
10- 4 in. 4
= 0.004 k/in.
Figure 6.12
Loads on grid: (a) nodal loads; (b) line load on element.
llie member Jscc Fig . 6.3(a)], and the second is a bending mome~t about they ' a~is. Mu idmum (or minimum) values of the nodal translations are 0 .1 157 m. and 0.08451 m.; 2 3 k-in . and -8. 181 x 10- k-in. 111111 lhosc for the moments arc 9.333 x 10
111
where the units arc US.
=
~ilh lhis ~al a we ran Program DYNAOR, 11.~ing DA MPR 0.05 and rcspousi• l'llle11lu11011s by Subprogrnm NORM OD . Translulions of' 11rnks 2 ond 3 in the z dirccliwi 111t• plolll·~l lu Fin , 6. 14(11), 1111d lhc 1110111c nl sAM1 1111dtl~,, 111 1h1 I. cud (nodc4)of'dcnil·III I 11pp1•111 111 p11,1 (h) ol 1hl· ll1111 rl'. T lw ll1sl ol lht•Nr 11111111r11I N iN111t,1qm• nhout llil' ·1 ' lllliN
J/o11 1lh, we shull disl'uss Progrum DYNJ\ST for spucc trusses, a?ai.n using J>YNJ\PF 1111· l'o111p11iiso11s. 111 till s l' IINl', tlw tllrt'l' types o f displace-
1•1,1w11111
274
Framed Structures
Chap. 6
275
Programs for Framed Structures
Sec. 6.6
NOOE 2
z 0 -10
j3
C
::; 0. 05
t
4
•-i2
(f)
z
/i
0
~ o.oo ,_
j1
T
<( ~
(f)
~ -0 ,05
a:: ,_ I
-0 . 10
N
(a)
X
Figure 6.13 Grid with step loads.
ments at a typical node j are translations in the x, y, and z directions, as shown in Fig. 6. l(e). In order to make the plane-frame structural data in Table 6.1 apply to a space truss, we must add Z(J) to each line of nodal coordinates. On a line of clement information, the moment of inertia ZI(I) is replaced by an identifier IP that indicates whether a third point p is necessary for locating principal planes of bending. If IP = 0 , the x '-y' principal plane in Fig. 6.4(a) is taken to be parallel to they axis. However, if IP = 1, the next line of data must contain the coordinates XP, YP, and ZP of the third point p. (This type of data is shown in Table 6.3 for space frames.) Last, the terms in a line of nodal restraints denote restraints against x, y, and z translations . Dynamic load data for space trusses are similar to those for plane frames, hut the meanings are different. For a space truss the initial displacements and wlocitics at a joint refer to translations in the x, y, and z directions. The thrcl' nodal loads at a joint are forces in the x, y, and z directions, as depicted in Fig. h. I .5(u). Also, the three sets of linearly varying line loads appearing in Fig . <, I Ci(b) require six parameters (BLl through BL6) for their definitions. Eq ui v nlt•nt nodal loads due to these line loads are easily obtained by extending Eqs. (•I . IO 4). To accomodate three components of ground acceleration , we must also mid the scale factor GAZ to the data table. 1,:xn1111>lt• 6.8 t•tpllH' C, IC, dt•pil ts II spun• 11 IINN with nine p11s11111t11 !'lt' llll'IIIN , h11v111g l'(lllll l •l ' IONN hl'I ti111111I llll'IIS 111111 r11111t111s1·d 111 hf1-•h sti1•11pt h 111,111111111 I h111 ~11111 tllll' i~ s11hj1·rtt·il to
C
-;- 0 .10 Q.
, ..:,f,
n
AMS
0,05
AH4
1-
z
w
I:
w
~
o.oo
8
4
T Cm s l
w z
UJ -0-05
z
0
I-
~ -0 . 10 I
0
z w (b)
Figure 6.14 Responses of grid: (a) displacements; (b) moments.
llu cc sets of double triangular impulses P(t) at locations indicated in the figure. For this
plllhlcm the physical parameters are 2
f
117 X 106 kN/m
A
9
X
10
4
m
111 whtl'h SI units 111t· impli1•d
2
p = 4.49
P.,m = 100 kN
Mg/m3
L = lm
276
Framed Structures
Chap.a
Sec. 6.6
Programs for Framed Structures
TABLE 6.3 Structural Data for Space Frames Type of Data Problem identification Structural parameters Space frame data (a) Nodal coordinates (b) Element information Coordinates• of point p (c) Nodal restraints
a Requued
when IP
No. of Lines
277 y
Items on Data Lines
1 1
Descriptive title NN,NE,NRN,E,G,RHO
NN NE
j2
t
J, X(J), Y(J), Z(J) I, JN(I), KN(l), AX(I), Xl(I), YI(l), ZI(I), IP XP, YP, ZP J, NRL(6J-5), NRL(6J-4), NRL(6J-3) NRL(6J-2), NRL(6J-l), NRL(6J)
I NRN
/i j3
L
"""
= I.
1
o -i1
~7'---X
y
(a)
z
z
Figure 6.16 Space truss with triangular impulses. b,2
Fifth, let us build upon all of the previous programs to describe Program DYNASF for space frames. With this type of structure, the six displacements at a typical node j consist of three translations and three rotations in the x, y, and z directions, as illustrated in Fig. 6. l(f). Table 6.3 contains the structural data required for space frames. As for grids, the shearing modulus G is added to the line of structural parameters; and each of the lines for nodal coordinates includes Z(J) , as for space trusses. Element information includes AX(I), X I(I), YI(I), 7J(T), JP, and (optionally) the l'Oordinn11·H of II lhird poiul ,, 1111 lornli111• ptinclpul pl111ws of'lx·nding. In thc lines
y
----x I UI
We used this data in Program DYNAST, with IP = 0 for all elements, DAMPR = 0.02, and called Subprogram NUMINT for numerical evaluation of responses . Figure 6. l 7(a) shows a computer plot of the applied force P (t); and resulting axial force-time histories for elements 5, 6 , and 7 are given in Fig. 6. l 7(b). Maximum (or minimum) values of the axial forces in these three elements are 325.6 kN, - 325.6 kN, and 175 .l kN, respectively.
{111
278
Framed Structures
! f!C. 6.6
279
Programs for Framed Structures
100
TABLE 6 .4
so
~
Dynamic Load Data for Space Frames No. of Lines
Type of Data
a.. ~ 0
s
c::
0
Dynamic parameters 10
15
20
T
I.L.
Cms)
8 -so _J
a.. a.. <
- 100
(a)
1
7
1 NNID NNIV
NNID, NNIV J, D0(6J-5), D0(6J-4), ... , D0(6J) J , V0(6J-5), V0(6J-4), ... , V0(6J)
Applied actions (a) Load parameters (b) Nodal loads (c) Line loads
1 NLN NEL
NLN,NEL J , AS(6J-5), AS(6J-4) , . .. , AS(6J) I, BL!, BL2, .. . , BL6
Forcing function (a) Function parameter (b) Function ordinates
200
V)
"Omit when IGA
IJ.J
~
0
ISOLVE , NTS, DT, DAMPR
Initial conditions (a) Condition parameters (b) Displacements (c) Velocities
Ground accelerations (a) Acceleration parameter (b) Acceleration factors•
.:!
Items on Data Lines
I 1 l NFO
IGA GAX, GAY, GAZ NFO K, T(K), FO(K)
= 0.
,,~ those for a space truss element. However, the equivalent nodal loads for a ~puce frame element must be obtained by first resolving the line loads into 1 umponents that are parallel to its axis and two principal directions. Using these 1111nponents and the solutions from Probs . 3.4-1 and 3.4-10 (for axial and tl1•1lural elements), we can find equivalent nodal loads in member directions. I hen the equivalent nodal loads in structural directions may be computed from l1q. (6.2-13). Finally, all three scale factors GAX, GAY, and GAZ must be 1,1 1vcn for ground accelerations.
0 IL
_J <('
~ -200
( b)
Figure 6.17
Space truss: (a) triangular impulses; (b) responses. y
'i°'
nodal ~estraints, the terms indicate restraints against three translational and 1 llt·c mtat1o~al displacements in the x, y, and z directions. Dynamic load data for space frames in Table 6.4 are more extensive lhun 101 Hn,Y, 0th.er type of framed structure. The six types of initial displacements uud Vl' 10~·1t1es rn the table consist of three translational terms, followed b ~hll'l' 1111111,onul terms. For nodal loads we have three force co dy . . . mponcnts an thrl'l' 1 11ortil 11l crn.n(mncnts 1n the .t , y , and z direc tions, us shown in Fig 6 f 8(•i) 'f'l . fhll'l' sl'ls of l111i:11rly vr11 y i11g Iinl' louds in Fig . (1 I 8(h) ait· chosl'n ;0 1111 ::.
,;l' tl;l. ~
fl'l1111n• 6. IK
I .oudN 011
N[llll'!'
lru,rn· (11) 110<1111 lo11dN; (b) line loads on clement.
281
Programs for Framed Structures
Sec. 6.6
E = 6.5 X 10 3 k /in. 2 L = 10 in .
p = 1.71 x 10- 7 k-s2 /in.
G = 2.4 x 103 k /in.2
Ix = 2/y
A = 0.01 in .2
= 21, = 1.667
X
5
10- in.
4
4
und the units are US. Using this data, we ran Program DYNASF, with IP = 0 for all elements, l>AMPR = 0.05, and Subprogram NORMOD for response calculations. The plots in tlig. 6.20 represent the forcing function for ground acceleration and time histories of the lnmslational responses at node 2 in the x, y, and z directions. Maximum (or minimum) values of Dji, Dj2 , and Dj 3 are 0.07213 in., 0.04685 in. , and -0.09977 in., respectively.
y
"'II)
l0
'
C
~----x
s
z
D I-
z
<(
0::
(b)
Figure 6.18 (cont.)
w -' w
0
u u
<(
200
400
800
60
l 0 00
l 00
T
ms
-s
Cl
K11.11mplc 6.9
z
:::>
The space frame in Fig. 6.19 has three rismatic . . st•cli
0 0:: 0
-1 0
I
X
(a)
y
0 .1 0
DJ3
C
2
('\J
w
o.os
Cl D
z
1<(
0. 0 0
Cf)
z 0
~ I-
-o. os
<
-' (/) % ;;5
:11
( b)
, l' ll,IIIII' II jll
- 0 , IO -
'lp 1111 1111 1111
wli lt
}• 111111111 11,, i Ii l lllltlll
DJl
l
282
Ch
'rnmncl Structurn 1
nI
6.7 GUYAN REDUCTION
The concept of · . matrix co11de111·01/011 11 , I mg the number of unknown ·d .. . •' I~ II Wl'll known procedure for rrd ~pp~ications no loss of accurac/::sl~1~~·:~r011ts 111• ''. stuti_cs problem . With NU is simply Gaussian elimination of d. I o m the ,~duct1on, because the mcrh analy · · · tsp acemcnts in m tri ti . s1~, a s1mtlar type of condensatio . a x onn. For dy11u111 bnngs m ~ ad~itional approximation. n was introduced by Guyan [41, whl Startmg with static reduction we w . . free displacements in the partition~d fo;te act10n equations of equilibrium fi
[:: ::][!;] -[~] SaADA
Solving for th
U, • 'l',w U11
+ SanDo = A8
(7b)
I "' 11 lor static analysis, this relationship is correct only when actions of type A 1111 11111 exist [see Eq. (3)]. However , Eq. (7a) follows the finite-element theme 111 "~lnve" and "master" displacements. Differentiating Eq. (7a) twice with 11 , p1'l'l to time produces
(8) I 111 tI1e purpose of reducing the equations of motion to a smaller set, we can form 1111 transformation operator
111 which 18 is an identity matrix of the same order as S 88 . Substituting Eqs. (7a) (8) into Eq. (4.4-1) and premultiplying the latter by Tl gives
111111
(10)
D;1 = S,U (AA - SABDo)
111 this equation the matrices S18 and Al still have the definitions given in Eqs. 1,1 nnd (6) . However, the reduced mass matrix M%8 is found to be (4)
in which
and
(.,)
Al= Ao -
s-
SBA M 1 A A From Eq .. (4) we see that Eqs . (2) have been (ll) ~~ly_the independent displacements in vecto r~uc~ to a smaller set involvi111l as m Eq . (5) is a modified version oft r .n: e reduced stiffness matri, reduced action vector A* in Eq (6) ?e ongma1 submatrix Saa. Also 1/1" 8 · th t contams tenns d'f · ' " a are considered to be equivalent loads f mo t ymg t~e subvector A,, Fu~ennore, Eq. (3) may now be view~ ?pe B due to actions of type A. required to find vector DA exactly fro s the back-substitution formulu T · m vector D ummg next to dynamic reducti B· undamped equations of motion for free od~' Jwe recall from Sec. 4.4 that the tsp acemcnts arc
Ml) -ISD 11
A (4 .4 I)
lll'W approxi murion lhur 1hr
(7a)
(9)
Substitute Eq (3) · t Eq . mo . (2b) and collect terms to obtain
Then assume m,
ol typ1• II , llli l111l11w11,
= AA
e vector of dependent displacements DA in Eq. (2a) yields
.
p1 111ll· t1l 011 ll1<1Sl'
IIJ
Here the subscript A denotes the d' 1 subsc · t B ti tsp acements that are t be 1· · np_ re ers to those that will be r . o . e immated, and th of equations, as follows: etained. Now rewnte Eq. (I) as two ~or
SMDA + SAaD8
213
C111yn11 Unclw 11011
ry pe A
llll'
M%o = TJM Ta= Mso + T1MAB + Mo;1T;10 + T io MMTAB
(11)
A~ mentioned before, all of the condensed matrices in Eq. (10) are approximate. II clumping is to be included in the equations of motion, a reduced damping 111111rix 0 also can be derived, which has a form analogous to Ml 8 in Eq. (11). When applying Guyan reduction to framed structures, we usually choose 111t111ions at the joints of beams , plane frames, grids, and space frames as the ,lqit·ndent set of displacements. However, the method can be used in a much 111111c general manner for various discretized continua. That is, any arbitrarily -rll'l'lcd set of displacements may be referred to as type A, while the remaining ,li~placements become type B. The trouble with this generality is that a good , hoice of "slave" and "master" displacements is not always obvious. Even with l111111ec.J structures there are cases when joint rotations are more important than t11111slations and should not be eliminated.
c;
l1 11111nple 6.10
1111' fixed beam in Fig. 6.21(a) is divided into three flexural elements, each of which has 1lw ~111ne properties£, I , p, and A. By Guyan reduction , we shall eliminate the rotations 111 11mb 2 and 3 and retain the translations. The reduction will be followed by a ,1l111111onnl annlysis that is compared against exact results. For 1h is cx111upk, the assembled structural stiffness matrix (without consideration
284
Fromod Structures
Sec. 6.7
285
Guyan Reduction
of restraints) is y
,----------- -------, / 6
, 3e
/
I / -6
I
s
s
= 2£/
e3
/
CD
ze2
-3e e
/
Sym.
I
r--------4---- ----I 12
I
2
I
3
I I I 1 2 1 I , '2" I I '-t:J I I --------+---------1------- ---l----------1 I I I I
3e 0 0
2
0
I I I I
0
0
0
0
0
o
4e
-6
-3€
3€
e
2
12
I I I I
O
4€2
I I I I
'3"
-3e
o
o
0
o I, 3e
e2
4
6
I - 6 I
5 6
i
6 -
(a)
I I I I I
'-:!.I
L------- - ~--- ---- - I
4
3e
2e2
7
I
1
8
I
(a)
L - - - - - - - - - - - - - - - - -- - 1
2
3
5
7
8
Similarly, the assembled structural mass matrix becomes
r ------ - ---------- ---~
l
/ 156 I I I I I
pM
2
22€
4€
54
13€
,-
/- Be
I I I
- 3€ l 2
I
CD
I - - - - -- - - --l- -- - -- - - -I 312 I 1 @
o
se l 2
Sym. ,
2
l
3 4
I I I
Ms = 420 L--·-- - -----f----- ------/--- --- ----f-- --- ---- I
o
o
/
54
13e l
312
0
0
/ -13f
-3€2 /
0
/
8€2
/
Q)
/
'----------4----------1 /
/ I
O
O
O
O
o
o
o
2 o !l___- _____ 13e -3e -22e _ _ _ ___ _ ___ _4€2 __ /
2
3
4
54
13€
156
/ !
5
6
(b)
7
- 1
7 (c)
8 Figure 6.21
8
(a) Three-element fixed beam; (b) mode I; (c) mode 2.
(h)
4e
2
Dushcd boxes in Eqs. (a) and (b) enclose the contributions of elements 1, 2, and 3, which urc drawn from Eqs. (3.4-24) and (3.4-26). Moreover, the joint displacement indexes for the problem [sec Fig. 6.2 l(a)] are listed at the right side and below the matrices. As the first step , we remove the first, second , seventh , and eighth rows und t·olumns from matrices S, and M,,, because displacements I , 2, 7, and 8 arc rcstrai,wd hy supports. Then the remaining 4 x 4 arrays arc rcarrnngcd to put the rotationul torn1N heforc the trans lutionul terms, as fol lows:
SAA SAD]
S=
[ SaA
Sall
2EI
=
7
Sym.
4
i
6
I
12
3
- 3e o i
-6
12
3
5
e2
4e2
--------j------0
4
3e
6
(c)
5
lU/
Fromod Struoturoe
186
-
pA I' 420
IJe I
()
13e
0
4
6
I
Cl11y1111
ti. /
llcuhmtlo11
"'h1•1l' L 3e. These values arc in error by I O.h 1% 1111d I 2.6%, respectively; and they 1111111titute upper bounds of the exact angular f'n:qucncics \51. The corresponding mode
4
Sym.
I
K('J
Ill ,
6
h11p1·s arc (k)
3
312
54
312
3
5
5 wh1l'11 appear in Figs. 6.21(b) and (c) .
1111llll
Eq . (c) the inverse of SAA is
8
4 -1]
e [
- 1 _
30£/ -1
AA -
4
of S from Eq. (c) into Eq. (5) prod • • Substituting this array and the other submatrices . J2£/ [ Ull 6 St11 = & _ E! = 6El [ 1 2 5e 1 4 5e3 - 11 16 5 <, l
2 -lJ _ [4 l]
~
16 -11] 3 3
5
whk h,i:~ the reduced stiffness matrix . I o reduce the mass matrix we form the tr . 'I\,, lo l!q. (7b) and substituting it into Eq. (9). ;~~~~rmat10n matrix Tn by evaluati11p TAB --
-
s-AA1SAB= -1 5e
1111d
Tn = TAB [ In
J
= I_
5€
[
1 -4
4] _1
1111 cc degrees of freedom indicated in the figure. The 3 x 3 stiffness matrix for unit values of D 1, D2, and DJ has the form
l I
(C)
11ml the accompanying 3 X 3 mass matrix is
4 1
(h)
MAB
5
M
5e
o
111
I
22L
22L
! 732
[El
\JpA
(n)
Ahm , the corresponding mode shapes are
.
'"'
(m)
(i)
l
- 0.5528
Jn
Sym.J
---------l----
l = 3.201, 15.14, 32.68 L 2
5
22.~l ,(,l)/1 I I J
l
8L 2
which M includes pAL for the mass of member 1. With these arrays we can set up 08 Nolve the eigenvalue problem, which yields the angular frequencies
wh11·h is the reduced mass matrix. theEqterms . . 111 th1· Using lrn111 or . (3 .6in4)matrices Tl . S111I and . f M* 1111 , ~e ~an set up the eigenvalue problen, . 1c angu a1 rcquenc,cs found hy this method arc "11 ,J
420
2
8L - 3L 2
1111tl
Wi , 2, 3
319] 3 [1696 319 1696 5 3
1
= pAL
88
111 this cnsc the submatrix 1/J in the lower artition of . . 1111· two rnmaining translational
Mlo = pAC 2100
ll.x111nple 6.11 l lgurc 6.22(a) shows a rectangular plane frame that has the same properties E, !,, p, and \1 l'or each of its three members. Let us find approximations to the fundamental angular ln·qucncy for this frame with and without elimination of the rotations at joints 1 and 2. l'o Himplify the analysis, we shall omit axial strains in the members, leaving only the
(g)
=-~e----0 1
Frequency coefficients /.Li for prismatic beams with various end conditions 1111• summarized in Table 6.5. In each case the beam is modeled by four flexural 1 kmcnts, and the results for the consistent-mass approach (with and without 1 ll111ination of rotations) are compared with those for the lumped-mass method (with elimination of rotations). The table shows that the consistent-mass model p1oduces much better accuracy than the lumped-mass model in beam analysis.
'
(
I)
= f<1>,
I hi• Hl111p1· ol
11111dr
<1>2 3) =
- 0 .5528 I.
I I~ ill11 HfrlllCd in Fig . (i.l.2(h).
- 1
0
-oLJ
(o)
- - ---------,
Frequency C oefficients µ ; for Pr ism atic Beams Mod eled by Four Elements
TABLE 6.5 Support Conditions
~
~
•
•
Mode
•
I
~ I
L
I
f=
3
I
% Error
CM- TO
% Error
LM- TO
% Error
9.870
9.872
+0.020
9.873
+0.030
9 .867
-0.030
~
•
•
•
~
I
I
I
2 3
~
•
•
•
•
II
+0.38
39.76
+0.7 1
88.23
90.45
+2 .5
94.03
+6 .6
83.2 1
- 5. 7
22.37
22.4 1
+0. 18
22.46
+0.40
18.9 1
- 15
62.06
+0.63
63. 12
+2.4
48.00
-22
~
•
• w =~ L1
ff. pA
•
~
12 1.9
I I
+0.83
22. 37
22.40
+ 0 . 13
61.67
62. 24
+ 0.92
123.5
122.4 22.4 1
I I
62.77
+ 2.2
3.5 16
3.5 16
124.8
+0.00
3.5 16
I I
+ 1. 2 + 0 . 18
39.19
86.84 22. 30
I I
-28 - 0 .3 1
+ 1.8
59.25
-3.9
+3 .2
97.40
-19
3.418
+0.00
-2.8
2
22. 03
22.06
+0. 14
22.09
+0.27
20.09
I
- 8.8
3
6 1.70
62. 18
+ 0.83
62. 97
+ 2. 1
53. 20
- 14
15.43
+ 0 .065
15 .40
I I
49.05
I - 1.8
I 91.53
I -12
I
~
I -0.73
39.63
120.9
I
I
39.48
.9
I
J.."IO
I
CM- TR
~ .67
I
frurl
2
Exact (5)
I
I I
15.42
I
2
I
I
3
I 104.2
49.97
I I
15.43 50.28
I 106.6 I
I
+0.065 +0.62
I
+2.3
I
50.56 110.5
I
I
+ 1.2
I + 6.0
- 0. 13
CM: Consistent Masses LM: Lumped Masses TR: Translations and Rotations TO Trm,.,IM.-,. Onl~
"',,
a;
m
'X
:....i
-: ~
C
':
::;
G)
g "Q ::.
C
-< I»
('";
:::. -0 o;
:,
"':.
aci'
-:
o' '"'
-:
0=:
O'Q - ·
3
!;
c. ::s
N
sCl>
~ ~
O
~
-
:, .;" t
ol 3 0 -·
:,:, a,
:,
..,
- -
"':.
::..
,z
,z
:,..
-
-
z
--
a
,. -
(IQ
5-
~
-
...,
o· (>
-
•
-
.::__i
.,
00
0\
~
...,
"O C:
C.
-
l..,
,-
~
0
-
-'
8-
r,
C:
$l.
o·
'<
:,
1----
~
8
l N'
.i "'
s:i i
'-,.'1-1>
\
a.
G
~
§:
o' ~
g
:i
1
~
3 0 & 0
l t
tJ
"'
f'
__l
r>
~
~
SJ"'~
N~
,-
.\
g,
t
"'
JC
"'g. 3
,--. "O .._,
I»
5. ;,<
~
290
Framed Structures
= 24EI _
Sto
36E/
3
Chap. 8
= 84E/ (q)
5L3 5L3 which is just a single term. Next, the arrays TAB and To are found to be L
= -s;J sAB
TAB
1
= _ _l_ [ ] 5L 1
(r)
Sec. 6.8
Constraints Against Axial Strains
291
In this equation the symbols D j t and D j 2 represent the x and y translations of joint j, and Dki and Dk2 are those at joint k. As before, the x and y direction cosines of the axis of the member are denoted by c., and Cy. Similarly, we write the axial constraint condition for zero elongation of a space frame member [see Fig. 6.5(b)J in the following manner: (Dit - Dk1)cx + (Di2 - Dk2)cy + (Di3 - Dk3)c,
and
(s)
_ ToT M
-
569 To = pAL 350
(1)
Now the eigenvalue problem gives ]~ w, = 3.215-/ 2
L
(u)
pA
Cornpari_ng this angular frequency against w1 for the unreduced 3-DOF problem we find the relative error to be +0.16%. '
6.8 CONSTRAINTS AGAINST AXIAL STRAINS
In Example 6.11 we omitted axial strains without difficulty because members of the plane fr_ame wer~ pe1:Pe~dicular to each other. However, for frames with me~be'.s onented ~rb1tranly m_ a plan~ or in space, the matter of neglecting axia l strams 1s not so simple. In this sect10n we will introduce axial constraints in plane and space fra?1es, pri~arily for the purpose of reducing the number of d~grees of freedom m dynamic analysis. The number of constraints introduced will be equal to the number of members m for which axial strains are to bl' neglected. By automatically selecti~g m of the joint translations to be dependent the rest of th~m, we ca? devise a method for reducing the number ol mdepend~nt_eq~ahons of motion. This reduction method may be combined with Gu~an e1Jmmat10n of joint rotations, which was described in the preceding section. 'v!e ~ssume that the frame to be analyzed is underconstrained, so that no c~mphcat1ons from redundant constraints (or superfluous members) need tw considered r6J .
?n
The axial cons~'.·aint condition for zero elongation of a typical member; in a plane frame !see f
(n,,
n11 k,
1
(n,.
n1 ,l1\
o
(1l
(2)
I !ere Dj3 and Dk3 are translations of jointsj and kin the z direction, and c, is the z direction cosine of the member axis. Assembling these constraint conditions into a matrix format for either a plane or a space frame gives
Using the latter operator in Eq. (11), we can reduce the mass matrix to the single term
* M BB
=0
C D1 = 0
(3)
The matrix C in this expression is called the constraint matrix, which contains only positive and negative values of direction cosines for the constrained members. This array is of size m x n,, where n, is the number of joint translations. The vector D 1 in Eq. (3) consists of only free joint translations , because no joint rotations are involved at this stage. Due to the constraint conditions, some of the joint translations in the vector 1)1 will be linearly dependent on others. To determine which translations are dependent and which are independent, it is necessary to investigate the rank of 111atrix C in a systematic fashion. The rank rand the basis (or vector space) of ( ' are found using Gauss-Jordan elimination with pivoting [7]. Although the 1ank of a matrix is unique, the basis is not. Therefore, the choice of dependent translations is arbitrary , and pivoting automatically produces the best selection. Thus, it becomes possible (in retrospect) to partition the matrices in Eq. (3) as lollows:
[C 11
C,2{~:]
=
0
(4)
111 this expanded form the vector D 1 represents r dependent translations, and D2 rnntains the remaining n; independent translations. Because no redundant con~t1nints are included, the rank r of matrix C will always be equal to the number ol members m. Therefore, submatrix C 11 in Eq. (4) is a square array of size 111 X m, and submatrix C 12 is of size m x n;. Multiplying the matrices on the left-hand side of Eq. (4) produces (5)
l, 1mwing that matrix C 11 is square and nonsingular, we can solve for vector D1 111 tl't ms of vector D2 . Thus,
(6) 111
whil' h llw
OJll'l ator
T1 2 is (7)
293
292
Framed Structures
Chap.
8
lluc. 6.8
Constraints Aga inst Axial Strains
TABLE 6.6 Member Information for Three-Member Plane Frame
During the Gauss-Jordan elimination h . identity matrix; and C12 is replaced b pr~~ss, t e matnx ~11 is replaced by an to an identity matrix I of order m / ·11 ~- If the operations are also applied this inverse to be the ~perator ' i wi e replaced by C ;-t We now define
T11
1
= C 11
Joint j
Joint k
Cx
1 2 3
1 3
2
0.7071 0 0.4472
(8)
and the augmented constraint matrix is
.
Ci
=
[C Im]
(())
To confirm ideas regarding the constraint . . frame in Fig. 6.23 for which th b . & ma~x: let us consider the plane ' e mem er m1ormat10n is given i T bl 6 6 T augmented constraint matrix [see Eq · (9)] ior & . n a e · . ho th"1s case is 1
c,
2
= [o.~001
o.101 -1.000 0
4
-0 .101 0 0.447
5
!
-0.101 1.000 0 : 0 -0.894 : 0
o 1.000 0
J
o 0 1.000
Cy
Member
1 4
2
0.7071 1.000 -0.8944
~hown in Fig. 6.18. After the Gauss-Jordan procedure is applied, the matrix in
11.q. (a) becomes 1
c; (a)
-[f
4
0 1.000 0
-1.500 0 -0.500
l'hcrefore,
The numbers above the columns of C indicate the translational displacements
5
2
T,1 = C 1/
1.000 -1.000 0
0 II 1.414 0 II 0 1.000 l 0
[.414 = ~
-1~18]
1.000 -1.000 0
-Lit~] 0
(b)
-1.118
(c)
-1.118
0.5P(t)
y
P(t)
1111<.I
t2
/
1
T12 = -C;-/ C,2 =
l
t P(t)
, - - ----t--- - \ - - - - - -1.
3
4
=
[~J
'" u111111h1•1 pl11111· 11 111111•
(10)
1111' llrst part of D consists of a null vector, representing member elongations 2 urc z.cro) ; and the second part contains the independent translations D .
D
'1111,•u
(d)
I whk h
I 11 11•l111c the vector D to D,, we write
11·1111111• t,,1.1
0 0.500
Thus, the dependent translations are found to be 1, 2, and 5, whereas the l111h.1pcndent translation is automatically chosen to be 4. Note that there is no 111•1,d to rearrange the augmented constraint matrix during this procedure. In preparation for a coordinate transformation associated with axial con~1111 lnts, we shall define a generalized displacement vector D, as follows:
D 3/ _ _ _ _ _ _ __J__ _
[500]
= T,D,
(l la)
(l lb)
294
I nc. 6.8
Framed Structures
In Eq. ( 11 b) the upper part represents the constraint conditions, and the lOWl'f part merely re~roduces D2. Because the generalized displacements are indepcn dent and constitute a complete set, there is also an inverse relationship in th~ form
D, = TcD
( I.? I
where 1 1 Tc= T, 1 = [C,1 -C,1 C12J = [T11 0 12 0
Ti2J 12
1
1 ]
]
(4.4-1)
( 14)
in :Which the subscr~pts rand t denote rotations and translations , respectively lJs1~g Guyan reduct10n from Sec. 6. 7, we eliminate the rotations in Eq. (14) and 1·cta111 the translations, yielding
M 1~D1 + S!Dr
= A;
( I ~)
where
s: = Sa + T~Srr
( I (1)
A t* -- Ar + TTrrA,
( 17)
M!
= Mu
+ T~M,r + M,, T,.r + T~M,, T,.,
( 181
'l'itl' transformation operator T,., in these expressions relates the rotations D to tht• trnnslntions Dr, as follows: '
D,
= T,.rD,
( Ill I
Whl'l'll
S,, 1 S, 1
·· s*rr D,-M* 11 D, +
A* + cTQ
(21)
1
111 which Q is a vector of axial forces in the constrained members. To perform the transformation, we substitute Eq. (12) and its second dl.lrivative with respect to time into Eq . (21) and premultiply by Tt to obtain M
D+ S D =
A
+
C Q
(22)
Writing this equation in expanded form, we have
By segregating rotations from translations, we can write this equation in thr cxpunded form
[M,M,,, M,M,, [D'ti, ] + [S,,S,, S,Su [D'Dr ] = [A'] Ar
~tiffnesses of members from the structural stiffness matrix. Then to satisfy ,•quilibrium at the joints, we must revise Eq. (15) to become
( I II
M~t:ix Tc_ is an operator that can be used to transform action equations from thl' ongmal displacement coordinates to the generalized displacement coordinatl'N for _axial ~onstraints. Notice that submatrices T11 and T12 are generated auto mat1cally m the Gauss-Jordan procedure described previously. . Let us now restate the action equations of undamped motion for fi'l'l' displacements at the joints of a plane or space frame as MD+SD=A
295
Constraints Against Axial Strains
()OJ
Tlw opt·1·11to1 'I~ lst•c Eq . ( 11)1 cun now ht· IISl'd to trun sforn1 the rcdut•t•d 1•q1111tH111s of 11101l1111 i11 llq . ( I Ci) to lht• /'t'1wrnll1t•d dlspl11t·t·1rn·111 rnordinhti•1, I) Ill J1q ( I()) 1!1•11111~1· 11~i11l l'OIISl111int~ JIil' Ill h1· 1111pc1s1•d , Wl' i,lwll 11111it 11x111l
M 11 [ M 21
~ J+ [~11 S21
M,2] [ M 22 D 2
~12] [ 0 S22 D2
J
[Ai] A2
+ [l']Q 0
(23)
Multiplying terms and rearranging the results produces (24)
1111d
Q = M12D2 + S12D2 - A,
(25)
I he barred matrices in these equations have the following definitions:
+ TT1 sf2 S22 = TL S i1 T12 + TT2 S i2 + Si'i T12 + Si'2
(26)
A1 = TT1Ai
(28)
S12 = TT1 st\ T,2
A2
= TT2Ai +
(29)
Ai
M12 = TT1 M~ T,2 + TT, Mf2 -M22 = T12M11 T * T,2 + TT12 M*12
(27)
(30)
+ M*21 T12 + M*22
(31)
1•q1111tion (24) represents a doubly reduced set of equations of motion that can Ill' Holved for the dynamic response of tJ.ie independent translations D2. Then the vn ·tor D2 and its second derivative D 2 may be substituted into Eq. (25) to ,h ll•1mine the vector of axial forces Q in the members. Next, dependent transl111 l1111s D I can be obtained from Eq. ( 6), and the rotations D, are found using Eq. , 11)) . Finally, other internal actions and support reactions may be calculated l1rn11 known relationships. If a damping matrix is included in the equations of motion, its reduction I "i111ilur to that of the consistent mass matrix. Other topics that could be 11111111dcrcd in this section are redundant constraints and nonzero length changes 111 111l' lllbcrs. I lowcvcr, these subjects arc more complicated and of less interest 1111111 thl· 111atlur or Hxinl c:onstrnints, us discussed here. Wht•11t·vt·r 11xi11l st rnins 11n· 0111ittl·d from :urnlyscs of plane or space frames, ,1 h1~11 11l 11l'l'll1'11l'Y ,~ h111111d t11 m·t·111 'l'hr 11l1 nilll·1111l'l' or such discrepancies will
296
Framed Structures
Sec. 6.8
vary from one problem to another. However, for most practical underconstrained frames, the loss of accuracy due to introducing axial constraints is likely to be negligible, except in the columns of tall buildings [8] and similar structures. Moreover, the numerical problem of ill conditioning due to combining large axial stiffnesses with small flexural and torsional stiffnesses is completely avoided. Of course, when the members in a frame are perpendicular to each other (as in Example 6.11), omission of axial strains is easily accomplished without the formal procedure of this section.
AL
~ ~20
M,,
- 8 288
6:
"
S = El, [ L2
"
3.879 -2.121
2
4
2.121
2.121
2.121
3.195
1
2
[ 1412 _ El, -2.121
S - -3 "
L
0.1564
L [
-I
-2.121 -1.585
J
6
2.121
- 2.121
2.121
T,, ==
- s;,. 1 s,,
0.6764
2.980
2.121
-2. 121
-1.692
-0.2301 - 0.3890
S,; ==
:
El,
U
[
_ pAL3 [ 8 M,, - 420 -3
M,,
p/\l}
•I W
r I 92 6 .444
<) ,
2 15 .56 9. 11) }
(h)
9. 192
\\2•1
5 9. 192] 3 5.7 12 6
(ll
4
5 Sym.11
0.808 0.391
0.565
2.552
-1.017
- 0.303
(n)
1.374
:
(o)
A: == {1, 1, 0.5, O} P (t) llq. ( 18) gives 2 [ 317 9 -3 1.46 M" - 420 81.91 1.733
4
to be
-2.189 -1.463
* _ pAL
-3] 3 8 6
(m)
111 11ddition, Eq. (17) yields
,111tl
6
si
9.940
Si mil arly, consistent mass matrices for the members are drawn from Eqs. (6.2- 15), nNNllmbled, and rearranged to give
3
-0.1787 0.2558] 3 -0.6374 0.2650 6
I !Hing Eq. (16), we find the reduced stiffness matrix
5
2.336
5
4
2
"'
(g)
(f)
t'hlln from Eq. (20) the operator T,, becomes
(f)
Syml
(k)
- 0 .04790] 0.2313
S,, = El, -0.04790
2 3
:
A, = {l , 1, 0.5, O} P(t)
5
4
291.2
For the purpose of applying Guyan reduction , we calculate the inverse of sub11111lrix S,, as
L (c)
(j)
8 300.8 62 - 1.6
== .!.[-0.7083
1.414] 3 1.414 4.617 6
L
11
, . p·1g .6. 23 , we see that the parts of the rearranged load vector are I•1om
6
_ El, [ 6.828 s -
Sym
[3~
A, = {O, O}
3
5
4
2
Example 6.12 For the three-member plane frame in Fig. 6.23, let us first set up the stiffness, mass, and load matrices for the six degrees of freedom shown. Second, we shall use Guyan reduction to eliminate the two joint rotations and retain the four translations. Third, by imposing axial constraints we will eliminate three dependent translations and keep the best single independent translation, which was found to be displacement number 4. Last, we shall calculate the response of the reduced system to a particular set of forcin1,1 functions. From Eq. (6.2-14), we determine member stiffnesses (without axial terms), as scmble them, and rearrange the results to produce the following submatrices of thl· structural stiffness matrix:
297
Constraints Against Axial Strains
4
5
Syml
289.1
(p)
1.152 262.0 62.84
6.302
293.6
:
Now we shall furth er reduce the stiffness , load, an~ '.11a~s matric~s to accoun~ f?r mcmhllrs. By rcnrrunpiog 1111d p,1rt1t1omng the stiffness matnx m
,1, 111 111111 strniuts 0 11 the I q (II) Wt' fi11d thnl
299 Framed Structures
298
2
* S11
e
9.940
(24) takes the form
Sym.] I
0.808
2.552
-1.017
I
1305pAL .. D4 420
(q)
1.374
2
:
+
22.41E/, L3
O D4 = 2. A
(a')
l'he eigenvalue problem associated with the homogeneous version of Eq. (a') yields the
5
* = El, S 21 L3 [ -1.463 0.391
Programs DYPFAC and DYSFAC
We now have all the matrices needed to find the response in Eq. (24) and the axial lorces in Eq. (25) for any forcing function. If the function P(t) is a step force A, Eq.
5
El, = L 3 -2.189
[
Chap,
Sec. 6.9
- O. 303] 4 = S *1/
(1)
11ngular frequency: (b')
4
* = El, S22 L 3 [0.565] 4
11nd the response to the step force Pi is P, L 3 EI,
the last of which is just a :ingle term. Similarly, for the load vector in Eq. (o) we haw
A, = {l, 1, O}P(t)
A; = 0.5P(t)
(I)
Also ,
Furthermore, submatrices of the mass matrix in Eq. (p) are
1
Mi,=
~fa
2
[~~;::6
l M*21 -- pAL[ 420 81.91
62.84
2
(c')
.. AL3w2 D4 = 0.08925-COSWt El,
5
289.1
1.733
D4 = 0.08925-(1 - cos wt)
Sym.]
~
293.6
5
A pAL
= 0.6437- cos wt
(u)
Substitution of D4, D4 , and the appropriate barred matrices into Eq. (25) produces
5
1.152 6.302] 4 = Mi2T
Q=
(v)
4 (w)
M;2 = ~fo[262.0] 4
(d')
Then the barred matrices in Eqs. (26) through (31) become
0.4442] ( 0.6454] 1.617 A 0.7370 P, cos wt ( -0.7714 - 0.6638
(e')
which are the time-varying axial forces in the members. To complete the example, we also compute the dependent joint translations, the joint rotations, other internal 1111;mber actions, and support reactions. However, these tasks are straightforward and
1 ould
will be left as exercises for the reader.
4
s,
2
= ~, [
~::~~]
-21.17
A, = P(t) [
4 1.:l ] - 1.118
~
4 EI S22 = e'[22.41] 4
11.9 PROGRAMS DVPFAC AND DVSFAC (x l
3
~
A2 = P(t)[2 .0] 4
(y)
3
4
79 1.31 1 574.3 2 799,7 h11
111 this section we briefly discuss programs named DYPFAC and DYSFAC for dynumic analyses of plane frames and space frames with axial constraints. These p1 ograms use Guyan reduction (see Sec. 6.7) to eliminate the joint rotations and lhl'll further reduce the equations of motion by imposing axial constraints (see '{l'l'. 6 . 8) . Response calculations for the reduced system are carried out using the 11m nrnl-rnodc method , as described in Chapter 4. Programs DYPFAC and DYSFAC both have the same outline, as follows:
4 pAI,
470 11 105 1 4
(/)
Outline of P1·0~1·nms DYPF AC and DYSF AC
1
1111'10' 111111 yN 1111' i111h- 11t•N I ' ') • 11111 I I I Il' llt lh' llll' lllhl'I 1111111!1111~. w h111l'IIH ll1th•x ,1
II1(11t'Nl' III ~ 1hr ltllll'( II 111h Ill 1111111 IIIIIINl11ll1111
lfrud 1111d wil h· Nl111l· tw11I tl11t11 II
S t1111 t11111l p 111 11111t• h· 1P1
301
:too
I 11111111<1
b. .loi111
m, 1101111111
1•111111111111
llYl'I Al
irnit 11V
I (\I
l'111111ll1111t,
C. Joint l'estnli11I N
2.
3.
4.
5.
6.
7.
d. Member 11110111111111111 Porm constraint lransf'o1111111io11 1111111 kc.:s a. Assemble constrnint 11111td x b. Find dependent translations c. Generate transformation matrices Generate, condense, and transform stiffness and mass matrict•s a. Generate member matrices b. Assemble structural matrices c. Eliminate rotational displacements d. Reduce matrices (due to constraints) Determine frequencies and mode shapes a. Convert eigenvalue problem to standard, symmetric forn1 b. Calculate eigenvalues and eigenvectors c. Write natural frequencies (cps) d. Transform, normalize, and write modal vectors e. Normalize modal vectors with respect to mass matrix Read and write dynamic load data a. Dynamic parameters b. Initial conditions c. Applied actions d. Ground accelerations e. Forcing function Calculate time histories of displacements a. Transform initial conditions and loads to normal coordinatl'N b. Determine time histories of normal-mode displacements c. Transform independent translations back to physical coordi,11111 d . Calculate dependent translations and rotations Write and/or plot results of response calculations a. Time histories of independent and dependent joint translation• and rotations b. Maxima and minima of translations and rotations c. Time histories of axial forces and other member end-actions d. Maxima and minima of axial forces and end-actions
Both of these programs perform the reductions optionally. Therefore, tw11 m•w structural parameters, [RO and IAC , must be added to the structural dut11 Within the logic of the program , if IRO ,f:: 0 the rotations arc eli minated; wul 11 IA( ' I O axial constraints arc imposed . Further details regarding these progn11111, 111t· pivcn in Rel'. 9. 1,:,m1111>1t, <,. IJ I •iHIIH'
(1
J,I(11) II l11Ht rnh·N 1111 1111tll•1co11s1rnl111•d pl11111• I 1111m• with I wl•lw cl1·H1l'l' N·of I 11•1
r
1
L
I'/.
1' (,)
G)
3L
12
/.
-l---..+----X 2L~
z
(al
P(t)
P,
t,
0 (b)
ix-member plane frame; (b) dynamic load. Figure 6.24 (a) S
aoa I rru1111d St1uotur 1111
.102
1'1111!1111111
c;h111 1111 1 tll'h
rn11111l~l• 11I I• ti tldl . .. • . . dom ( 12 DOF) . 'l'I1i.l. I1111111• . I Y i onn~ctcd pr1s mt1t1c me111h1:1N' 1111 which have the sa me vah1us ol /• ' I 111 thu .1 di rection at jo ints 2 u11d ,'1' , ' ~ I >y111111 tl c lorces p (1) and 2P (1) !Ill' nppll Assuming th at the fram; is steel , '".11 t II t 1111• v111111 tion of P (1) appears in Fig. 6, '·ltl 1 ' Wl ww ll ll' ll1ll11wlng numerical values ,0 r 1• pan1111r11 1
I ';"''
E = 3·0
X
104 k/'m. 1
A = 30 in. 2 L = 50 in.
fJ
7 .35
X
t2
110111
10- 1 k-s 2 /in. 4
I, '"' 1.0 X 103 in.
10 k
of the th1 cc l'IISl'S W1 Hl I 1\1111 1111 II MpOIIHl' C\IIVC for Case (b) is indistinguishthat fo r Cus1: (11) . 111 11dd l1h111, till' H'HJlllt1Se curve for Case (c) differs by only 11 11111 umount from thu olhl'IH (1•11nt Is 11pprnxl mate ly - 3.3% at the first peak). Thus, 1111 h•vcl o f accuracy shown hy 1hls L'xumple seems sufficient for practical purposes. 1lih
REFERENCES
4
= 2t1 =
Weaver, W. , Jr. , and Gere, J. M. , Matrix Analysis of Framed Structures, 2nd ed., 35 ms
whure US units are implied. . T hi s structure was analyzed b Pro ra . wtlhout reduction (12 DOF)· (b) ·tyh 1· g_ m_ DYPFAC three times, as foll ows t111 •Ii . . ' w1 e immat10n of r t t' (8 ' ~ m1nut1on of rotations, followed b reduc . o_a 10ns DOF); and (c) wllh 6.25 shows time histories of respofse t due to axial constraints (2 DOF). Fip111 or e independent translation D4, as comptth ii
Vnn Nostrand Reinhold , New York , 1980. Weaver , W., Jr., and Johnston , P. R., Finite Elements for Structural Analysis,
1
Prentice-Hall, Englewood Cliffs, N. J., 1984. Yoshida, D . M. , and Weaver, W. , Jr. , "Finite-Element Analysis of Beams and Plates with Moving Loads," Int. Assoc. Bridge Struct. Eng., Vol. 31, 1971, pp.
:~on_
179- 195 .
D,(i n.)
0.2
G) and@
l, 1965, p. 380. , 'l'imoshenko , S. P., Young, D. H ., and Weaver , W., Jr , Vibration Problems in 11
/1'11gineering, 4th ed. , Wiley, New York, 1974. Weaver , W. , Jr. , and Eisenberger, M. , "Dynamics of Frames with Axial ConHtraints," ASCE J. Struct. Eng., Vol. 109, No. 3, 1983, pp . 773- 784. Ocrc, J. M., and Weaver, W., Jr. , Matrix Algebra for Engineers, 2nd ed. ,
!\rooks/Cole, Monterey , Calif., 1983. II Weaver, W., Jr. , Nelson, M. F., and Manning, T. A. , "Dynamics of Tier Buildings," ASCE J. Eng. Mech. Div. , Vol. 94, No. EM6, 1968, pp. 1455-1474. 1 Eisenberger, M., "Static and Dynamic Analysis of Plane and Space Frames with 1 Axial Constraints," Technical Report 48, J. A. Blume Earthquake Engineering Center, Stanford, Calif. , Oct. 1980.
PROBLEMS r,,:.t• l.
Suppose that a uniformly distributed force by>(t) acts on half the length (from x' = 0 to x ' = L/ 2) of the plane frame member in Fig. 6.2(a). Find the equivalent nodal loads at joints j and k for both member and structural directions , assuming that
Xjk
= 4 and
Yik
= -3.
td•.2, Let the concentrated force Py•(t) on the plane frame member in Fig. 6.2(a) act at the point where x ' = 3L/4. For both local and global coordinates, determine the II l,
equivalent nodal loads at points j and k, with Xjk = Yi k· t A moment M, (t) is applied at the midpoint of the plane frame member in Fig. 6 .2(a). If XJk = I and YJk = - 2, calculate the equivalent nodal loads atj and k for both member and structural directions.
f1,L•4, A lri angul ar distribution of force ~b 2(1) acts in they' direction on the plane frame 111c1nhcr in Fig . 6 .2(a). rind the equivalent nodal loads in local and global
o.~
d \rl·CtillllH rm 111 1 und Y/4 4' r, ,,l,,!I. 'l'w11 cum•1•11ti11l1•cl l11H'l'S P ~·(I) 1111· 11pplk d ut the third points of the plane frame
304 I t 11111ml Str 1101111111
r11crnhc1· In Fig h '(II) c ,111 1 directions usii1u I 11 q111 1 ,tit Ill 1111d11l louds in memhcr and ~,1111f JI 11111 I ,1 I 6.2-6. Assume that two 1111 ,, 11 • 1 111 1I) ' " I ut tlw q1111rter points of the pl11111 Ir member in Fi l 11 1 g . 6.2(u). lktt•1111111r i·q11l v111l-11t nodal loads in Joe· I I I . coord mates for x1k = YI*· u u11t 11 • 6. 2·7. A parabolic distribution of force /J (t) . . frame member in Fig 6 2(a) F .. 21s applied m the x ' direction on 1hr pl,11 I d . . . . or XJA and YJk = - I find th . I oa s m member and structural directions. ' e equ1vu t•11t 111 I '
1
ti ''," •
'
"
e
6.3-1. Let a uniformly d'st 'b t d f; the grid member /n ;i u e orce b, (t) act on ~alf the length (L/2 s ,\ ' I I and global directions, ti!:.~:)~ ~al:~a:e e~u~valent nodal loads for both I, 6.3-2. Assume that a concentrated f; P. ( ) . 1k • • the grid member in Fig 6 3(o~ceF z t I~apphed at the point wherex ' IIII · • a1 . or x 1k - 3 and y I d · alent nodal loads in both me b d j~ -: , etermme tht· ,·qui 6 3. . . m er an structural direct10ns. ,. 3. The gnd member m Fig. 6.3(a) has a . x ' = L/4. Find the equivalent nodal 10:~~entbMb(:) applied at the point wh using x jk = YJk· s or ot ocal and global coordi11111
6.J -4. Repeat Prob. 6.3-3 for a moment My(t) in they direction 6 ..l•S. Rep~at Prob. 6.3-3 for a moment Mx(t) in the x direction. <,.J .6. A tnangular distribution of force (I - 1:)b () . . 1 1 member in Fig 6 3( ) F _ !> acts m the z direction on tlw "• Id · · a · or XJk - -4 and y - 3 b · " loads in both member and struct 1 d' I* - ' o tam the equivalent 111111,II ura coor mates.
lk peat Prnh. 6.•I ~ wllh 1111 11111111,•111 111·tl11g in the y' direction at the point I ' /, /5. Ass ume that a unifo1111ly diNtilhuted force b (t) acts on a space truss member in the di rection shown by Fig. 6.4(b). Using the relative coordinates Cj k = {2, 2, 3} nrtd c1,, = {O, 5, O}, delermine equivalent nodal loads in member and structural di rections. fl ~ H. Repeat Prob. 6.4-7 for a triangular distribution of force [bi(t). fl
~
t. Let the space frame member in Fig. 6.5(a) be subjected to a uniformly distributed
fl ~ L. fl
I.
fl ~ 4. fl 8 ~.
fl ~ <,. fl II 7.
fl , H.
fi
(1.4-2. Repeat Prob. 6.4-1 for a un;fo~ly ~istributed ~ segment where L/ 3 s x ' s 2L / 3 . orce b,·(t) applied ove1 flu
fl
for the three unrestrained displacements. Reduce these matrices by eliminating lhe rotations and retaining the translation. Then find the angular frequency of vibration for the remaining system, which has only one degree of freedom.
ly 1~ 1
~/4i
l
11
force by·(t) over its whole length. Find the equivalent nodal loads for both local and global directions, assuming the relative coordinates c1k = {3 , 2, 4} and Cjp = {- 2, 4, - I}. Repeat Prob. 6.5-1 for a uniformly distributed force b,·(t) over half the length (O s x ' s L/ 2). Suppose that a concentrated force Py,(t) is applied atx ' = L / 4 on the space frame member in Fig. 6.5(a). Using the relative coordinates c1k = {2, 3, - I} and c1,, = {-1 , 2, 2}, calculate the equivalent nodal loads in member and structural directions. Repeat Prob. 6.5-3 for a force in the z' direction at the point where x' = 3L/ 4. A moment My•(t) acts on the space frame member in Fig. 6.5(a) at the point where t' = L/3 . Determine the equivalent nodal loads for both local and global direclions, assuming the relative coordinates Cjk = {O, 3, 4} and Cjp = {O, 0, -2}. Repeat Prob. 6.5-5 for a moment in the z ' direction applied at the point where r' = 2L/ 3. Assume that a uniformly distributed force bx(t) acts in the x direction on the space frame member in Fig. 6.5(a). With the relative coordinates Cjk = {I, -2, 2} and c1,, = {- 3, 4, O}, obtain the equivalent nodal loads for member and structural directions. Repeat Prob. 6.5-7 for a triangular distribution of force (I - t )b1(t ) in the z direction.
I I. The two-element prismatic beam shown in Fig. P6.7-1 is fixed at point 3 but free to rotate at point I. Construct the stiffness and consistent mass matrices S and M
(1,4-J. Assume thut a concentrated force P,,(t) act ' member in Fig 6 4(a) For both b s at x - 2L/ 5 on the space 11 11~~ equivulent nod~! ioads. with th me~ _er and st~uctural directions, find th, c//, { I , I , 3}. ' e reat1ve coordmates c;k = {I , I , I} 111111
(1.4-4. Hl•peut Prob. 6.4-3 if the force acts in the ' I' 1/,/5 . Y direction al lhe point wh,•11 ,,.4-!i. I i·t II moment M ·(t ) be upplied at . t ' / 111 Fig . (>.'1 ( 11) . C'11lc111t111· the l'q11iv111~:~:11,m:l11I lt,4/1 on !he SJ)lll'C truss .nll·111h1·1 /l 101 th,· 1l'i11fl w 1,11111 11 1111111.~ 1• ' ) 111 N 11 lornl 11ncl 1,1 loh11l d111·ctlt11iN • 11 , • 1, 1111d 1• , { I , l, 2),
306
Prnhl111111t
CD 1-
/
'
3L •
4/·1-
1: 11'1111111• l't,,7, I
0
5+ 6/~ 3
X
307 3015
Frnmnd Struoturna
f'rohlnm1
0
6.7-2. Figure P6.7 2 shows u two 1•lrml'lll pliN11111tl1· lw11111 thnt is free to trunNhlll not rotate) at point I nnd is llx1•d 111 polut I Ass1·111hle the stiffness anti rnn•I mass matrices Sand M for the th11·1· 11n11•s1111incd displacements, and rctlu~o by eliminating the rotation and keepl111-1 the trunslations. Solve the l't"t'IIY problem to obtain angular frequenc ies and vibrational mode shapes tor th duced system.
y
y
Figure P6. 7-4
. . P6 7-5 has the same properties E' P, A, and I, for The plane frame sho.wn I~ Fig . . t the stiffness and consistent mass each of its three pnsmatlc me~~e;:~;;e; ~ shown, neglecting axial strains .in matrices S and M for the f;ur d b liminating the rotations D1and D2, while the members. Then reduce an e . the angular frequencies and mode ietaining the translations D3 and D4. etermme shapes of the remaining system .
J6
z Figure P6.7-2
r.:;--,
03 \
!..--,_.....,;\V::::::3:__--1~ 4 6.7-3. The simply supported beam shown in Fig. P6.7-3 is divided into two flc,iu elements with equal properties. Set up the stiffness and consistent mass mutrl S and M for the four displacements that are unrestrained. Reduce these mutrl~ by eliminating the rotations and retaining the translation. Then find the angul frequency of the remaining system, which has only one degree of freedom .
/3 y
D,
y
,+, ~J;;
/
I:
CD e
3L •
Y.,. L
0
5+ 3 X
6/~ :1
z
F igure P6.7-S
. F
Figure P6.7-3
6.7-4. A two-element prismatic beam has no restraints whatsoever , as implied in liill P6. 7-4. Construct the stiffness and consistent mass matrices S and M for the Hix unrestrained displacements, and eliminate the three rotations. For the thrc1· '" maining translations, calculate the angular frequencies and mode shntK'S.
1i P6 7~6
has 2El and 2pA for elements and . A for their properties . Construct I und 2; whereas, elements 3 and 4 hav~ ' S M· and eliminate the rotations the stiffness and consistent mass matnc~\. anD a~d D Calculate the angular U, , 0 2, and D,, while keeping thehtra~s a •.o~~g ;ystcm s~eglecting axial strains li cquencies and mode shapes for t e ,cmaim , 111 the members.
fl 1 ,,. Assume that the plane frame shown m
JOB I 11111111<1 H1111011111111
308
hnp
n
1rnhl1111111 1
n I n·111H in !lig. P6.7-8. l)ctcnnine the A 1•,1id with tlll l'l' i•q1111l 111 t, 11111th ,111 ,11~1\:1~11 !'or the three displacements at joint HtlffncHti lllld 1.'0IIHIHh ' lll lllll-11 111111 111111, I t'on and calculate the angular , II ki'llll 1hr trunsf a h1 , embers in this • ' Then clln11111111• 1111' 1111II 111111, structure, "· p ,r each o t e m 1 lrcqucncy of lhL· i1·1h11•1•d ~YH ~O and E = 5G/2. ussume that 21, 2 /1, ,/ , ,.N ,/ '
U
y
(1)
"·
1
1,
l'II\
G)
o ..
3
z
• - - -- -- - - - -
03/
o, / - - L ~
1- -
l
\
0 4
-----x
031
CD y
5 '/
/
/--'
.,
-
L
/
X
4
Figure P6.7-6
(1.7-7. Figure P6.7-7 shows a grid consisting of two prismatic members with 1•q1111I properties. Assemble the stiffness and consistent mass matrices S and M frn lh three di splacements indicated at joint 2. Reduce these matrices by eliminali111-11h rotations and retaining the translation. Find the angular frequency of the ro11111l11 ing SDOF problem, assuming that Ix = ly = J, Tg = L/10, and E = sen z
X
/ Figure P6.7-7
\
Figure P6.7 -8
D2
0
,__/
3
I'
7
tlll llNIIOII
wl
IIIL·
111ul
displacement
311
t.111111 111 Irr I 1111111111n
Vl'l'lo1 IH
(3)
u - {u, v, w} I 11t lltl•1more , the 6
3 Stra in-displacement operator d has the form
X
a
Two- and Three-Dimensional Continua
d
7 .1 INTRODUCTION 1
=
7 ,2 STRESSES AND STRAINS IN CONTINUA
a ay
0
0
0
a az
(4)
0
a ay a ax
-
. . ak from Eqs. (3.2-2) and (3.2-3). which the derivatives are t den t ·n relationships for an isotropic maaddition we develope stress-s rm . In ' such relations · h"ips can be wntten-. >I< As before ,, , /' 1/ . ' (5) CT=EE. 111111
rc the stress vector is CT
= {CTx ,
· E was given by Eq. (3.2-lO). our discussion to problems in two For the moment, Let us specia ize. ly those terms depicted in Fig. ,l1 111onsions, where the stress vector contams on I l(u). Thus, in the x-y plane we have (7) ·
=
=d
{
· p·1g. 7 · 1(b) ' which re the strains indicated m to these stressesda. fi .t simal eiement in the x-y plane. Hence' hows a displaced and deforme m m e 11 11 i,train vector is
' ' 111 t'csponding
111 Sec . 3.2 we introduced the topic of stresses and strains in three dimension strain-displacement relationships with respect to Cartesian corndl these relationships will be expressed in matrix form as E
(6)
,1 11d the 6 x 6 stress-strain matrix . l'
1111<1 dclined 1111 ll' s. Now
E.
111
( 11 either static or dynamic analysis. The strain vector in Eq. ( 11 '
(
.,
= {Ex ,
(8)
Ey, ,'xy}
which
U
au
E 1
IH
110
0
a a ay ax a 0 az a 0 az
CT
lo
0
111
Tht framed structures discussed in the preceding chapter consist of only sh.-1111 (or one-dimensional) members. However, we also wish to analyze solids 1111d Nl1 uctures composed of two- and three-dimensional finite elements. In this ch111, fl'I' we shall deal with two-dimensional continua in states of plane stress 11ml plnnc strain LI] as well as general and axisymmetric solids. Our emphasis will h(.) upon the isoparametric formulations leading to the most commonly l1Nl1d t•lc:mcnls 121. Computer programs cover dynamic analyses of plane-stress 1111J ph11w ·Strain problems, general solids, and axisymmetric solids . In all cases lh di screti zed continua are assumed to be linearly elastic with small strains 1111'1 displacements.
which applies
0
ax
=ax -
av
eY =iJy
au
av
--+ ay -ax
(9)
,'xy -
•• Rc•r. 2·' und for generally an-
~ l'nr o, thol(OHlllly u11lM ot,opic (m·1/1111r1111/1•) i1111ll' t 1n1N, St e
11,,111111/r· 11111lllrit1I N, Nl'l'
l {(I I , "
J12
Two- and Three-Dimensional Continua
I
oc. 7.2
313
Stresses and Strains i n Conti nua
I'hese derivatives constitute strain-displacement relationships in two dimen~lons. As before, they may be written in matrix form as shown in Eq. (1), where lhc linear differential operator d becomes
a ax
yx
a ay a a ay ax
(10)
0
d=
·-J
0
dy dx
1111d y
L,
the vector u is
u
'
= {u,
(11)
v}
When a thin plate is loaded with forces in its own plane , the resulting two-dimensional problem is called plane stress . Assuming that the plate lies in lhc x-y plane, we can impose the following conditions of stresses and strains:
(a)
=
'Ty,
=
1'zx
=0
€ , =I=-
0
Yyz
= 'Yzx = 0
(12)
111 particular, note that the strain normal to the plate is nonzero. Using these nmditions in Eqs. (3. 2-5), we find that i)V
dy If-- ~ ,w
V I ily dy
E = -----=E 1 - V2
l~ : ~ 1 0
0
(13)
1- V -2-
'l'his is the stress-strain matrix for plane stress in an isotropic material. In uddition, the normal strain in the z direction is (14)
The case of plane strain arises when a long prismatic solid has a constant l't>ndition of loading normal to its axis. In this instance, the solid can be analyzed 11~ an infinity of two-dimensional slices of unit thickness, as illustrated in Fig. I , l (c). However, the conditions given in Eqs . (12) must be changed to
V,V
i!u U
I
i!x
dx
(ll)
(11) Stresses und (b) struins In two dimensions.
0
1'yz
=
T,x
=0
'=z
=
')'yz
=
Yzx
=0
(15)
The first expression indicates that the stress
314
Two- and Three-Dimensional Continua
E - (I
+ v)~I
2v{ : v
-
I
>rvl
y y
y'
I
x'
Th0 latter equation gives the stress-strain matrix for the case of plane strain in 1111 isotropic material.
i'
Now we shall consider rotation of axes for stresses and strains, starting in three dimensions. For this purpose, the stress vector in Eq. (6) may be recast into the form of a symmetric 3 x 3 matrix, as follows:
(J'
315
Stresses and Strains in Continua
Sec. 7.2
= [~:
X
X
(J 8)
z z'
Tzx
(b)
(a)
In which complementary shearing stresses are included. Then the rotation-of uxes lransformation for stresses can be stated as ( I 'JJ
1n th is equation the 3 x 3 matrix O'' is similar to O' in Eq. (18), but it cbntait1H Ntl't'SNllS in the directions of inclined (primed) axes shown in Fig. 7.2(a). Thl' I X J rotation matrix R in Eq. (19) has the form
y
y y'
A1 1
R
=
Yp
A2 1
(20)
[
x'
,\31
111 the lultcr matrix the terms / 1, m 1 , and so on are slightly more efficient symbolN Im the direction cosines A11 , A12 , and so on, that were explained in Sec. 3.fi Nolu lhul lhe rows of R contain the x, y, and z components of the unit veclor11 I' , .I', und k' shown in Fig. 7.2(a). Similarly, the strain vector in Eq. (2) may be restated as the 3 x 3 sy,11 111t1l1 k matrix
E
=
€x 'Yxy ')'yx €y
'Yxz] 'Yyz
[ 'Yzx
')'zy
(21 l
€,
which in<.:ludcs dependent shearing strains . Again, the rotation operation is
i' j~
0
~
X
X
Id)
(c)
Figure 7.2 0 rthogona1 aXes-three . . dimensions: (a) inclined; (b) principal-two dimensions: (c) inclined; (d) pnnc1pal. .
111 this equation the symbol nl principfil normal
O'p
. 1matrix (or spectral matrix) represents a diagona
stressesu:lr ~'
l]
(24)
(2.11
whld1 the 3 X 3 matrix e' is similar to E in Eq . (21), but for inclined 11xt•11 Wt• cun obtain principal normal streS.\'<'.I' und I heir clircclions us the solut i1111 111 1111 ,•ll{1·n1 ul11,· prohl<•111 141. Using thut nmthod )'iws 111
1
er,,
1{1, er H /,
(.! I)
.1 fl (23) denotes the rotation matrix for the 111 nddilion, the symb~>l ~,, ; ~~) This mntrix is the transpose of the nor1,,/m·liwl rm•s shown 111 hg. ·,2 , · , ,111111/t· d ,·IR1'111•,•,·1or 111,11ril c1,N I hnl IK,
.1111 Two- 011d 1111011 Dlrnonslonal Continua
Chap, 1
Rp = X (2') lnwhichthc rowsof R,(orc l . f) Simi larly . .' l o umns o. N are normalized to have unit lengths ' prznczpa . normal strains may be calculated as . Ep = RpE R] In tlu s expression the symbol Ep stands for a di
(2(11
Sec. 7.2
Stresses and Strains in Continua
Therefore, the stress transformation matrix '(,. is proven to be the transposed inverse of the strain transformation matrix T,. The inversion implied by Eq. (33) is not actually necessary, because expansion ofEq. (19) shows that the parts of T" bear the following relationships to the parts of T,:
. agonal matnx of normal strain-
l'1ll'rcsponcfing to the stresses m ·
e' = T,e (27) l lcrc the strains e and e' are in the form ofE ( 2). d .l'lm/11 tronsformation matrix T, in Eq. (27)qis mstea ofEq. (21). The 6 X ,,
TO' = ['t.11 T0'21
TO'l2] To22
=
[IT,,, 2 T,21
2T,12J Tm
u' = E' e'
(35)
Next, we substitute Eqs. (27) and (32) into Eq. (35) , producing
TO'
/ .I2
(34)
As before, the subscripts 1 and 2 on the submatrices of TO' refer to normal and shearing stresses, respectively. Now the transformation of E' to E can be accomplished by first writing stress-strain relationships in the primed coordinates as
/7 Ii
·r: •
317
2/1 /2
T;;', and use Eq. CT = TJ E 'T.:E
(36) (33) to find that (37)
Thus, we see that
2/2/3
E = T;E'T,
(38)
which represents the transformation of E' to E. The reverse transformation is
E' ='(,.ET~
(39)
For the purpose of specializing axis rotations to two dimensions, we form n symmetric 2 x 2 matrix of stresses, as follows:
(40)
This matrix can be used in Eq. (19) to determine similar stresses u' for the Inclined directions shown in Fig. 7.2(c). For this case the 2 x 2 rotation matrix
ls (oe')TO'' = ' l'h1111
ohf11l11
OET
( 10)
sin (} cos (}
substitute the transposed incremental form f Eq
° · (27) into Eq. (30) to
J
(41)
111 which the rows contain the x and y components of vectors i' and j'. FurtherI l11111't'.
( 11,
we conclude that
er'
= 1;,o
wh1•11•
'(;, • 'J' ;- I
1110rc, the symmetric 2 x 2 matrix of strains becomes
E= [Ex Yxy] 'Yyx
(42)
Ey
whklr <.:ur1 h1; applkd in Eq. (22) to find strains E' for the inclined axes. By solving II Nl'l'Olld Ol'(kf' t't/•l·nv11llll' prnhlcm , we cun also obtain prin1 lp11l 11011111d 1-1111•:,i,t•1, 1111d Nl111i111, in two di11111111,lo111,. Thus, tht• l'igt·nv11h1t• prob-
318
Two- and Three-Dimensional Continua
Sec. 7.3
319
Natural Coordinates
lem associated with the 2 x 2 stress matrix in Eq. ( 40) is
,\;l);
(O' -
0
=
(i
= 1, 2) Tl = 1
In this equation the symbol A; denotes the ith eigenvalue of O' , and ; represents the corresponding eigenvector. To find A; we set the determinant of the coefficient matrix equal to zero , as follows:
IO'
-
A; I I = 0
(44)
or
o-y T~ A;
I=
0
(45)
Expanding this determinant yields
AT - (o-x
+
y
0-y)A;
+
0-xO'y - T;y
=0
ll
=
11
=
i;
=
0
1
L.
(46)
From this quadratic equation, we obtain the roots
i; = 2
- 1
ll = - 1 (a)
(47) 3
which constitute the principal normal stresses o-p 1 and o-P2, They act in the directi ons of the axes Xp and YP shown in Fig. 7.2(d). By substituting them into the homogeneo us equations [see Eq. (43)] , we can find the eigenvectors, which (when no rmalized) contain the components of the unit vectors ip and jp. The strain transformation matrix T, in Eq. (28) can be specialized to two din1ensio ns by removing the third, fifth , and sixth rows and columns to obtain
~~=~+
T. = [---:: - - - - :: I - - - - --- - --42l,l2 2m 1m2 / l,m2
l2m,
Also, the stress transformation matrix T" becomes
T"
=[
If2
l2
mf2
m2
I I I
2/, m, 2/2m2
------- -4- - - --- - - -
J
i; = 1
(48) 2
J
s= - 1 y
L.
(49)
11/2 m,m2 I l,m2 + l2m,
which is related to T, in accordance with Eq. (34).
s = --12
ll = - 1 (b)
Figure 7.3 Natural Coordinates for quadrilaterals: (a) straight; (b) curved.
7.3 NATURAL COORDINATES
111 lllis Sl't'ti on we de fine natural coordinates for q1111drilatcrals and hcxahedrn in prnp11rnt io11 for· dl'Yl'lopment of isopurnnwt, k 1•lt·1111•11t 11 having these shapes . Wt· li1·1il11 with 11
pl une are xH
= Hx1 + X2 + X3 + X4)
Ys
= HY1 + Y2 + y3 +
y4)
. (1)
( ) arc the x and y coordinates of po mt 1, . 111 these averaging expressions, Xi, Yi .. , .. ·1 the same as the N t ti I tl w gl•n1111.llrk t'l'llll' I' is no t ncccss,m Y . 1111d 1-10 on. o l' Ill Al i I th . llHllll' un· till' di11w11.1·im1f<•,1·.1· I l l' llll Old
111 th1• q1111dtil11h•11il
t\
NII S HlW II
I
l
321
1:10
lwn "'"' ll11nn 111111,1111111111111
11,111m1/ 1·11<1rdl,u111•.1·
t and T/
Ill poi111 /.:, it is i111portant l<~
Alli ru·
Cu 11 t1 111111
I
i','.H'.~ 1' f111 >j1 '1 ls1•w rn.onlinutcs huvc
1("~
'1111:. / ,'J
<:hni,
!hell' rn 1)!111
pol111 o11 lhc clement. We see t~' I ml Ihr Ii dlt l'l'fions arc different ut l'Vl'I V 1011 ' I, T/ . I ulong edge 3-4, an~ ~ 7dgc 1-2, ~ I _along l'd1,tr pol11!1011 111 both the 1: and d" . ,dong cdge 4-1 · With linear i111r 1 !> r, 1rect1ons form la . f 1 . unywhcrc on the quadrilateral become ' u s or ocatmg a generic po1111
J_ / "
x=
I
4
f;x;
4
y = 2, f;y;
1=1
111
,
s
(
,,
. I
N11t11111t Cow dl1111lot1
11or this arrangcmcnl, terms in the coefficient matrix are easily obtained by differentiating Eqs. (2). This array is called the Jacobian matrix J, which l'Ontains derivatives of the global coordinates with respect to the local coordinates. Thus, we have
J=
,
•
•
g)(l - r,) g)(l
+
r,)
f2 = !(I + [)(1 _ f4 = HI - [)(1 +
111
= x,g
Y,ri
X,')
4
= L f;,gx;
112 = Y,g =
L f;,gY; i= I
i=l
r,) r,)
(8)
[x,g Y,g]
Terms in this 2 x 2 matrix are
f 21 = X ,T/
( I)
= 'v
= Y.') =
122
+.
~Jl,T/X I·
L AriYi i= l
i=I
Casting these expressions into matrix form yields (10)
= DLCN
J
The array DL in this equation contains derivatives with respect to local coordinates, as follows: DL
= [ !1,g fi.g /3,g
!4,{
!1.ri Ari Ari Ari
(ti)
(9)
4
4
1
af a71 ar, ax af ar, ar, ay
=
122
4
I Ill Sl' I 1111ct1ons give the global coord· t f . l!~1111s of lhc natural coordinat H ma es o a pomt on the quadrilaternl 111 hypl11·holic), the local coordina~es~ gao~ever, because Eqs. (2) are bilinear (rn ol Iill' global coordinates x and n 71 cannot be expressed directly in te111111 W ·11 y. c w1 need derivatives of the function " dl11plun•111cnt relationships The h . s J ! through f4 to use in stra in 1 for differentiation off([, r,) with u•spl'l'I lo x and y gives . c am rue
af = af ag + ax ag ax af = af ag + ay ag ay
112]
121
which the geometric interpolation functions '~e
!1 = Hl / j = !(1 +
[111
[-(1 =4 1
TJ) -(1 - g)
J
(1 - r,) -(1 + g)
(1 + r,) (1 + g)
r,)J
-(1 + (1 - g)
(11)
and the matrix CN consists of nodal coordinates in the arrangement
01'
[J.x] = [S.x 'Y/,x] [J, g] !,y. S,y 'Y/.y !.ri
5
wh1•n• the s m .· ( ) ,, . · Y bol f. x means differentiation off ·th I 1•1111s 111 lhc coefficient matrix of E 5 w1 _respect to x , and so on 1111, 11111,hlc lo solve explicitly " t: q.d( ) ~e not readily available, because Wl' 1 ior !> an r, m terms f d '""'' I1ll' opposite approach and differenti . o x an y. However, if w,• 111lt• produces ate/with respect to sand r,, the chain
af = af ax + ag ax ag af = af ax + a,, ax ar,
af ay ay ag af ay ay ar,
f,1
f:: ::1 X3
X4
((l)
By comparing Eqs. (5) and (7), we can see that the coefficient matrix in the former expression is the inverse of the Jacobian matrix. Using the formal 1 definition of the inverse, we can obtain from J as
r
where Ja denotes the adjoint matrix of J and quantity is calculated by
\JI = l,1122 ('/)
(12)
Y3 Y4
-112] f1 1 =
1)1
f = [.x., Y.cl [J., l IJ.,, x,,, Y.,, /1,.I
=
CN
-
Ji1l12
m 1 [
Y.ri
-x, 71
-y,g]
(13)
X,g
IJI is its determinant. The latter
= X,gY.ri - X,riY,g
(14)
To determine the derivatives of all of the functions with respect to x and y, we cun apply nq. ('l) ll'Pl'lllctlly . Hence,
322
Two- and Three-Dimensional Continua
J-1[fi·fi.7]s]
fi.x] = [ fi.y
(i
= 1, 2, 3, 4)
Sec. 7.3
(15)
323
Natural Coordinates
numerical integration for both straight-sided and curved quadrilaterals, as explained in Sec . 7.4. . Now let us consider a h exa hed ron with straight. edges, as depicted in Fig. 7.4(a). At the geometric center (point g) the coordmates are
Altogether, we have (16) The matrix DG given by this expression consists of derivatives of Ji with respect to the global coordinates. That is,
DG =
[fi.x fz.x .h.x f4.x] ii .y fz.y h.y Ay
1
Xg
=
1
8
8L
X;
Yg
1
8
L Yi 8
=-
Zg
i=I
1=)
=
8
8 L Z; i= l
T)
(1 ) 7
3
Evaluating terms in DG, we find ]
Dc11 = 4 / J / [ -(1 - YJ)l22 + (1 - [)1 12] 1
Dc12
= 4 / J / [(1
I Dc13 = 4 / J / [(1
- YJ)l22
+ (1 +
[)1 12]
+
{)112]
+ YJ)l22 -
(I
y
)-. z
5
I
Dc14
= 4 /J /[ -(1 +
YJ)l22 - (1 - [)112]
1 DG21 = 4 /JI [(1 - YJ)l21 - (1 - [)l1i]
(18)
la)
1 Dc22 = 4 / J / [ - (1 - YJ)l21 - (I + [)111] 1
Dc23
= 41 JI [ -(1 +
Dc24
= 4 / J / [(1 + YJ)l21 + (1
YJ)l21
+ (I + [)Ji i]
1
- [)111]
lly this approach we are able to solve for De numerically. Figure 7.3(b) shows a quadrilateral with curved edges that may follow q1111drutic functions, cubic functions, and so on. Regardless of the complexity ol llwsu functions, the natural coordinates g and YJ play roles similar to those for 11 quudriluteral with straight edges. The form of the 2 X 2 Jacobian matrix 1l111111i11s the same, even though the functions to be differentiated are of higl1t11 order. Because of the appearance of the determinant of J in denominator posi 11011s, Wl' 11suully cannot integrate terms explicitly to obtain stiffnesses, consis lli11t ou,sscs , und equivalent nodal loaclli. lnstcnd , ii hl•comes necessury lo 11s1•
z
5
~
=1
lb)
Fl11111·1• 7.4
Nuturul coordinurcs for hcxnhcdrn : (11) straight; (b) curved .
(19)
I w11 nrn I 1111 n.- I>l111111111lur111I Cu11tlr11u1
Chnp
wht'll' ,,, I''.' "."d i, 1~1·c the 0 1rt<:siu11 l'11111dl1111h•11 of poi nts J thrnugh 8. 'l'lw fi ,,,., 11lN11 Hhows t/1111,•11.,·101,le.1·.1· natural coor,lln,,r,•,,· (.. '>'J und r Nol<: lhal 1: 11 fm•t• l l 7 ( ' linear b· b 11 • >: 77 . J on face 3-4-8-7, nnd so 0 11 .' With interpolation i 11 th I.ti
~~ 'I• und ( d1rcct1 ons, the location of any point in the hexahedron may be wrlflrn s
s
= 2_ fix;
X
y
8
= 2, f;y;
z
= 2, f;z;
i=l
( .'ll I
i= I
111 whk h the geometric interpolation fu,nctions are
/, ./'i
ff{ I - {)(1 - T))(l - t)
!2 =
.f'..
HI + g)( I + 77)(1 HI - g)( I - 77)(1 +
.Ii
ff{ I I s)(I
Ii = k(I - {)(1 + 77)(1 - C) 16 = ko + t)(1 - TJ)(1 + c) f s = k(I - f)(l + 17)(1 + ()
{)
?)
+ TJ)( l + {)
ko +
t)(1 - 11)(1 - c)
Ill
il-(
,.:1
325
Nntw 111 (.;oor dltrntuu
(1 + 77)(1 - () (I - 77)(1 - () (I ~)(I - I,') (1 + {)(1 - () -(1 + g)(l () I - {)(1 - () - ( I - {)( l - 77) - (1 + g)(l - 77) -(1 + g)(l + TJ)
(1 - 77)(1 + () - (1 + 17)(1 - () -(1 - T))(l + t) (1 - g)(l - () -(1 - g)(l + {) -(1 + g)(l + () (1 + g)(l - TJ) (1 - g)(l - 17) - (1 - g)(l + 17)
I,')]
+ 17)(1 + t) -(1 + ~)(l + (1 - g)(l + () (1 + {)(1 + () (1 - g)(l + 17) (1 + g)(l + 17)
(1 ('II
\ lso, the matrix CN becomes an 8 x 3 array of nodal coordinates. Thus,
llt'l'IIIIN.c th<:sc interpolation. functions are trilinear, the local coordinates i 1 d ' 1 I ' II . ' d. . x' y' an z. ' ucc. 11ncns1ons the chain rule for differentiation with res ect to lhr 11111\rntl rnordrnatcs for a hexahedron leads to the foll · 3 P owmg X 3 l aco/,/, 111 111 ,,r,'II : 1111d , 1111rnot be expressed rn terms of the global coordinates
1
= [;~: ;~: ;~:] = [:: h,
1 32
x,,
] 33
r·: :::] YJ
z.,
(24)
() l,
CN
=
X1
Yi
Z1
-~2.
Y2
.~2.
Ys
zs
r
Xs
1
(25)
The inverse of the Jacobian matrix may be expressed as (26)
1
'f't 1111s in thi s matrix are found by the differentiations indicated. Thus, S
1 12
= 2, f;,€Yi
I I
i= l
8
.!2, - ~ fi.r1X1 .I 11
==
2, f;,€z;
= 2, f;,.,,y;
= 2, J;,.,,z;
i~ I
i= I
8
8
8
I
I
1 32
==
2, f; ,,Y; i• I
Wl' use Eq. (16) . In this case, the matrix
s
123
I I
L !t.,x,
l'o find the derivatives of all of the functions with respect to global coordinates, DG consists of the following terms:
i= I
S
122
where the symbol Jarepresents the adjoint matrix of J, and IJI is its determinant.
8
11 3
] 33
(2 1)
= 2, J;,,z;
/4,x fs ,x Ax / 1,x fs.x] !1.y !i,y /3,y !4.y fs ,y f6 s !1.y fs.y f i,z f2.z
Az /4,z
(27)
fs ,z /6.z h,z f s.z
i• l
AN hl•IOi l',. lh.csc calculations may be arranged in the matrix format given b p 1 ( I !I) In lh1s 111slance, the n:iatrix Dl is the following 3 x 8 array of derivt~i~1 will, H·s1wc1 to local coordmates: "
!
u,
_ [/1,x A x Ax
DG -
t.,]
.Ii., .Ii., .li., A, 1~., /6,, !1., .Ii ,1/ .Ii.,, /1.1, 14.,, 1~.r, A,, Ari ls.ri .Ii., .Ii., /i,, ./~., 1.~., Ji1,( !1., .IH.t
A hexahedron with curved edges appears in Fig. 7.4(b). Geometric interpolation functions for this solid may be quadratic, cubic, and so on. The natural 1•oordinates g, 17, and ( will still be used for such a hexahedron in spite of its 111c11tcr complexity. The 3 x 3 Jacobian matrix in Eq. (22) remains the same, 11llllough the functions to be differentiated are of higher order. As for a quadril1111•rt1l with straight or curved edges, it becomes necessary to use numerical l11tl·~ration to evaluate stiffnesses, consistent masses, and equivalent nodal l1111ds.
'I
328
lw11
111111 11111111
l>l1111m11lo1111I Cc111tl111111
111
/JI
327
N11111n1 lo11l l111u11111tlo11
7.4 NUMERICAL INTEGRATION
The process of computing the valuc ol
Ix =
l
'l
x,
II
dt•flnitc integral I see Fig . 7 .5(11ll
/(x) d x
t IJ
from a set of numerical values of the integrand is called numerical integ, 111 h 111 15]. The problem is solved by representing the integrand by an interpol11l l1111 formula and then integrating this formula between specified limits. Wht•11 1111 plied to the integration of a function of a single variable, the method is rcl1•1111I to as mechanical quadrature. If interpolation formulas for numerical integration are polynomial~ 111 sufficiently high order relative to those assumed for displacement (or 0111111 functions , the integrations will be exact. Otherwise, the process of nunw1 h 111 integration introduces an additional source of error into finite-element analyi.1 The most accurate quadrature formula in common usage is that of 01111• which involves unequally spaced points that are symmetrically placed. To upplv Gauss's method, we usually change the variable from x to the dimensiouh ~ coordinate g with its origin at the center of the range of integration, as sh11w11 in Fig. 7.5(b). The expression for x in terms of g is
x
= H(l
-
t)x, + (1 + t)x2]
( 'I
Substitution of Eq. (2) into the function in Eq. (1) gives
J(x)
= cp(g)
(a)
(
,, (b)
Also,
( II Then substituting Eqs. (3) and (4) into Eq. (1) and changing the limits 111 integration yields
3 2
Gr111.1·.\" sformula for determining the integral in Eq. (5) consists of summing lh, wcighted values of (g) at n specified points, as follows:
(c)
(hi
In this cx.pression ~ is (he location of the integration point} relative to the t·t·111!·1, 11 1w·i~liti111< fa ctor f'or point j, and n is the number of point.1· ul whic:h c/,( ( I is to lw t·nku luted. Yult1es of these pururnt•(l\rs nn: listed in Table 7. I.
N1 is
Figure 7.5 Gaussian quadrature.
1111
l\,\111 111111 tl111111 Ul11111111 lo1111I cw111!11111
<
11111
Nu111L·1h;ul intcgrntinn hy, ,, 111111 , I 'fl' t '. ,111 ,111,1rlm1t11·1· IS CXIICI for poly11011111tl I ' . ,H IS , 011f y 11111' i11(r)tl111i . • . 111tcgration of a linear functi , I oo point rs required f'or tlw
Ml
/
,II
,,
dl'gl'l'C 2 11
,·,11
und so on. For example, con~;~ ~.. ~:1. p11111t ''.' \nccd_ed for a cubi c poly1111111111I ing in Fig. 7.5(c). First let n I ~t111111,111r11 , 11,11,~ uncT ·tron W 3 - g2upp1·111
=
'
t
lit
g, = 0
/? 1
mm
able 7 .1
= that
=2
2
~ 1?1c/,(~1) j
±(;
2
0.5773502692
1.0
3
0.7745966692 0.0
0.5555555556 0. 8888888889
4
0.8611363116 0.33998 10436
0.3478548451 0.6521451549
5
0. 9061798459 0.538469310 I 0.0
0.236926885 1 0.4786286705 0.5688888889
6
0.9324695142 0.6612093865 0.2386191861
0.1713244924 0.3607615730 0.4679139346
0.9491079123 0.7415311856 0.4058451514 0.0
0.1294849662 0. 2797053915 0.3818300505 0.4179591837
7
Y2 l X2 f(x , y) dx dy
l
X\
r
=x+
8
y
=
(8)
xi + yj
g is
lhc rate of change of r with respect to
ar ax. ay . - = - 1 +-J ag ag ag
(9)
iJr
ax -
dT)
- dTj ilTJ
iiy
-
\
d-ri
0.1012285363 0.2223810345 0.3137066459 0.3626837834
I
~
iJr
r
+ - dTJ dT)
y
'l'hl'll from Eq. (6) we have
1E = R, ((;,) which is approximate. Next ' let n
(;, =
- (;2
=- ~=
I knee, we lind from Eq. (6)
= (2)(3) = 6
= 2 and
. f rom Table 7. J: o btam X
-0.5773 .
R,
= R2 =
dA
dT)
y
dT)
0.9602898565 0. 7966664774 0.5255324099 0.1834346425
(7)
llnwcver, this integral is more easily evaluated if it is first transformed to the 1111t11ral coordinates for a quadrilateral. We accomplish this by expressing the llull'I ion f in terms of g and 1/ and using the limits - 1 to 1 for each of the l11tl·grals. In addition, the infinitesimal area dA = dx dy must be replaced by an 11ppropriate expression in terms of d(; and dr,. For this purpose, Fig. 7.6 shows 1111 infinitesimal area dA in the natural coordinates . Vector r locates a generic pnlnt in the Cartesian coordinates x and y, as follows:
R; 2.0
gD
. .) = 5.333 .. . , hlrh is exact. Now let us apply Gaussian quadrature to quadrilaterals in Cartesian coor1ll1111tcs. The type of integration to be performed is
YI
0.0
t ( 1)(3 -
I
I=
l
~O
(1)(3
= (2)(2.666
TABLE 7.1 Coefficients for Gaussian Quadrature n
329
N1111111tl1111t 1i1111u1111lt111
I
z li'IKllr(! 7.6
Infinitesimal area in natural coordinates.
i!r
-
i)~
d~
:no AINo ,
I Wt1 1111il I 1111111 l>lt11,1111l111111I <.1111tl1111n tltl'
rntl' ol chungc ol
I'
With
1111,111·1 I
I'
lo t/ IN
1111
d , I I iJ y I ll tJ ii 'Y/'
/Ir
o'YJ
t ,tll'S
ot' drnnge of,. wit h respect to , ,
== ( :~ d g x
:~ d 11) · k
==
(ax~_ ax ay) dg d 11 ag a11 a11 ag
=
I;,,,,,s
Y.s I dg d11
Y.,,,
==
IJI dg d11
(18)
t t l the symbols a, b , and c denote
(I
The expression in the parentheses of Eq. (12) may be written as a 2 • dctcrminant. That is,
dA
uml 'arc
ax ay az or _ - i + - j + - k ag - ag ag ag a ay az ar = ~i + -j + -k a'Y/ a11 a11 a11 a +ayj + -azk ar _ 2-i ~ - ai ai ai
I I 11
Substitution of Eqs. (9) and (10) into Eq. (11) produces
dA
T/ ,
~
When multiplied by dg and d'f/, the derivatives in Eqs. (9) and (10) fo1111 tW 11dj11ci;nt sides of the infinitesimal paraHelogram of area dA in the figun: I hi 111 cu may be detennined from the following vector triple product
dA
( I'/)
d I v,I I ~k
ar (19) ar a dg C = -d~ - _! b = - d11 ai a - ag a11 . . as the edges of the infinitesimal pai::al78 lfo.isc vectors are shown m_ Fig. . be determined from the following !11lcpiped of volume dV. This volume may Vl'Clor triple product:
(I I)
111 which J is the Jacobian matrix given in Eq. (7 .3-8), and IJI is its determina11t ThuH, the new form of the integral in Eq. (7) becomes
4
( I •II Two successive applications of Gaussian quadrature result in n
I ==
n
L L RjRd(g, TJd IJ(g, 1/k) I
( ISl
where Rj and Rk are weighting factors for the function evaluated at the point ((;J, 'Y/k). Integration points for n = 1, 2, 3, and 4 each way on a quadrilateral arl' illustrated in Fig. 7.7. Next, we tum to hexahedra in Cartesian coordinates, where the type ol Integral to be evaluated has the form
I
==
iz2 JY2Lx2 f(x, y, ' I
YI
z) dx dy dz
(b) (a)
4
4
(16)
Xi
lkfore integrating, we rewrite the function in terms of the natural coordinates
,, 'Y/, and Cand use the limits -1 to 1 for each of the integrals. Furthermore, we must replace the infinitesimal volume dV == dx dy dz by an expression involving tlg, d'f/, and d{ Toward that end, Fig. 7.8 shows an infinitesimal volume dV in the natural coordinates. Also depicted is a vector r, which locates a generic point ln the space. Thus,
(d) (c)
. -l·(b) n = 2;(c)n =3; Figure 7 .7 Integration points for quadnlateral: (a) n - ' (d) n = 4 (each way).
lw11 n111f
1111 1111
Olt1111111h1111,1 Co11t l111111
111 I II
V
ltmp11111111111tln Ou11drll11t11111I• Im 1'1 11 1111 Strose end Pl ane Strain
333
1, ISOPARAMETRIC QUADRILATERALS FOR PLANE STRESS II )(
b
AND PLANE STRAIN
lloilc c lement is isoparametric if the same interpolation formulas define both tlt1• ).lt:Ometric and displacement shape functions. Such elements satisfy geo-
as well as displacement compatibility conditions. If the geometric interp11l11tion functions are of lower order than the displacement shape functions , the , h 1111.mt is called subparametric. On the other hand, if the reverse were true, the 1 l11111t:nt would be referred to as superparametric [6]. However, most commonly lf~l'd finite elements are either isoparametric or subparametric. Because isop11111mctric elements are usually curved, they tend to be more suitable than 1thpurametric elements for modeling curved geometric boundary conditions. Figure 7. 9(a) shows the rectangular parent R4 of the isoparametric quad1 //,11,•r al Q4 , which appears in Fig. 7.9(b). Conversely, the rectangle may be 11 111sidered as a special case of the quadrilateral, for which the natural coordi111tll'S gand 7J are orthogonal throughout the element. For either figure the generic dl~placements at a typical point are 111111Ik
-
I
j
/
I /z _ _ _ _ _ _ _J.,/
u
= {u, v}
(1)
X
N,,dal displacements indicated in Fig. 7.9 consist of x and y translations at each 11rn lc, as follows :
z Figure 7.8
Infinitesimal volume in natural coordinates.
(2)
dV = (a x b) . c = / J / d( d7J d(
=
ax ay az ag ag ag ax ay az aTJ aTJ aTJ d( dTJ dt ax ay az
11,11· both elements we assume the displacement shape functions 4
u <1111
in which J is the 3 x 3 Jacobian matrix d /J / . . . revised form of the integral in Eq. (16) 'b:~omes IS Its determmant. Hence, lhl1
fl fl
/JI d(dTJd(
I, 11111
= f;q;
1
=
n
.Z: .Z: .Z: RjRkRtf(Sj, T/k, ?,) 1J(Q, T/k , ?,) 1
(4)
111 which f.r
n
l = J k= I j= I
(i = l,2,3,4)
(.! I)
Three successive applications of Gaussian quadrature yield n
(3)
i=I
111 these expressions the functionsJ;,!2 , Jj , andf4 are the same as those in the v.1•0111etric interpolation formulas given by Eqs. (7.3-3). Therefore, the Q4 1 lt•ment is isoparametric, and previous statements regarding differentiation and l11h•µration of functions hold true. Equations (3) may also be written in the matrix U;
f1!((,TJ,t)
4
V = Lj;V;
i= l
ac ac ac I=
= ""ifiu;
(.' 'I
Integration points for n = 1, 2, 3, and 4 each way number I 8 27 IcI respectively. , , , unc ,. ,
= [O 1 OJ 1 Jri
(5)
I h1• !'Cncric displacements u1 in Eq. (4) represent translations at any point due q, al node i. As n furth t:r effi ciency of notation, we can write 1111• l'unction .fi us h1lht• displacements
./i
1(I
I £11H I I r111)
(6)
t.14
lwr, n111I
11111111
l>lt11n11alrn 111 1c0111r, 11111
ni
/ ,h
I opn11111u1ttlc 011udill11h11 nl11 for 1'1111111 btro1111 1111d Plono Str11l11
336
whl I l' (7) q6
T~ --+---:_-u~J-,.
Vnlucs of
-----
----Lli_____:J2tq~ q3
i-+--~~~~~2a~~~~~-/
I
2
3
4
g,
-I
1
1
-1
1/i
-1
-1
I
I
i
91
L,1~ ~tq-,
'Y/, for this element are listed in Table 7.2 TABLE 7 .2 Nodal Coordinates for Element Q4
Tdt 4
2b
g1 and
Similarly , the strain-displacement relationships for element Q4 may be 11111cisely expressed as
~; = B;q1
(i
a ax 0
B1 = d f1 = TJ
a ay
0 a f1 ay
[f;, /,]
=
0 f;,y fi .x
a
(9a)
ax
lfrfcrring to Eqs. (7.3-17) and (7 .3-18) , we see that the submatrix B1 can also h written
q,_
B1
-+
+
G2
K
y
Dc11
=
[
0
J
0
Dc21
DG21
Dc11
(9b)
Next, we express the stiffness matrix for the Q4 element (with constant lhlckness h) in Cartesian coordinates as
G1
l•'lgurc 7.9
(8)
whore
(a)
L.
= 1, 2, 3, 4)
=
h
L
BT(x,y) E B(x,y) dx dy
(10a)
Ilowcver, in natural coordinates this formula becomes (10b) (b )
1111d
with two applications of Gaussian quadrature we have
Element Q4: (a) rectangular parent; (b) isoparamctric countc1p111t.
n
K
'=
h
II
L L R1RkBJ.kE Bj.k IJj.k I k I J I
(10c)
338 l•oi, 11111111111,t, 1111nd1llnl111nl• 1111
In this expression for nun11:r 1rnl 11111•pr11111111, llw 1u1ttrix n1,4 und the
LfT
1'1111111 t.11 111111 nrul l'ln1111 h1111l11
337
lhrn•tw e, liqs . ( I Oh) 1111d (III,) 111111plll y to
K
"""
(x, y) f(.x, y) d.x dy
rr
u,(~. TJ)E B(g. 'Y]) dg d'Y]
I
(14)
11111
or M = abph
f, f fT(g, TJ)f(g, TJ) dg d'Y]
(15)
i
I h,· resulting 8 x 8 matrices are given in Ref. 2 and need not b~ repeated here. 11,u,lly, the equivalent nodal loads in Eq. (12b) also take the simpler form
or II
M
= ph L
IJ
L R1RdJ.k
Pb(t)
j
~ .k J Jj,k
f,f,
= abh
fT(g, 'Y])b(g, 'YJ, t) dg d'Y]
(16)
k E I j=I
I ,111nple 7.1
in which p and h are assumed to be constant. In addition, equivalent nodal loads due to body forces may be statl·d 11 follows: Pb(t)
=h
L
fT(x ,y)b(x,y, t) dxdy
or
.
111' 11 vc
.
numerically the consistent mass term M24 for the Fig· / 'l(h), using Gaussian integration with n = 2 each way. Assume that p an 6 aa;~ , 1t1INlunts and that the coordinates of nodes 1, 2, 3 , and 4 are (3, I), (8, 2), (6, ), 1 ',~). respectively . . . · ( G T I Eq ( 11 c) we must set up the network of four mtegrat1on pomts or auss o appG2, Y G3 · , and' G4 indicated in Fig. 7.9(b). In part1cu · Iar, ~or n -- 2 we have l'•illlls) GI, 11, Rk = 1 (from Table 7 .1), so that isoparametric Q4 elemen~1~
M24
= ph
2
2
L L
(fit.k(/it.k IJj.k I
(a
)
k=l / • I
I lw l'unctions/1 andfi are needed for evaluating the term M2 4 becau_se of the arrangement 111ntrix f shown in Eq. (5). Substituting/1 andfi from Eq. (6) mto Eq. (a) produces
or
111 n
Pb(t)
n
= h L L R1RkfJ.kb(t)j,k IJJ,k
h J
M24 = f!._
k=I jE I
}6
where both Pb and b are functions of time. Except in special cases, the integrals in Eqs. (IO), (11) , and (12) must h performed by numerical integration to obtain approximate results. However, 1r the element is rectangular, direct explicit integration may be used, for cithl'r Cartesian or natural coordinates. Also, line loadings with g or 'YJ constanl 11111y be handled by explicit line integrations. Of course, if the body forces consist of point loads, no integration is required at all. By specializing the quadrilateral Q4 to become the rectangle R4, we nm derive explicit terms in its stiffness and consistent-mass matrices. For thi s cu~" the Jacobian matrix and its determinant become
J= [
~
~]
JJJ = ab
in which a and bare half the width and height of the rectangle in Fig.
( 11) 7. 9(11)
2
2
L L [(1
- g2)(1 - 17)2] j,k IJj,k I
(b)
k - 1 j=I
111 Implement this formula, we first calculate the Jacobian matrix using Eq, (7 .3-1 0) , as l11llows:
(1 - 11) -(l +t)
l[
=4
(1 + 11) (1
+
t)
-(1 + 17)] (1 - ~)
38 21] 6 6 2
9 - 17 - 3-~
~]
5 (c)
1111'11 the determinant of J is
IJ I
H39 I I;
(d)
417)
1<1 11111 'l'nhlc 7 . 1 tlw vn lues of' { und 11 nt point Cll nr'l·
I/V3 and
-
I/V3, and so on
336
lw11 1!1111
1111 1111
IJl11101111lon11I Continua
l oc. 7.5
lsoparametric Quadrilaterals for Plane Stress and Plane Strain
1
for !hi.) otl11.)r i11tl.)gralion poi11tN l vul11ull11M 1111• 11,1111M in Eq. (b) at each of lht• 11111 integration points and summing lht• r1•Nt1II N yl11ldN I . I .I<) pit
Mi,,
which can be finalized using numerical values for p and h. 4
Now we shall consider a higher-order quadrilateral element that is b111.i 1l on quadratic geometric and displacement shape functions. Figure 7. IO(a) doph I the rectangular parent R8 of the isoparametric quadrilateral Q8 illustrat(•d In Fig. 7. IO(b) . We may consider the rectangle to be a special case of the qu11d1 I lateral, for which the natural coordinates gand r, are orthogonal throughoul lh element. In addition, edges of the rectangle are straight, and nodes 5 throu~h H are located at midedges of sides 1-2 through 4-1. For both elements R8 and <)H the nodal displacement vector is q = {qi, qz, ... , q16} = {ui ,
V1, . . . ,
vs}
V
i=J
.ft
= i (1 + fo)(l +
.ft
= f (I - ( )(1 + 'Y/o) = HI + (o)(l - r, 2)
f;
'Y/o)( - I
•-u
---
y
LX
8
= 2, f;v;
+
(o
+ 'Y/o)
6
gl
I
I 15
q1-
2
2a
r
(a)
( I H)
i= l
2
----
I I
8
= 2, f;u;
where
t
2b
q2
8
3
V
(1 7)
which contains x and y translations at each node. We also assume the fo!Jowi11 j1 quadratic displacement shape functions: u
7
(i
= 1, 2, 3, 4)
(i
= 5, 7)
(i
= 6, 8)
(,
,,,
'l'hll values of 9 and 'Y/; required in these formulas [see also Eqs. (7)) are glvrn i11 Table 7.3.
for element QB we take the geometric interpolation functions to be lht1 sunie as the displacement shape functions in Eqs. (19). Physically, this
11w111111
Ihut the natural coordinates gand r, are curvilinear, and all edges of the elc111c•111 hc•co,nc quadratic curves [7]. Thus, we locate any point on the quadrilatrn,I (including point g) by the formulas 8
X
= 2, .ftx; i= l
8
y
= 2,f;y;
(.!II)
l'lllment Q8 is seen to be isoparametric . Formulations of stiffnesses, co1111i~ lc1111 111usscs , and equivalent nodal loads for this element are very similar to thoNt 1111 t.1lt.1 111ent Q4 given earlier. Table 7.4 contains the necessary shape fu11<:lior11, 111111 Iheir derivatives with respect to gand r,. Numerical integration also follow~ lh1• s111m· pattern as before, even though the lornl coordim1tcs arc curwd . 1111el
q 1 - . __ _ _ _ _ _ __
t
q2 Fl11111·r 7, IO
Elunu.inl
(Ii)
Qll: (n) i 1:ct11111111l111 p111 \1111 : (h) iN01111111111tltric cou111crp11rt.
339
J40
-
tw 'l'Alll ,1!. 7, I
-
Nmlitl t
111111
11111111 I lh1111111 l1111nl < 111111111111
111111ll11nl1••
tor
2
(·,
- I
,,,
- I
lr.11•111N1I ~
6
()
- I
- I
/ ,II
111,
1'11111111111 l>YNAl'li 1111 1'1101111,trnH nnd 1'111110 S1111ln
QH
TAIILE 7.5
Slrudur11I Dutu for Program DYNAPS
-
7
Type of Datu
0
0
341
hlem identification 1ctural parameters
Items on Data Lines
l 1
Descriptive title NN , NE,NRN,IPS,NEN,E,PR,RHO,H
l'ln ne
TABLE 7.4
-
I
I
-3
Shape Functions and Derivatives for Element Q8
Ji (I
stress (strain) data n) Nodal coordinates b) Element information• c) Nodal restraints
No. of Lines
~( I - 71)(-{ - 71 - 1)/ 4
Ji.,,
fi.t (2{ + 71)(1 - 71)/ 4
( I - m211
( I I- ~( I - 71)({ - 71 - 1)/4
(2{ - 71)(1 - 71)/4
(I+ m2r1
(I + m 1 + 71)({ + 71 - 1)/ 4
(2{ + 71)( 1 + 71)/ 4
o + m211
( I - ~( I + 71)( - { + 71 - 1)/4
(2{ - 71)(! + 71)/4
o - m271
(I - §2)(1 - TJ)/2
- g(I - 71)
- (I -
(l + ~( I - 71 2)/2
(I - 712)/2
-( I I ' )11
7
(I - f)(J + 71)/2
-[(I + 71)
(I -
H
( I - ~( I - 71 2)/2
-(I - 71 2)/2
2
4
-5
"
Dynamic Load Data for Program DYNAPS
Type of Data
€2} /1.
- (I -
• Jlor sequences of node numbers, see Figs. 7.9(b) and 7. IO(b). TABLE 7.6
C1
~ )11
7 .6 PROGRAM DYNAPS FOR PLANE STRESS AND PLANE STRAIN
Now we shall discuss Program DYNAPS for dynamic analysis of thin plnll·~ Ill plunc stress or prismatic solids in plane strain. For this purpose it is assumed tha H ttivcn continuum has been discretized using either Q4 or Q8 elemc111,, a dt•st·ribcd in Sec. 7.5. All such quadrilaterals in the analytical model huw th 1111111c thickness h, and the material is taken to be homogeneous and iso1111pl Table 7.5 shows preparation of structural data for Program DYNAPS . I hi lnhlc is simi lar lo Table 3.2 for plane trusses, described previously in Ser. I N 1lowcvcr, the second line of Table 7.5 contains four additional structural p11111111 1•k1s . The first of these is an indicator for plane stress (IPS = 0) or plarw ,111,ln (IPS I). Next is the number of element nodes NEN for the quadril11lt•111I Wht•rc NEN = 4 for Q4 elements and NEN = 8 for Q8 elements. The otht•r 1w1 sl rurturul parameters included are Poisson's ratio PR and the thicknt·~, II hrr lhcrmore, clement information now consists of the element number I 111111 ll(l(k numbers one through four for element Q4 and one through cip.hl fur t·lt•111t·nt Q8. Prcpuration of dynamic load data appenn, in Tublc 7.6, which is si 1111l111 111 Tuhk 4. 1 f'or plane trusses (sec Sec. 4. 10). l lowt·YC.1r, the loud p111·11111t·ll•1s NI I
J, X(J), Y(J) I, IN(I, 1), IN(I, 2), ... , lN(I, NEN) J, NRL(2J-I), NRL(2J)
NN NE NRN
Dynamic parameters
No. of Lines
Items on Data Lines
1
!SOLVE, NTS, DT, DAMPR
I
Initial conditions (a) Condition parameters (b) Displacements (c) Velocities
NNIV
NNID, NNIV J, D0(2J-1), D0(2J) J , V0(2J-l), V0(2J)
Applied actions (a) Load parameters (b) Nodal loads (c) Line loads• (d) Volume loads
I NLN NEL NEV
NLN,NEL,NEV J, AS(2J-l), AS(2J) J, K, BL!, BL2, BL3 , BL4 I, BVI, BV2
( lround accelerations (a) Acceleration parameter (b) Acceleration factors Forcing function (a) Function parameter (b) Function ordinates
NNID
l l
l NFO
IGA GAX,GAY
NFO K, T(K), FO(K)
"Pertains only to element Q4 (NEN = 4). For element Q8 (NEN = 8), we need three node numbers (J, K, L) and six force intensities (BL!, BL2, ... , BL6).
,111d NEV in Table 7.6 must be explained. If NEL * 0, at least one line load length) exists on the edgejk of an element. The types of line loads 1111 quadrilateral elements are illustrated in Fig. 7 . 11. For the Q4 element a ll111•11rly varying line load is specified by the force intensities BLl through BL4 1111,t arc listed in the table and shown in Fig. 7 . 11 (a). On the other hand , the Q8 , ll'111cnt in Fig . 7. 11 (b) hus two comprnwnl~ of quadratically varying line loads, ,h llrll'd by the force intensities BLI 1h1011gh BL6 (sec the footnote below the 111hll t lmt·c per unit
1
).
341
I W11 1111d lh rnn lll111111111lc11111I (;01111111111 111
I II
1•11, 11 111111 l>YNAI' 1111 l'ln11n
1,111111
nnd f'luno Strain
343
Ihm• v11ry1ng · st,·esst•s o., 0 1, ·111., 1111t I
I• 111n1>le 7.2 11 lllll 7 . 12 illustrates the cross section of a machine part that is s~bjecte? t~ a state of Ill · Th tis divided into Q8 elements and has a parabohcally d1str1buted step I' nm· Stram. e par h' J the physical hmd on edge 1-5 , with maximum intensity bx at node 3. I n t 1s examp e p,1111111ctcrs are: y
y
L.
b,,
5
la)
8
b,s
/2 4
t
b,, /1
·-
3
i
2
y
L.
b,,
j
lb)
Figure 7.11
Line loads for quadrilaterals: (a) element Q4; (b) element QB.
If NEV =I= 0, at least one element is subjected to volume loads (fol"l'l' prr unil volume). Here the notation implies that element I may have a unifiH11il y distributed force BVl in the x direction and a second uniformly distributed f«H l 13 V2 in the y direction.
In Program DYNAPS we take n = 2 each way Lo locale point s 1111 numerical integration. At each of the four points lhe computer evul1111t t•s 1111
n, IC'lic1u·1• 7.12
l'lnm• Hlcnlc1 with ()H ,,1,111tc.llllN.
11
344
Two- and Three-Dimensional Co ntinua
E = 207 X 106 kN/m 2 R2
=
2Ri
= 0.02 m
v = 0.30
p = 7.85 Mg/ m3
h=Im
bx= I kN/m
Chap. 7
for which the material is steel and the units are SI. We ran this data with Program DYNAPS, using IPS = 1, NEN = 8, DAMPR _ 0:°5 ,. and NUMI~T for the solution method. Translations of node 5 in the x and 1, directions appear m the computer plots of Fig. 7 .13(a). Also, the normal stresses SX SY, an? SZ near node 29 are plotted in Fig. 7.13(b). Maximum values of the nodai translations are 0.06505 mm and 0.04775 mm· and those 1"or th I 56 20 MP ' e norma stresses an• ' · a, 11 6 · 8 MPa, and 51 . 89 MPa. (The maximum shearing stress accompan in the normal stresses near node 29 is less than a twentieth of SX.) y ~
Sec. 7.7
7.7 ISOPARAMETRIC HEXAHEDRA FOR GENERAL SOLIDS
Figure 7 .14(a) illustrates the parent rectangular solid RS8 of the isoparametric hexahedron H8, shown in Fig. 7.14(b). In the former element, the natural coordinates g, 1/, and ( are orthogonal everywhere. For either element the generic displacements at a typical point are
u
= {u, v, w}
(1)
Nodal displacements consist of x , y, and z translations at each comer node, which fill the vector: q
DJ l E
345
lsoparametric Hexahedra for General Solids
= {qi , qz, q3, ... ' q24} = {u1, Vi , W1,
. . . ' Wg}
(2)
Trilinear displacement shape functions may be expressed as
,._I:; 0, 06
u
Cl
=
I
v = Ifiv;
J;u;
w = Ifiw;
i=I
i= I
~ 0.04
8
8
8
lf)
w
(3)
i=l
where
I-
<
f;
(/)
=
k(1 + to)(l +
T/o)(l
+ (o)
(4)
z
;'. 0. 02
nnd
I-
<
T/o
-' (/)
z < 0. 00 o:: I-
o.os
o. l 0
0 . 15
0.20 (ms)
(a)
~
l 20
a.."' :i::
~
SY
l 00
(})
N
w an Cl
z
60
U; =
z
<
f;Q;
(i
f;
= [~
= 1, 2, ... , 8)
(6)
111 which
~ 40 (/)
~ (/'/
(5)
TJ;TJ
The formulas represented by Eq. (4) are the same as those given previously in llqs. (7 .3-21). Values oft;, 1J;, and (; required for Eqs. (5) appear in Table 7 .7. For the parent rectangular solid RS8, explicit integrations are feasible, and Hliffnesses for an orthotropic material were presented by Melosh [8]. Furlhcrmore, consistent masses for this simplified element are easy to derive and were given in Ref. 2. Turning now to the more general isoparametric H8 element [9] , we take lhc geometric interpolation functions to be those that were given earlier as Eqs. ( I .3-20). Because the geometric and displacement shape functions are the same, lhc H8 element proves to be isoparametric. Equations (3) can also be stated as lhc matrix expression:
0
er
=
20
w
0
Ct".
·CJ) 0
( lil 1"1111111• 7, 11
lfr~llllllNl N Il l pJ11111 Nflllill I ~11111pl!1 fll Nll1p 1111111
~
0
(7)
~]/; 1
Ai, lwl'ore, the generic displacements u1 in Eq. (6) denote translations at any p111nt due lo the displ11cc mcnts q, at node i. l1111ddilio11 , Hlnlin displut.'l'llll'lll n•l11tionships 11111y be written efficiently as: t I
ll1«f1
(/
I , 2.
• 'H)
(8)
I w11 n11d 1111 nu l>lmonsionai Continua
Sec. 7.7
lsoparametric Hexahedra for General So lids
TABLE 7.7
,,
Nodal Coordinates for Element HS
i
g;
7/i
C,
1
-1 1
-1 -I
-1 -1 - I -1 1
2 3 y
)-, v,
.f
/
I• -
347
1
1
4
-1
5 6 7 8
-I 1 1
I -I -I I I
-1
1 1 1
where 0 fi .x 0 0 f;,y 0 0 0 f;,z f; ,y Ji,x 0 0 J;,z f;,y f; ,z 0 f; ,x
• ti
B;
(a)
= d f; =
0 D G2i
0 0
0
D G3;
DG2;
D Gli
0
0
D G3;
D G3;
0
DG!i
0 0
(9)
DG2i DGli .
Terms appearing in the submatrix B; were discussed in Sec. 7. 3. Next, we express the stiffness matrix for element H8 in Cartesian coordiIU\tes to be
K
=
fv BT(x,y,z) E B(x,y,z) dx dy dz
(10a)
In natural coordinates, this equation becomes
v,
I
und three applications of Gaussian quadrature give
5 i
6
/ w,
r • -
(b)
1•'1,curl! 7.14 111111 .
n
K
• II
rt
R j RkR1B[k,1E Bj,k,l
IJj,k,/ 1
(10c)
l=l k= l j= l
' l'his formula for numerical integration implies that the matrix Bj.kJ and the determinant IJ j,k,1 I are evaluated at each integration point, where the coordinates lll'l'
Element 118: (11) parent rectangular sol id; (b) isoparamctric countllf
11
=II I
(~, 'Y/k, (1). Similarly, the consistent mass matrix for element H8 is
L
M
p
M
,,f1F, (r'u.
fT(x,y,z) f(x, y,z) dx dy dz
,,.l ll'(/ , 1/,tl l ,IU , r1,[)l,ll,/,,t1l
(1 la)
(1111)
340
lw11 111111 I l11u11,l)lrrn11111fw111I Co111l1w11
mi,
7.1
lsopuromotrlc Hexahedra for General Solids
349
or Considering first the term M, 1 , we have from Eq. (15) 1l
M =PI
"
I I
ft
Rj RkR,ffk.l lj,(., j,JJ,k,/ 1 (a)
/= I k= I j = I
in which p is assumed to be constant. Furthermore, equivalent nodal loads due to body forces have the f'cu 11
Ph(t)
=
L
fT (x , y , z) b(x, y,z,t) dx dy dz
wlll'rc 1
J; = kO -
( I 111)
g)(l -
TJ)O - ()
(b)
1
,11hslilution of Eq. (b) into Eq. (a) produces
or M,, =
a:~p LL L
(1 - g)2 (1 - T/)2(1 - ()2 dg drJ d(
= Sabcp
(c)
27
or 11
Pb(t)
=L
n
1'1 occcding in a similar manner, we can find all of the terms in the first column of matrix M, us follows:
n
L L RjRkR1ffk,1b(t)j,k,I IJj,k,d
l = I k= I j = I
(I
1 1
J
pV (M)col.1 =
where both Ph and bare functions of time .
216
{8, 0, 0, 4, 0 , 0, 2, 0, 0, 4, 0, 0 , 4, 0, 0, 2, 0, 0 , 1, 0 , 0, 2, 0, O}
. F?r the rectangular solid RS8 , the Jacoblan matrix and its deterniirumt specialize to
J =
r~ ~ iJ 0
0
(d)
111 which the volume V = 8abc.
JJj = abc
( Ill
C
I(ere _the constants a, b, and c are half the dimensions in the g TJ , 1 ~ 1111 d_1rcct_ 10ns, as shown in Fig. 7. 14(a). Thus, Eqs. (10b), (lib), and (l,2h) 111 simplified, as follows:
Now let us examine a higher order hexahedral element that is formulated
11Ning quadratic geometric and displacement shape functions. The parent rectan»11lur solid (element RS20) is illustrated in Fig. 7.15(a), and its isoparametric 1 ounterpart (element H20) appears in Fig. 7 .15(b). For the rectangular solid the 1111tural coordinates g, TJ , and ( are orthogonal, and nodes 9 through 20 are lol'nted at midlengths of the straight edges. Both elements have the nodal 1ll11placement vector
Q = {q,, q2, q3, , , . , q60} = {u,, V1, Wt , . . . , W20}
( II) lllld
(17)
1h111 contains x, y, and 111t•nl
z translations at each of the 20 nodes. Quadratic displaceshape functions for these elements are 20
u=
Ullcl
P,,(t)
= abc
r, r. f
I
20
J;u;
i= l
20
v = Ifi v;
w=IJ;w;
i=l
i= I
(18)
wh~·rc I fT(g, TJ, ()b(t, TJ, ( , t) dt dTJ d(
( lr1)
11:x111u11fo 7.3 Ass1 u11i11g .1hu1 P _is constant, let us derive the terms in the first column of the l'OIININII r1111Ns mutn x M for the rectangular solid clement RS8 shown in Fig 7 J '1(u) JI II 111 111 J>111poN1\ w1· tWL'd functi ons /, through};i given hy liq . (4) wi th vu lu~•s ; ,. ,: • 111k<111 lwn , '1'11hl c 7.7, ' · 1 ,.,, l/1, _11111 L•
i"
/,
H I + so)O + 7Jo)(l + (o)(to + T/O + (o - 2)
I,
,t( I
I,
Ht
/1 .t ( I
+ TJo)(l + (o) TJ 2)( 1 + (o)( I + so)
- g2)( I
(i
= 1, 2,
... , 8)
= 9, 11, 17, 19) (i = 10, 12, 18, 20) (i
(i
= 13,
14, 1.5, 16) ( 19)
II o
IWtt 1t11ol 1111 1111 lll1111111nlo1111I Co11ll111111
"' / .U
Pro111 u111 l>YN/\1,ll
1111 (11111111111
351
Solids
1'AU1.11; 7 .H Nodnl Coordinates for Element H20
10
V
)-,
20
7 8 9 10
v,
I
. er
I
/
5--------e----..JI 6
•
II
-1
- 1
11
0
1
-1
1 1 -1 -1 1 1
- 1 1 1 -1
- 1 -1
12 13 14
- 1 -1
0
-1
-1
0 0 0 0
- 1
1 -1 0
0 1
1
-1 1
1 1 1
- 1
15 16
- 1
1 1
17
0
- 1
18
1
0
-1
19
0
1
1 1 1
- 1
20
-1
0
1
!1.{ = kO + 2g + I >I' X
T/
+ ?)(1 - 71)(1 - ?)
course, numerical integration is required for this element.
2
v, 5
/
i
t • -11
(b)
ll'll(Ut't' 7.15
1 1
111 this instance, the faces and edges of the element are quadratic surfaces and 111 ves, as indicated in the figure. Terms in the Jacobian matrix for element H20 are the same as those given 111 Eqs. (7 .3-23), but with the upper index 8 changed to 20. Furthermore, the 1qu utio ns for element H8 will pertain to element H20 if the number 8 is changed In 20 in appropriate locations. Derivatives fr.,, and so on, required for the d«•vclopment of element H20 are easily obtained and need not be tabulated. For 1•K11mple,
V
l'Ollll(l\lJ)Url .
- I
TJ;
1
3
l
t,
{,
- 1
i
t,
T/1
w,
(al
)-
I
2 4 5 6
2
2a
t,
1
-----, 17
I
Element H20: (a) parent rec tangular solid ; (b) isoparamclrk·
Y11 '.111·s ,0 1' §,, 'Y/1 , un~I (, for these formulas arc listed in Table 7.8. Ex 1,111 11 111111,1111,i1011s a.re poss ible for the subparametric parent clement. I 120 clement 1101 in p·,g · 7· I5(b) , we use gco1111•11 I1 I I • hir I ·lhc 1sopurumctric · . 111 ipo 1tl1<'.1~ lunc11ons th al arc the same as the displacement shape f'unctionN 111 1
•\'" ,/'I 11 I I
20
Y
~ .l,v, I
I
20
z
}: .fi z1 I
I
( 1111
1,8 PROGRAM DYNASO FOR GENERAL SOLIDS
111 this section we describe Program DYNASO for dynamic analysis of general H111ids. Before using this program, we must discretize a given solid using either I IH or H20 hexahedral elements (see Sec. 7. 7). The material of the solid is .tHN umed to be homogeneous and isotropic. The manner of preparing structural data for Program DYNASO is shown 111 Table 7. 9. Included among the structural parameters is NEN, which is the 1111mber of element nodes for each type of hexahedron. If NEN = 8, the hexalu•dru are H8 elements; and if NEN = 20, they are H20 elements . Therefore, r ll'lllent information consists of the element number I and node numbers 1 011ough 8 for element H8 and 1 through 20 for element H20. Table 7. IO gives preparation of dynamic load data for Program DYNASO. l•11 sl, we sec that there arc three possible initial displacements, initial velocities , 1111d applied forces for each node, Also , lhc clement loads implied by the p111m11ch.:rs NEI., NEA , und NHV ll' q11111• Home explanation . If NEL :/= 0, at h•111111u1t• li111.• loud (lon'l' pl' I' uni I h-n~tlt) 1•111 ~tH011 lhc l'dgl' jk of' an c lement. For , 1«•1111•111 I IH II lt11l'1til y v11 1yi ,11• 11111• Ii 111d ii, d, f1111•d hy tlw 1'011·1• i111rnsitks Ill. I
Jb2
Iwo 1111«1 I '111111 l>l11111111lcm11I c·0111'111111
TABLE 7. 9 Type of Data Problem identification Structural parameters General solid data (a) Nodal coordinates (b) Element information• (c) Nodal restraints a
'111c. / .8
Structurul D11tt1 for Program DYNASO No. of Lines
Items on Data Lines
1 I
Descriptive title NN,NE,NRN, NEN, E, PR, RHO
NN NE NRN
J, X(J), Y(J), Z(J) I, IN(I, 1), IN(I, 2), ... , IN(I , NEN) J, NRL(3J-2), NRL(3J-l ), NRL(3J) .._
For sequences of node numbers, see Figs. 7. l4(b) and 7. I 5(b).
TABLE 7.10 Dynamic Load Data for Program DYNASO Type of Data
No. of Lines
Dynamic parameters
I
!SOLVE, NTS, DT, DAMPR
Initial conditions (a) Condition parameters (b) Displacements (c) Velocities
l NNID NNIV
NNID, NNIV J , D0(3J-2), D0(3J-l), D0(3J) J, V0(3J-2), V0(3J-l ), V0(3J)
l\ppl icd actions (11) Load parameters (b) Nodal loads (c) Line loads" (d) Arca loadsb (c) Volume loads
l NLN NEL NEA NEV
NLN,NEL,NEA,NEV J, AS(3J-2), AS(3J- l), AS(3J) J, K, BL!, BL2, ... , BL6 J, K, L, M, BAJ , BA2, . . . , BAl 2 I, BVJ, BV2, BV3
( lround accelerations (11) Acceleration parameter (h) Acceleration factors
=
Items on Data Lines
Fmt'i ng function (11) Function parameter (h) Function ordinates
lhr ough BL6 . The first four are as shown in Fig. 7.1 l(a) for element Q4, and rhc last two pertain to the z direction. On the other hand, the H20 element may huvc a quadratically varying line load, specified by BLl through BL9 (see the llrst foo tnote) . The firs t six force intensities have the same meanings as in Fig. I . I I (b) for element QS , and the last three are for the z direction. IF NEA -::f:. 0, at least one element has an area load on one of its surfaces. I•'ig ure 7 .16 shows the types of area loads for hexahedra. For element HS such loads on face jklm are defined by 12 numbers, of which the first four (BA l through BA4) denote force (per unit area) in the x direction, as indicated in F ig. l . I 6(a). The next four (BAS through BAS) pertain to they direction, and the last lour (BA9 through BA12) apply to the z direction. It is assumed that each component of area loading has a bilinear variation over the surface jklm. On the other hand, we take a biquadratic variation of area loading on a surface of the 1120 element. In this case the loads are specified by 24 intensities (see the second l'ootnote). The first eight represent force (per unit area) in the x direction , as depicted in Fig. 7 .16(b). The next eight are for the y direction, and the last eight 11rc for the z direction. Volume loads BVl, BV2, and BV3 on both types of hexahedra simply l'Onsist of uniform intensities of force (per unit volume) in the x, y, and z directions. Also note that GAZ is included in the list of acceleration factors , as for any three-dimensional analysis. In a manner similar to Program DYNAPS, we use n = 2 in each of three ways to locate points for numerical integration. Thus, there are eight such points, at which the time-varying stresses CTx, CTy , CT,, 'Txy , Ty,, and Tzx are determined in Program DYNASO. Again, we had to use n = 3 each way for terms in the consistent mass matrix of element H20 to retain sufficient accuracy . 11:xample 7 .4 'l'hc tapered cantilever beam in Fig. 7. l 7(a) is doubly symmetric, and the parabolas with 11pcxcs at the support determine the rate of taper. This beam is made of reinforced l'Oncrcte and has a rectangular impulse of magnitude Pi and duration t 1 applied in they direction at its free end. Physical parameters are given as follows: E = 3.6 x 103 k/ in. 2
J l
I NFO
"'I t•11111nh only ro clc_mcnt HS (NEN = 8).
353
1•,011111111 DYNASO for General Solids
IGA GAX , GAY , GAZ
L
NFO K , T(K), FO(K) For clement H20 (NEN
nodr numbe1h ond nine force intensities . "Fm L' k1111.ml 1120 use eight node numbers and 24 force intensities.
= 20)
wc need rhrl'l' '
=
= 120 in.
11
= 0.15 Pi =
400 k
p
=
2.25 x t1
=
10- 7 k-s 2 / in.4
5 ms
where the units are seen to be US. Figure 7 . I 7(b) shows discretization of a quarter of the beam into two H20 ele111cnts, with nodal restraints imposed for symmetric and antisymmetric deformations. To define the geometry of this simple network, we need only state that the x-coordinates of nodes 9 through l 2 are 9 0 in. , those of nodes l 3 through 20 are 60 in. , and those of nodes 1 1 lh rough 24 arc 30 in. We used the foregoing data in Program DYNASO with NEN = 20, DAMPR = II. I0 , 1111d solu1 io11 hy Suhprogrnrn NORMOD (wilh NMODES 12). Figure 7. I 8(a)
354
Two- and Three-Dimensional Continua
Sec. 7.8
Program DYNASO for General Solids
355
k y
y
0.1L
)-, z 0.3L
y
(a)
(a)
y
j2
t
Sym.
25
21
j3
13
•· - i,
/
I
1
4
29
/
32
12
/20
24
Typ.
Sym. (ll) (h )
li'l1111r1• 7. 16
t\rcu loud~ for hcxuhcdrn: {u) t•lcnll'nt IIH; (h) clcnw111 1120
Figure 7. 17
(a) Tapered c11111ilcvcr beam; (b) 1-120 clements.
t wu 111d lli11m 1>11 1111 111 1011111 Co11t l111111
."'
Ill
~.Lt
lnop111 1111111t1 ii I hHIIMIII N 1111 A,<11y11111111tllo tlolld•
367
,h pil'IH II qu111 l(:r ol l hl• 11•1•t1111Httl111 l111p11I N1' 111 nodc 6; and they-translation at that point I plotll'd in Fig. 7. I8(h), Nh11wi11g II muximum value of 0.09514 in . at t = 3. 75 ms. \ 1~11 . the normal stress SX ncur node 25 is plotted in Fig. 7. 18(c), for which the 111l11i111u111 value is - 1.040 ksi at the same time .
10 0
Q.
;! 50
'-
(.0
~ 0
I
'
2
0
'
4
6
2'
<-so
,0 ISOPARAMETRIC ELEMENTS FOR AXISYMMETRIC SOLIDS
I
8
\ 11 nx isymmetric solid is defined as a three-dimensional body that may be d1•vdoped by rotation of a planar section about an axis. This type of body is _,1111ctimes called a solid of revolution. Cylindrical coordinates r, z, and () p111vide a suitable reference frame, as illustrated in Fig. 7 .19. We assume that th1• body is axisymmetric with respect to the z axis and that a typical finite 1 l11111cnt is a circular ring. This ring element may have various cross-sectional h11pcs, but we will deal only with isoparametric quadrilateral sections. Allhough nodes are shown as dots on the cross section of a ring element, they are m•1ually nodal circles. If the loads on an axisymmetric solid are also axisym111c1ric, we may analyze a representative cross section as if it were a two-
l-
o
<.
3 - 100
-
(a )
C ~
0.10
(.0
~ o.os
0
7
'..t,
Z,V
I
0, 00 2
6
7
8
I
T
Cl
;- -o.os
s)
<
-J U)
~
-o. 10
,0(b)
I
>-
r
1--- -. I , 00
IJ'l
---
o.so
(\J
~
o.oo
2
0
4
8
7
~
10
T Cms)
-o.so
I
Iii
:;,:
~ - 1 .oo
I 1z
(c)
!figure 7.18 Tapered beam: (a) loud; (b) displucemcnt ; (c) stress.
Sec. 7.9
lsoparametric Elements for Axisymmetric Solids
369
Two- and Three-Dimensional Continua
358
dimensional problem. At first, only axisymmetric patterns of loads will h~ considered, but nonaxisymmetric loads also will be covered later in the section . For any point on an axisymmetrically loaded ring element, the gencrk displacements are u={u,v} (I)
Figure 7 .20(a) shows the cross section of element AXQ4, which derives its characteristics from the quadrilateral element Q4 in Sec . 7. 5. Bilinear displacement shape functions in matrix f are the same as those for element Q4, and the strain-displacement submatrix Bi becomes
Tr~nslat~ons u an? v occur in the rand z directions, as indicated in Fig. 7.1'> With ax1symmetnc loads, the translation w in the () direction is zero and tlu shearing strains 'Yr& and y,8 are also zero. However, the figure shows f;ur typl'~ of strains that are nonzero, as follows: (" )
Relationships between these strains and the generic displacements in Eq. ( 1) seen to be Er
_ au - ar
€9
111 r
av
= az
'Yrz
=
au
av
dZ
+ ar
a ar 0
d=
1
r a
az
(i = 1, 2,3 , 4)
0
r
(7)
fi., .h,r which is obtained by using the operator d in Eq. (4) on submatrix ( from Eq. (7 .5-5). The radius r in Eq. (7) is found as (8)
In addition, the derivatives/;,r = Deli , and so on, are given by Eqs. (7.3-18) , except that r and z replace the coordinates x and y . The stiffness matrix for element AXQ4 may be formulated in natural coordinates as
0
K =
a az
ff 1
<•I)
0
= 21T
Jo27T BTE B \ J \r d(J dg d17 1
L, rl
BTE B \ J \r dg d17
(9)
Similarly, the consistent-mass matrix is
a ar
M
Sym .
1J
+ 11)(1 - 2v)
L
Bi=
i= l
For an isotropic material , the stress-strain matrix is
(I
fi.,
4
In this instance the nonzero term 1/r in the third row of matrix dis a multipl1r1 of u, not a derivative. Corresponding to the strains in Eq. (2), the four types of nonzero strcslil'II depicted in Fig. 7. 19 are (J' = { CTr, er,, (T9, Tr, } ( S)
E
0
r = "'Zfir;
These relationships are embodied in the differential operator
E =
0
21r(r + u) - 21rr u = =21rr r
( I) E,
.h,r
,,
1J
1J II
1J
0
0
= p
rl L, rl rl
L27T fTf \ J \r d8 dg d17
= 21Tp
fTf \ J \r dg d17
(10)
/\lso, equivalent nodal loads due to body forces are
Pb(t) =
L, rl r1T
= 21T
rl rl
fT b(t) \ J \r d8 dg d17 fTb(t) \J\rdgd17
(11)
( ) 1J
0
I -
2 11
2
'l'ht M•I X •I 111111 y i1; H01tll111 tu till' I X I 111utii x 1111 pl1111,• Nlnii11 In l\q ('/ ,). 11)
Numerical integration is required to evaluate Eqs. (9), (10), and (11). The cross section of element AXQ8 appears in Fig. 7.20(b) . Its properties nrc sitnilar to thrnic of the quadrilateral clement Q8 in Sec. 7 .5. Biquadratic di splnL'l'tlll'III ~l111p,· t 1111l'lions in 111alt ix f H1·c the sam e as for clement Q8.
360
Two- and Three-Dimensional Continua
Chap. 7
Sec. 7.9
361
lsoparametric Elements for Axisymmetric Solids
Expressions for B;, r , K, M, and ph(t) are similar to those for element AXQ4 given by Eqs. (7) through (11), except that i = I , 2, ... , 8.
z \
\
Example 7 .5
\
Find the consistent mass term M 35 for the axisymmetric solid element AXQ4 in Fig. 7 .20(a), using Gaussian numerical integration with n = 2 each way. Let the coordinates of nodes 1, 2, 3, and 4 be (10,2), (14, 1), (15,5) , and (11 , 4), respectively. From Table 7 .1 for n = 2, we have R1 = Rk = 1, so that the numerical form of Eq. (10) gives 2
=
M 35
21rp
2
L L (ht.k(J3)j,k IJ,.k lrJ,k
(a)
k=I j= I
• ---.... u . i '
ks
In this case the bilinear functions h and h are needed to evaluate the term M35 because of the arrangement of matrix fin Eq. (7. 5-5). Substituting Ji and.fi from Eq. (7 .5-6) into Eq. (a) yields (b)
r
I
To apply this formula , we first calculate the Jacobian matrix from Eq. (7. 3-10), as follows:
(a)
(1 - 'l'/)
(1
+ 'l'/)
+ g)
(1
+ g)
-(I z
' '\
-(1
+ .,.,)] 14 10 42~1 g) [ 15
(1 -
11
(c)
I
3
Then the determinant of J is (d)
llvaluating the terms in Eq. (b) at each of the four integration points and summing the results produces (e) M35 = 21.061rp = 66. l5p
I
5
2
•-u.'
which can be finalized using a numerical value for p.
i
k==ser----..1__0- - r I (b)
Turning now to nonaxisymmetric loads , we can divide them into two sets / I 11 . The first load set is symmetric with respect to a plane containing the axis or revolution , and the second is antisymmetric with respect to that plane. For l'onvenience, the r-z plane is taken to be the plane of symmetry. Fourier tl/'composition [12] of the symmetric loads form harmonic terms produces Ill
Figure 7.20 Quadrilateral sections: (a) element AXQ4; (b) element AXQ8 .
b,.
=
I
Ill
b,j cos JO
b,
=I
J• O
J• O Ill
ho
L IJ111 sin JO I
o
b,j cos JO (12)
Two- and Th ree-Dimensional Cont inu a
362
Chap. 7
Sec. 7.9
lsoparametric Elements for Axisymmetric Solids
363
where b,j, bzj, and b8j are functions of rand z only. When} = 0, we have be = O; and Eqs. (12) become the case of axisymmetric loads. Otherwise, j = 1, 2, ... , m represent cases of nonaxisymmetric loads that are symmetric with respect to the r-zplane. Figure 7.2l(a), (b), and (c) show the first harmonic loadN for the r, z, and directions, respectively. If the loads were antisymmetric with respect to the plane of symmetry, the functions sin Je and cos Je would bl· interchanged. Generic displacements for nonaxisymmetric loads must include the trans lation w in the e direction . Thus,
e
u = {u, v, w}
(IJ)
(a)
and we must also have ,',8 and r',o in the strain vector, as follows: E
z
(14)
= { Er, Ez, Ee , Yrz, Yzo, Y,o}
Strain-displacement relationships developed by Love [13] are Er
= "1
au ar
=
E,
av az
1 aw
= -; + -; a8
av r ae 1 au aw w -va=-- + - - r ae ar r
au+av -az ar
rrz -
u
Ee
aw az
1
'Yo= - - + z
( 15)
r
(bl
I lcrc we see that the radius r appears in the denominators of several expressions From Eqs. (15) we can form the operator d as
d=
a ar
0
0
0
a az
0
r a az 0 1
a r ae
1
0
a ar 1 a r ae
a
r ae
a az
a or
0
( I hi
0
Figure 7.21
,.
Tlw slrcss vector for nonaxisymmctric loads musl contain r, 11 and lollows:
er
!11,, 11,, 11 11 ,
11
z
,
1 ,,, 1111 )
7', 11 , 11 '1
( 1/)
Nonaxisymmetric loads: (a) b,1 cos 0; (b) b,1 cos O; (c) bo1 sin 0.
Stress-strain relationships are easily extended to cover six types of stresses and llw corresponding strains. For example, if the material is isotropic [see Eq. (6)] , Wt' odd / 1\, - H,16 H/2( I + v) to form a 6 x 6 matrix E. Tlw response of Hll nxisymmclric solid to II series of symmetric , harmonic lo11ds rnusists of 11 1w1 il'S of symnwt, it·, h111111011k, pentric displ11ccmcnts that 11111y Ill' t'X PH'HIH'd 111'
364
lwo- 1111d l 1111111 l>l111nr11l111ml Lo11tl11u11 n1 II/
=L
u
Uj
/ ,
1'1011111111 l>YAXSO for Axleymmotrlc Solids
10
cos J(:)
V
j• O
=L
VJ
cos JO
IINl11g llqs . (2 1) in Eq. (22) , we find that
j• O Ill
L
w =
M
w1 sin}(:)
j=O
Again, i~ the l?a~s were antisymmetric with respect to the plane of syn11111·1, v ~he funct10ns sm J() and cos JO would be interchanged. Applying the opcr111111 d m Eq. (16) to Eqs. (18) expressed in terms off, we find a typical partilio11 uf the strain-displacement matrix to be
(B;)1
Ji., cos}()
0
0
Ji.z COS J()
0 0
Ji- cos}'()
0
}.Ji - cos}'()
Ji., cos J() .Ji Slllj . '() -1-
0
r
=
Ji., cos}() 0
.Ji -1r
r
r
. '()
i,r -
;
f, {, f" f, L
where k = 2 for j = 0, and k = 1 for j (k = I) appears as a consequence of 2,r
1
2
cos J() d() =
0
(j = 0, 1, 2, . .. , m) 2, . .. , m. The latter co1111111111
sin 2 J() d(J = 1r
( 'II
i11 which CJ
()
()
,}J
I
fT f IJ I r dg dYJ
(24)
fTcJc1b(t)j
I
IJ I rdOdgdYJ
fTb(t)1 !JI r dg dYJ
(j
= 0,
1, 2, .. . , m)
(25)
= {b,j,
bzj, be1}
(26)
Finally, the stresses for each harmonic response are (j
= 0, 1, 2, . .. ,
m)
(27)
t >I course, such stresses, as well as nodal displacements, must be added at the , 11tl of the analysis.
/ ,10 PROGRAM DVAXSO FOR AXISVMMETRIC SOLIDS
( "l
0 cos JO
1
77
b(t)j
~Ill}
Similarly, the consistent mass matrix for each harmonic set of symn11•1111 di splacements becomes
rco;/°
Lf
( 111)
= 1,
[21T
Jo
k1rp
whore
BJE B1 JJI rd() dg dYJ
BJE B1 IJ I r dgdYJ
= k1r
f, f, f = f f =
hr
(ii fi) . '()
0
Slllj
( I!))
Ji.z sin j()
=
1liiN l'ormula is the same for j = 1, 2, ... , m, and it does not change for ,1 111tsy111111etric displacements. Thus, to determine frequencies and mode shapes 1111 nny value of j, we use the (variable) stiffness matrix K1 from Eq. (20) and 1111• (constant) mass matrix M from Eq. (24). However, in the latter equation note 11!111 k = 2 for the case of axisymmetric vibrations, where j = 0. Equivalent nodal loads for each harmonic set of symmetric body forces 111k1• the form
p,,(t)1
where i = 1, 2, ... , nen and j = 0, 1, 2, ... , m. An element stiffness matrix for each harmonic set of symmetric displ111 ments may be written in natural coordinates as
~=
365
Ill
(j
0, I , 2, . . . , 111 )
(
1
11
I , t us now consider Program DYAXSO for dynamic analysis of axisymmetric ~ollds with axisymmetric loads. We assume that such a solid has been discretized 11110 ring elements AXQ4 or AXQ8, which were decribed in the preceding Nt•1·tion. As before, the material of the solid. is taken to be homogeneous and IHOll'Opic. With very few modifications , Program DYNAPS in Sec. 7.6 can be convm tcd to program DYAXSO. For example, the subprogram in DYAXSO that H1%:rates the element stiffness matrix is practically the same as that in Program I >YNAPS. However, when calculating stiffness terms there is multiplication by 1 11 r instead of h. Within the logic of that subprogram , the computer must , vnluatc not only BJ.k and IJ1. k I, but also ,i. k at each numerical integration point. 'll 111ilur comments also apply to the generation of consistent masses and equiv11lt111t nodal lontls. St rncturnl dnt n lor l'rngrum DYNAPS (s1·1• 'l'uhlc 7.5) must be altered to 1H'l 1111111 for llw l11l I 11 11 11 !ht• rnnt i,11111111 to ht• 111 11 d y:1.1•tl is 1111 nxisynrnwtric solid .
366
Two- and Three-Dimensional Continua
Chap. 1
Sec. 7.10
367
Program DYAXSO for Axisymmetric Solids
The structural parameters JPS and H must be deleted, and the nodal coordinat~N x and y are replaced by r and z. The only significant changes in the dynamic
loud
data for Program DYNAPS (see Table 7 .6) are that line loads become area loatlN and the acceleration factors GAX and GAY must be replaced by GAZ. As in Program DYNAPS , we take n = 2 each way to locate points for numerical integration on the quadrilateral sections. At each of the four points th~ computer evaluates the time-varying stresses a,, a,, a 8, and Trz. Once more, w~ needed to use n = 3 each way for terms in the consistent mass matrix of elemt•nl AXQ8. Example 7.6 An axisymmetric titanium valve head is discretized using seven AXQ8 elements, 11shown in Fig. 7 .22. Acting on the lower surfaces of elements 2, 4, and 6 is an explosivt1 internal pressure p,, which is resisted by the valve seat (or restraint) at node 34. Enrh edge of the finite-element network is divided into equal lengths between the nodes on thnt edge. For this problem the physical parameters are E
= 1.7 x
104 k/ in. 2
L
=
v
= 0.33
0.25 in.
p
(p, )max
=
= 4.20 x
z I
I
~ ~Stem L
1 I 2'
l---~~~~ 6L ~~~~~~·1
, I 3 _ _ _ _ _ _ _ _ _ _ _ _ -,
I I I
I
10- 1 k-s2 / in.4
I I I
1.53 k/ in. 2
b, = p,e~·
(Ill
llcrc the symbol e ~· denotes the transpose of a unit row vector in the direction of 11 ', which is normal to the tangential direction g' at the surface. The vector eT), may be fo11111I by first calculating a unit row vector in the direction of g' as 1
I
\
- iI
and the units are US. For this example we need to calculate equivalent nodal loads due to pressur,• I' on a surface of an axisymmetric solid element AXQ4 or AXQ8. Considering any of tl11 four surfaces of such an element, we can find the components of p, in the direction~ ol r and z, as follows:
e,, = -[r,f
I
-
9
"
(!})
C
I
"\ I
Restra int
29
ti 2L
~~2~ 7
k::::'.'.~ -/~ /
Z,f ]
I I I L
I ///
~30~ \ __,____ r
1
where c = Y(r.d
+ (z.,,)2
(d
Second, from known orthogonality relationships, we can write the unit normal vcctrn In the form eT), =
L
- [ -z.,. r .f]
4L
l~//
~r::;i;es~:face
C
Then the equivalent nodal loads on surface nodes become
Figure 7.22 (ii
In this lormuln the dctrnnlnnnt of .J' is
Valve head with AXQ8 elements.
,11111
IW11 ,11111 I limn Dl1tw1111lo1111I Cot1ll111111
369
1111111111111 ""
Pi, = 2np,.
" V) V)
w
0:: a_
0, 50
0::
w ....
z
o.oo
so
I 00
r
200
sr
::. so 40
I")
w C1 30 0
~
z
0::
i:'_j 20
ri .
z V)
w I0
V) V)
w
0:
'
V)
0
H.
200 r
(m s\
( b)
1r1gurc 7.23
r df
(h)
1', f
REFERENCES
(m s l
'
"' -" ..,
I
11111N on these equivalent nodal loads are automatically determined by Eq. (h) for 111, ~sure in the positive sense of the normal direction "f/ '. We processed the foregoing data with Program DYAXSO using NEN = 8, 11/\ MPR = 0.05 , and Subprogram NUMINT for responses. The computer plot in Fig. I 'l(a) gives the variation of internal pressure acting on the valve head, and Fig. 7 .23(b) i111ws time histories of the normal stresses SR, SZ, and ST at the integration point near 1111dl· 34 . Maximum values of these stresses are 40.04, 34.54, and 50.67 ksi at time 80 ms .
w ~ I ,00
°
f r1' [-z.f] - I
y
11,
Valve head with internal pressure: (a) load; (b) stresses.
IJ'I =
I(),
r.e· If/
- z.e· C
11
C
'I'hls. ''.oh.ll'lninant tr~nsforms arc length instead of area because its second mw iN 11111 1111111:1.u l lo have unrt length. Expanding the determinant produces
IJ' I = :.[(r.d + (z.(') 2'1 = c S11hNfil111i11g liqN. (a) , (d) , 1111d (g) into Eq. M , wt• find 111 111
111
I' II
Timoshenko, S. P. , and Goodier, J. N. , Theory of Elasticity, 3rd ed., McGraw-Hill, New York, 1970. Weaver, W., Jr., and Johnston, P. R., Finite Elements for Structural Analysis, Prentice-Hall, Englewood Cliffs, N.J., 1984. Lckhnitskii, S. G., Theory of Elasticity of an Anisotropic Body, translation from Russian by P. Fern, Holden-Day, San Francisco, 1963. Gere, J. M., and Weaver, W., Jr., Matrix Algebra for Engineers, 2nd ed. , BrooksCole, Monterey, Calif. , 1983. Scarborough, J. B., Numerical Mathematical Analysis, 6th ed., Johns Hopkins Press, Baltimore, Md. , 1966. Zienkiewicz, 0. C., The Finite Element Method, 4th ed. , McGraw-Hill, Maidenhead, Berkshire, England, 1987. Ergatoudis , B., Irons , B. M. , and Zienkiewicz, 0. C., "Curved Isoparametric 'Quadrilateral' Elements for Finite Element Analysis," Int. J. Solids Struct., Vol. 4, No. 1, 1968, pp. 31-42. Melosh, R. J., "Structural Analysis of Solids," ASCE J. Struct. Div., Vol. 89, No. ST4, 1963, pp. 205-223. Irans, B. M., "Engineering Applications of Numerical Integration in Stiffness Methods, " A/AA J., Vol. 4, No. 11 , 1966, pp. 2035-2037. lkgatoudis , J., Irons, B. M., and Zienkiewicz, 0 . C. , "Three-Dimensional Stress Analysis of Arch Dams and Their Foundations,"-Proc. Symp. Arch Dams (Inst. Civ . ling., London) , 1968, pp. 37-50. Wi lson, E. L. , " Structural Analysis of Axisymmetric Solids," AIAA J., Vol. 3, No. 12, 1965 , pp. 2269-2274. Snkoln ikoff, I. S. , and Redheffer, R. M., Mathematics of Physics and Modern il'118i11eering, McGraw-Hill , New York, 1966. Love, A. E. H., The Mathematical Theory of Elasticity, 4th ed., Cambridge UniverNily Press , Cambridge, 1927.
8 Plates and Shells
371
I h11111111t IOI l'lnt111 111 ll11mll1111
1111, 11.~
1,2 ELEMENT FOR PLATES IN BENDING
11 IN possible to specialize an isoparametric hexahedron (see Sec. 7. 7) to become ,, plulc or a shell element by making one dimension small compared to the other two. This type of modeling was introduced by Ahmad et al. [2] and applies to ,11111lyses of both thick and thin plates and shells. For analyses of flat plates, it I~ 11lso necessary to restrict the other two dimensions of the modified element to II,• in a single plane. This section is devoted to the specialization of the isop111·111netric hexahedron H20 to become a plate-bending quadrilateral called ele111,•111 PBQ8. While an H8 hexahedron could also be specialized, the resulting 1111ight-sided quadrilateral would not be suitable for conversion to a shell ele11111nt later in the chapter. Figure 8. l(a) shows the original H20 element, which has quadratic interpolation formulas defining its geometry . In order to understand the constraints lll'l'ded to convert it to a plate-bending element, we first form a flat rectangular ~i,li<.I by making the natural coordinates 1/, and ( orthogonal and the ( dh11ension small. The resulting element appears in Fig. 8. l(b) as the rectangular 1111rent PQR8 of element PBQ8 before constraints. Note that groups of three 111dcs occur at the comers, while pairs of nodes are at midedge locations of 1 l'ment PQR8. By invoking certain constraints, we can convert each group and p11'r of nodes to a single node on the middle surface, as shown in Fig. 8. l(c). lht.: nodal displacements indicated at point i in that figure are
t,
8 .1 INTRODUCTION When a plate is subjected to forces applied in the direction normal to itN own plunc, it bends and is said to be in a state of flexure. For this type of prohh 111 we deal with flexural and shearing stresses and strains that are somewhat 1111111 ogous to those in a beam. However, the analysis of a plate is more compl11·111 ~ because it is two-dimensional; whereas a beam is only one-dimensional. On the other hand, a shell is three-dimensional, and its analysis is ,·vitn more difficult than that of a plate. In shells we must consider not only flt•x11111I 1111d shearing stresses and strains, but also those associated with membrn11,· 111r i11 plunc) deformations. Finite clements for dynamic analyses of plates and shells will be h1111r1I upon those for general and axisymmetric solids from the preceding chuph 1 These specializations will automatically include the effects of shearing dl'lut 111111ions and rotary inertias , as in Mindlin's theory of plates [1] . . Computer programs in this chapter perform dynamic analyses of pl111,•11 111 bending , general shells, and axisymmetric shells. All structures that we nn11ly1 urc nssumed to be composed of linearly elastic materials with small struim, 111111 lllspluccmcnts . Guyan reduction (see Sec. 6. 7) is used in the plate and Nhrll programs to eliminalc the nodal rotations and rl.ltain the translations.
ll/0
5
z
)-, X
16
2
3
la)
Figure 8.1 Spcd11H1111l1111 111 ht• K11hl·dron: (a) element H20; (b) rectangular parent PQR8 of clllll1l'III l'll<)H 111•11111• l 1t11Hl11il11tH: (c) constrained nodal displacements.
111 ,
u.7
373
I 11111111111 1111 1'1111011 111 llondlno 3
(l
.t
.t
) 'l
/
I
j
------ 2
h; 2
l
3
;t
k
~.u lb)
21 )-, _l z
le)
1
t1
-3
i
/
2
/
h;
Figure 8.1 (cont.) qi =
{qil, qiz,
2
qi3} = {wi , ()xi, Oyi}
(i = 1, 2, ... , 8)
where ()xi and Oyi are small positive rotations about the x and y axes. Relationship between nodal displacements at a comer of element PQR8, a midedge of P()l{ H and a node of element PBQ8 can be seen more clearly in Fig. 8.2. The two lyp of constraints to be introduced are:
k
k
7/t
------ 5
8
4/j 6
9
the [ direction. 2. Normals to the middle surface remain straight (but no longer non11111) during deformation. Using these criteria, we can relate the nine nodal translations in Fig. 8.2(11) 111 the three nodal displacements in Fig. 8.2(c) by the following 9 x 3 consm1111t
Figure 8.2 Nodal displacements: (a) comer of PQR8; (b) midedge of PQR8; (c) node of PBQ8.
Similarly, the six nodal translations in Fig. ~.2(b) ~e related to the three nodal displacements in Fig. 8.2(c) by the constramt matnx 0
matrix:
0 0 1 0
0 0 0 0
hi 2
Cai= 0
J/'1
0 0
0
0
0
1
0
0
0
0
h, 2
0
()
()
0
hi 2
0
(
)
hi 2
0
hi 2
0
Cb,=
hi 2
(c)
(b)
la)
I . Nodes on the same normal to the middle surface have equal translation Ill
k
0
0
hi 2 0
0 0
(3)
hi 2
0 0
· of size · 6 x 3 · If we were to apply each of these constraint· matrices in · h 1s wIuc 1 t lour locations , we wou ld be ublc to reduce the number of n~dal d_1sp acemen s 60 lo (8)(1) 24. Instead of following th1~ path , how(4)(9) + (4)(6) I iom we will pursue 11 11101t• diil'l'I lorn1ulat1 , ·on o f e 1ement PBQ8 m a manner ,•vcr, "imilur to that in Rd. I, . · · 1 Figure 8 . I show Nt•h•uu 111 l'IIIJH, ol t 011Nl11nt thickness h , with its neutra
l'l11t111 1111d Hh11ll1
1111. 11.2
376
I h11111111t Im l'lnl1111 In lh111cll11u
1111k pcndcntly of w and arc not related lo it by differentiation , as in a beam. In lhlNcusc the displacement shape functions may be displayed in the matrix form
0
0
0
-,-
1
0
)I, V
f;
4
8
=
lJ
'X, U
i /
h 2
3 6
fAi
y,
0.,
=
8
= 2., f;y;
i= I
0 lJ f;
[1
0
-1
0
(7)
0 0
0
f.I
h
= fA. + ~Y-2 fo· I
I
(8)
nnd
8
y
(6)
0
O O OJ 0 0 0 f;
N111 fuce Iyin~ in the x-y plane. Its geometry is defined to be the same as thut I1,1 t'lll111cnt Q8 m Sec. 7.5. Thus,
= 2_, /;X;
1, 2, . . . , 8)
l'hcn
Figure 8.3 Element PBQ8.
X
=
(i
f;
0
l'o isolate terms in submatrix f; that multiply (h/2, we let
'-
"
t •-e .
/!.2
(z
= 0)
(7 .5- 20)
(9)
i=I
where
fi = W +
fo)(l
= f(l -
{2)(1
f;
Clcncric
+ 7Jo)(-1 + + 110)
{o
+ 110)
(i = 1, 2, 3, 4)
= 5, 7) (i = 6, 8) (i
(7 .5- 1'))
'l'he formula in Eq. (9) will prove useful for obtaining the consistent mass matrix nnd equivalent nodal loads for element PBQ8. The 3 x 3 Jacobian matrix required for this element is Xg
J =
displacements at any point off the neutral surface are u
We assume that w,
0.,, and
= {u, v, w}
[
! II
Oy vary quadratically over the element, so thul
where z,,
Y,€
Y.7) 0
X ,7)
0
=
z{)y
=
z
V
=
-z()x
=
8
z,,
and so on.
L f;Oyi i= I
i= I
'l'he inverse of J becomes
-z"' I'()xi L, Ji
{,x
71,x
{;
~y
i= I
8
w = 2.,fiw; /• I
wht•i:t· z. (~,(~· I1~ ll~ese expressions the displacement shape functions.ti IUl' 1h, N11111t us I hose 111 f!qs, (7 .5- 19) . Note thnl tht· rotations o, und ()1, urt· l'lmNrii
(10)
= h/2 and
8
u
OJ 0
J I = J* =
[
(11)
where (,z = 2/h. We need cc1·tnl11 dl 1ivutlv1•N with respect to local coordinates, which are mllccted into the l'ollnwlup I I 1111111 ix:
J/6
I 1l111un ond Sholle
U,(:
v,,
h '2,fi,l)y;
w,, 8
u,,,,
h ( 2fi.110y;
=I
v ,11 w,r,
- (1/i,/Jxl
u.,
v.,
h 2f;Oy;
w.,
377
Elomo11t l or Plates In Bonding
As with the submatrix f 1, we can isolate terms in B1 that multiply (h/2, as
f i.,w,
h -( f; . 11 0x1 2 h - J;ex, 2
i=l
'luc. 8.2
lollows: f;,r,W/
0
0
0
0 0
0 0
BAi
=
Bs;
0
0 0 0
=
-f; 0 0 f;
b;
Tra~sformation of these derivatives to global coordinates is accomplished usin~ the 10verse of the Jacobian matrix, as follows:
0
0 0
a;
0
0 -b; -a;
a;
0 b;
0
0
0
0
(17)
'l'hcn [
: ::
U,z
:::
: ::]
J-1[:.·: :.·: :::]
=
V,z W,z
U,(
8
=I
i= I
V,(
BI
= BA + l
W,C
h h ( a10y1 -( a10x1 2 2 ( 2h b10y; -( 2h b10x1 f; Oy1 -J; ()xi
h (-BB' 2 I
(18)
h '2BB
(19)
und
a;W;
B = BA
b;W;
( I 11
+
E4uation (19) will be convenient when finding the stiffness matrix for element
l'BQ8.
0
Stresses corresponding to the strains in Eq. (15) are
i11 which
(20)
a;= Jf,f;., + lf2fi,11
b;
= lf1fi., + lf2J;, 11
(l•ll
The five types of nonzero strains to be considered for element PBQ8 Ex
U,x V,y
Ey
E
=
'Yyz
U,y + V,x V,z + W,y
'Yzx
W,x
'Yxy
=
Then the stress-strain matrix for an isotropic material becomes
11111
1JI
E= ( I ~l
(1
+
E
v)(l - 211)
+ U,z
lly inspection ~f the second version of this strain vector, we can assemble 11 11 /lh purl of matnx B from terms in Eq. (13) as 0
0
h (2a;
0
h - (-b · 2 I
0
0
h - (2a,
h (2b;
B,= b,
- J,
0
(/1
()
,r,
(i = 1, 2, ... , 8)
Sym.
JI
1 -
JI
I - 2v
(21)
0
0
0
0
0
I - 2v 2(1.2)
0
0
0
0
2
1 - 2v 2(1.2)
l'his matrix is similar to that in Eq. (3.2-10) , but the third row and column (1•orresponding to u, and Ez) are omitted . Also, the last two diagonal terms are 111 vided by the form factor 1. 2 to account for the fact that the transverse shearing Nlrcsses produce too little strain energy [4]. We may write the stiffness matrix for element PBQ8 as
K
= J' 1 J' 1
r
- J',J',
/',(11, 1 c''.11,,)
I
1
H 11:
HI.JI di; d11 d( 1
1•:(n,,
,~ nu)I Jld{dTJdt
(22)
3/8
l1lnl111
1111d
ShollN
Ch•P
In this expression the malrices B11 and n11 . array contains onJ t h urc hoth 01 .~lzc 5 X 24, but tht• lull Y erms t at are multiplied t rJ /2 1 . . through the thickness of the element gives )Y .,, , . ntegrat10n of Ilq (
!
2
+
K = f , f/2BJE B11
BJE Be)JJJ dg dTJ
which must be evaluated nume . II I . multiplied by h/2 from the thir~~;:~f n this process the factors~ and hi(<• II hand h 3 /12. Thus th fi t f IJ_J [see.Eq . (10)], producmg the lm·1o , e rs part o matnx K m Eq (23) · d shearing deformations, whereas the second .. 1~ ue t~ transvrr deformations. part is associated with Jlc1111,1I The consistent mass matrix for element PBQ8 becomes
M
111
11.~I
.1 hns 1111ll y 111 pllll'l' of 1, /2; and b, in Eq. (27) now has the 1111•11ning of forcc pcr unit 11n•11 . Note that this body force causes no equivalent 1111dnl moments. After finding the lime-varying nodal displacements in the vector q(t), we 1 1111 evaluate stresses at any point in each element, as follows: (29) (J(t) = EB q(t) 111 this cxprcssion
I •or best accuracy, these stresses should be calculated at the numerical integra111111 points (5].
1,1rnmple 8.1 1'111' the rectangular parent of element PBQ8 (after constraints are imposed), find the 111111;istent mass terms M11 and M22 , In the first case, Eq. (25) specializes to
=pf, f, f, fTfJJ I dg dTJ d( =Pf
1
f
1
f
Mll = pabh
r(
i(r + Ci r + cirs) I 11
fs
In this equation the matrices f and f are b th
. h11s only terms to be multiplied b B 'h/2 oI of s1~e 3 X 24, but the scrond thickness produces y · ntegratmg Eq. (24) ,through lh
_
1
2fif11
1
(a)
d{ dr,
M1 , = pabh 16
f1 f -I
i
(1 - t)2(1 - r,)2(-1 -
g-
r,)2 d{ dr,
(b)
- 1
1'111forming the integrations indicated in Eq, (b) results in M11 = ipabh
_ fl JI (
M - P _
LL /T
111hstituting the shape function/1 from Eqs. (7.5-19) into Eq. (a) yields
JI dg d,,, d(
11
379
1'11111111111l>YN/\1'111111 l'l11h1• 111 llm1cll11u
h2 fJfe) +6
JJ / dg dTJ
(c)
which is simply a fraction of the total mass. Similarly, in the second case we have
Again, the factors 2 and h2/6 are multi lied b / 2 part of matrix M consists of transl f p . Y,h from IJI. Hence, the hnl 1 rotational (or rotary) inertias. a wna mertias, and the second part givt1 Equivalent nodal loads due to bod fo using only matrix f as follow , y rces on element PBQ8 are calcu l11ll'd
M22
=
fl fl /T
pab h3 _ 12 1
_
d{ dr,
1
1 = -pabh 3
96
(d)
S.
A,
which has units of mass moment of inertia.
Pb(t)
=f =
1
f
2 f1
f l fJb(t) I JI dg dTJ d(
I
f
11,3 PROGRAM DYNAPB FOR PLATES IN BENDING
rJb(t)JJI dgd,,,
in which
.
b(t)
= {O, 0, b,}
( 171
und b: is force per unit volume in the z dir . , lhc factor h/2 from IJI and rewn·t E (e2c6t)1~n . Alternatively' we may cxll m I e q. m the form P&(t)
= {, {,
fJb(t)/JI
dgd,,,
( 'HI
Wc shall now describe a computer program named DYNAPB for the dynamic 111111lysis of plates in bending, which uses element PBQ8 from the preceding ~t'l'I ion. This program is constructed by modifying the part of Program D YNAPS p111t11ining to element Q8 (see Sec. 7 .6), because the geometric and displacement 11h11pe functions are the same for both. However, in program DYNAPB the 11111trices K, M , and p,lt) must he handled according to the expressions devel11pt·d in Sec. 8.2. Table 8. I shows pH' p11111ti1111 of sln1ctural data for Program DYNAPB. I '11111paring this d11t11 wllh 1l1111 l11 '1'11hll' I 'I, Wl' sec that the structural parameters
380
1
l'htht• n11el Hhnll• TAUi.iC H. 1 NII 111'1 111111 1111111 for Type of Data
1'1•0141·11111
Problem identification Structural parameters
Items on Datu Lines
Plate (a) (b) (c) a
bending data Nodal coordinates Element information• Nodal restraints
l 1
Descriptive title NN,NE , NRN,E,PR,RHO, H
NN
J, X(J), Y(J) I, IN(I, 1), IN(I, 2), .. . , IN(! , 8) J, NRL(3J-2), NRL(3J-l), NRL(3J)
NE NRN
!PS andd NfEN are omitted. In addition, there are three possible nodal restnunt mstea o two per node. · Dynamic load data for Program DYNAPB is given in Table 8 2 1-1, .• ?ave three possible initial displacements, initial velocities and ~o-datt ~ mstedad of/wo per node (as in Table 7.6). Line loads act in th~ z direction u~:1,•,: I an e ge o an element and ma~ vary_ ~uadratically. Therefore, we require lh1 t1 ;~de numbers and three for~e mtens1hes (per unit length) to describe them IN'1 g ..7. l !(b)J. Also, ~uadratically varying area loads (force per unit area · rl 111 z direction) need eight force intensities for a complete de~criptio~ N~ 1
TABLE 8.2 Dynamic Load Data for Program DYNAPB
Initial conditions (a) Condition parameters (b) Displacements (c) Velocities Applied actions (a) Load parameters (b) Nodal loads (c) Line loads (d) Area loads Ground accelerations (a) Acceleration parameter (b) Acceleration factor
_
381
fllgu rc 8 .4 shows half of a square, symmetric, simply supported plate that is divided into ,,lght PBQ8 elements. Also indicated in the figure is a moving load P that travels in the I' direction along the centerline, where nodes 5, 8, 13 , ... , 37 are located. We wish lo determine the translational responses of the plate at node 21 due to the load moving 111 constant velocity and constant acceleration [6]. This problem is analogous to Example 6.5 for a moving load on a simplyNllpported beam divided into four flexural elements. However, the displacement shape l11nctions for the PBQ8 plate element are quadratic instead of cubic. Therefore, the Junctions given by Eqs. (7.5-19) must be used in Eq. (a) of Example 6.5, and y,(t) 11Jplaces x,(t) in both Eqs. (a) and (b). Of course , we need to extend Program DYNAPB to handle one or more moving loads, as described previously for Program DYNACB. Physical parameters in this example are E = 69
106 kN/ m 2
X
V
a = 0.1 m
p = 2.62 Mg/ m 3
= 0.33
h = 0.025 m
P = 20 kN
lor which the material is aluminum and the units are SI. We ran this data on the extended j3
No. of Lines
Dynamic parameters
1110111 11111 DYNAl'II 1111 l'lnt n1 111 lln111ll1111
11.xumple 8.2
For sequence of node numbers, see Fig. 8.3.
Type of Data
11.:1
PIH, 7. I6(b) J. Finully, nolr th11l ll1l' llt'l'l' lrrution factor GAZ is for the z direction 1111ly, As in Program DYNAPS, we lake n = 2 each way to locate points for 1111111crical integration. With a plate-bending element, the computer evaluates the rl111c-varying stresses a:,, O'y , Txy , ry,, and T,x at each of the four integration points. lll•cause the displacement shape functions are quadratic, we found it necessary 111 use n = 3 each way for terms in the consistent-mass matrix of element PBQ8.
l>YNAPII
No. nl LillCN
1111,
t
Items on Data Lines
z
I • - j2
/
ISOLVE, NTS, DT, DAMPR j1
I
NNID NNIV
NNID, NNIV J, D0(3J-2), D0(3J-l), D0(3J) J, V0(3J-2), V0(3J-1) , V0(3J)
r------------7---v /
Sym .
1 NLN
NEL NEA
NLN,NEL,NEA J, AS(3J-2) , AS(3J- l) , AS(3J) J , K, L, BL!, BL2, BL3 I , BA! , BA2, ... , BAB
I, //
_L~ I/
8
p
/
y(t)--413
I
Typ. 21
j_
//
/
IGA GAZ
Forcing fun ction (a) Function parameter I NFO (b) Function ordinates NFO _ _ _ __ _ _ ___J__ _ __J__ K, T(K), FO(K)
-------=-
Typ.
11'111111·11 tl.4
Simply-supported plate with moving load .
382
Platos and Shells
Chap.
e
Sec. 8.4
Element for General Shells
version of Program DYNAPB , using Subprogram NORMOD to obtain responses. Coni puter plots of Di, at node 21 arc given in Fig. 8.5. For the case of constant veloci ty (VOP = 153.7 mis), the plot shows a maximum translation of 2.320 mm; and for constant acceleration (AOP = 59.05 x 10 3 m/s2 , with zero initial velocity), we have 11 maximum of 2.034 mm. Their ratios to the static deflection of 1.472 mm (due to the loud applied gradually at node 21) are 1.576 and l.382, respectively. As for the beam, lh1· values of VOP and AOP used in this example both give travel times equal to llw fundamental period of the plate, which is 5.206 ms.
383
1 0 0 0 1 0 0
1
0 0 0
1 0
0
0
0
0
0
Ca;=
h; 2
YOP CON Sr.
0 0 0
hi 2 0
0
1
0
0
0
0
0
1 0
h; 2
0
0
0
0
0
0 AOP CONST.
0
h; 2
2,00
E E
~
(1)
I .SO
When this 9 x 5 matrix is compared with Eq. (8.2-2), it is seen that columns I and 2 have been added. Similarly, the constraint matrix Chi for a midedge node of the rectangular parent becomes
-
( \J
w
Cl 0
z
.._ I . 00
1 0 0
<
0
-, Cl
0 o.so
0
Cb;= o.oo 3
4
T
5
Figure 8.5 Translational responses at center of plate.
8.4 ELEMENT FOR GENERAL SHELLS
In this section we specialize the isoparametric hexahedron H20 to becorn1: 11 curved quadrilateral element for the analysis of general shells. Development of lhc slw/1 element SHQ8 is similar to the technique used in obtaining c lcnw11l Pl3Q8 (sec Sec. 8.2) for plate bending. However, the constraint conditions 1111 111odilied because two additional translations, u1 and v1, occur at each node of 1111• Nlwll eh:menl. Thus, the constraint matrix C111 for a corner node of 1·h e rectn11g111i11 fllll'l'III clL·111cn1 Isc-e Fig . R. I (b) l has two more 1:olumns thun before, us f'olfowi,
h; 2
0 0
1
0
0
0
0
0
1 0
0
0
h; 2
0
h; 2 0 0
h;
(2)
2 0 0
which is a 6 x 5 array that can be compared with Eq. (8.2-3). With five displacements at each of eight nodes, element SHQ8 has (8)(5) = 40 nodal displacements. . As with the plate element PBQ8, the general shell el~ment SHQ8 will be formulated directly (2, 3]. Figure 8.6(a) shows the geometnc layout of element SIIQ8, in which the coordinates of any point are
h• [ l L !, IX;l Y1 + L fd-:j_ g
I I
%1
3
8
;] m3;
I I
n 3;
(3)
'l'ht· i11lll1pol11lio11 '1111Plt111111 / 1 11p111•111i11g In Eq . (3) nrc given by Eqs. (7.5-19).
3114
' 110,
11.4
3811
I ltt1111rnt lor lln11n111I hnll•
111111,, w1.; sec th ul thl· th1r~t1l'HN I, 11111y vmy quadratically over the element. In 1uldltion , th1.; terms /11, 111 11 , und 1111 11n• the direction cosines of a vector V3; that IN normal to the middle surl'uct.l und spuns the thickness h; of the shell at node I Figure 8.6(b) shows this vt:ctor, which is obtained as (4)
z',{;
/,W
,.,,,
\i' ,,/
)- y,v
y
, ./
2
t1oints j and k in the figure are at the surfaces of the shell. In a computer program 1•lther the coordinates of points j and k or the direction cosines for V3; must be Hivcn as data. Generic displacements at any point in the shell element are taken to be in the directions of global axes. Thus,
/~v, (a)
u,
u = {u, v, w}
C<,~
...
(5)
l )n the other hand , nodal displacements consist of these same translations (in Hlobal directions) as well as two small rotations a ; and /3; about two local
tnngential axes x' and y', as indicated in Fig. 8.6(a). Hence,
q;
z',~
= {u;,
v;,
W;,
a;, f3;}
= 1, 2, . .. , 8)
(i
(6)
[ v'
UJ = I8f; [U;J + I8f; 'ih fJ.; [a·] V
W
V;
t=l
I
W;
(7)
/3,
, =I
In this formula the symbol µ ; denotes the following matrix:
fJ.;
=
J
-/2;
Iii
-m2;
mli
-n2;
nli
(8)
[
( 'olumn 1 in this array contains negative values of the direction cosines of the Hccond tangential vector V 2;; and column 2 has the direction cosines for the first tangential vector Vii, as shown in Fig. 8.6(b). These vectors are orthogonal to the vector V3; and to each other, but the choice for the direction of one of them Is arbitrary. To settle the choice, we let
(9) Figure 8.6
(a) Element SHQ8; (b) nodal vectors.
'l'hcn (10)
Ill' V II is parallel lot.\ , 111 l(q . (I)), thl' lntlCI' is replaced bye,.] Figure 8 .6(b) shows locul generi c trunslnt11111H 11 1 111111 ,,, (i 11 llll' directions of V11 and V2 1) due to the 11od11I rotations /1 11111d ,~,. 11•M111·1 tl Vl' ly Tlwlr vulucs arc
386
Plates and Shells
=
v'
Sec. 8.4
1 0 f;
=
0
and so on.
0
1
0
0
0
1
/!!z 2 11
h;
h;
-(2 m2; ( m1; 2 h; -(2 n2; (2h; nli
f;
(i
= 1,
2, ... , 8)
(12)
The inverse of J becomes
11nd C,;
~ G! f f f}
~G
0 0 0 0 0 0
Then
J
-l2; 1, l h• -m2; mli -;j_J; -n2; nu
1
=
J* =
f y TJ.y
V,g
v,,, v.,
(13a)
8
=I i= l
fi.s
0
Ji.,,
0
0
0
0 0 0
fi.,
w., w,11
0
w.c
O
0
fi .,, 0
0 0 0 0 0 0
- C/;,gl2; - (f;,,,l2; -f;l2; -(f;,gm2; -(f;J,f1 m2·I
0 0
Ji., Ji.1) O O
f;
= fAi +
(fs;
(14)
O 0
U, y
f
= fA +
(fs
J= We find the derivatives in matrix 8
x., = I I I
x,s
1
~ I I
Y.c
I
( I<, l
Ji,,, 1,
... -4
I I
-f;m2; -Cfi.,n2; -(fi.,,n2;
(Ji.,n1; (fi. ,,n1;
-}in2;
Jinli
~e:f
:·' I
J* 0
...
ore.
W;
(18)
h; -a· 2 I ~{3;
0 J*
,'1
(19)
11
W ,(
U ,x
a;
0
U,y
b;
U,z
C;
V,x
0
0 0 0 a; 0 b; 0 c, 0 0 (11 () /11 () t'1
V,y
v .•
I
~'
V;
Multiplying the terms in this equation, we obtain
J from Eq. (3), as follows: 8 h, fi.,x, + I !i.d2'"
H
\,,,
z,,, z. 1
U;
(f;,,mli Cfi.11mu fim1;
W ,z
z,s]
Y.t x,,, Y.,,
[x,
. . .
( 15)
The last of these formulas will later be used to derive the consistent mass matrix und equivalent nodal loads for element SHQ8. The 3 x 3 Jacobian matrix required in this formulation is
Cfi.iu CJ;,,,l1; f;lu
Transformation of these derivatives to global coordinates requires that the inverse of the Jacobian mla:~x be aprp;~ed
und
(17)
t.y
[ f , 'T/,z (.,
We need certain derivatives of the generic displacements [see Eq. (7)] with respect to local coordinates. These derivatives are listed in a column vector of nine terms, as follows:
u., u,,, u.,
(13b)
t,x 'T/,x {x]
r
In order to isolate terms in submatrix f; multiplying (, we let
f,,
387
h• -(~a;
Contributions of these terms to the generic displacements at any point are given by the second summation in Eq. (7). As for the plate element, the displacement shape functions in Eq. (7) may be cast into the matrix form
h -(-;j_l2;
Element for General Shells
,,, I, ,, (. J I ,, 111
whldl
8
=I
l• l
0 0
W,1
()
IV,1,
()
II', ,
()
0
0
. -d;l2;
d;lu
-e;l2; e;l1; -g;l2; g;lli -d;m2; d;m1; - e;m2; e;m1; -g;m2; g;mt; - d1n21
d1n1;
('1 1121
<11n11 {<11111
R1 ll i1
U;
V; W;
a;
{3;
(20)
388
Plates and Shells
Sec. 8.4
(21)
Element for General Shells
As in a plate element, the normal stress a,, and the strain '=z' have been omitted. Then the stress-strain matrix E for an isotropic material becomes the same as that for element PBQ8 in Eq. (8.2-21). To relate local strains in the vector E' to global strains in the vector E , we can use the 6 x 6 strain transformation matrix T, in Eq. (7.2-28), as follows: E'
E
U,x
€y
V,y
€,
=
=
'Yxy
'Yzx
(28)
+ V,x V,z + W ,y W ,x + U,z
(22)
Noti?g the second ver.sion of this strain vector, we may construct the ith part ol matnx B from tenns m Eq. (20) as
B,=
a, 0 0 b, 0 C;
0 0 b, 0 0 c, a, 0 c, b, 0 a, (i
-d,121 -e1m 2;
d;lli e1mli -g,n21 g,nli -e,L2, - d,m2; e,lli + d,m1, - g1m2, - e1n21 g,m1; + e,nli -d,n2, - g,12; d,n1, + g,Li,
=
l, 2, .
=
BA;
Matrix B' will contain only five rows, due to the deletion of the third row of T,. Now we are ready to formulate the stiffness matrix for element SHQ8 using matrix B', as follows:
L, f f = f f f, =
I
I
I
(24)
Submatriccs B111 and Bn; are composed from Eqs. (21) and (23), but the actuul dc111ils are omitted. Altogether, we have (2~)
which will be useful when determining the stiffness matrix for the shell ch.lllll' III The following nonzero stresses in the directions of primed axes will ho l'ons idcrcd:
1111d
(29)
I
'8)
+ (Ba;
In these expressions the vector (J 1)norm. denotes the first row of the Jacobian matrix normalized to unit length, and so on. When calculating stresses in local directions, it is also useful to have
K (2))
.Similar to the plate element, we can isolate terms in submatrix B, that multiply ( lo find
B;
(7.2-27)
w,, U,y
'Yyz
= T.-E
However, the third row of matrix T, must be deleted, because E,· is not to be included in vector E'. For the purpose of evaluating T, at an integration point, we need the direction cosines for vectors Vi, V2 , and V:i at the point. This may be done with the following sequence of calculations:
For element SHQ8 we consider six types of nonzero strains, as follows: Ex
389
(B')TE B' df dY] d( (B~ + tBs?E(B~ + tBe)
1JI df dY] dt
I Iere the matrices B ~ and Be are both of size 5 x 40, but the latter array contains only terms that are to be multiplied by (. Integration of Eq. (30) through the rhickness* of the element leads to
K={,
f
[2(B~?E B~ + HBe?E BeJ 1
IJI df dY]
(31 )
The remaining integrals in Eq . (31) must be evaluated numerically, using two integration points in each of the gand YJ directions [5] . In this process the factors ) ond } are multiplied by h;/2 from the third row of IJI, and Be also contains rite same constant. Thus, we effectively obtain the factors h, and hl / 12 in the two purls of matrix K. The consistent mass matrix for element SHQ8 is
(32)
llw r01 1·csponding slruins nrc
lt1,•,
C30)
t\•,
Y,•,·, Y, , Yl','I
• 111-11111•ll l y 1111q1111ll11111l11111111h 1hr lllll ~111 "'· tr1111~ 11111111111,
J 1111111111111111, 111
111·11ln 1rd
JOO
I lt,11 a 1111d Sholl H
•1t11i. U.b
Rcc:ull lhut the matrices f11 and fJJ arc bolh of r,1111 I •10, but the scc:on<.l hus 11111 lcnns to be multiplied by (. Integrating Eq . ( ll) th rnugh the thickness y1t•ld
M
=pfIfI[2flf11 + HEruJ /JI dt d71
TABLE 8.3 Strncturul l)utu for Program DYNASH
= =
fff 1
2
I
ff 1
I
( I ~)
1
rib(t)
/JI dtd71
( II)
( I'\ )
As for the plate element, these body forces do not cause any equiyalent 111ul11I 111omcnts. After the time-varying nodal displacements in the vector q(t) have lwrn obtuincd, accompanying stresses in the element may be calculated for Im nl (primed) directions . That is, (t)
=E
General shell data (a) Nodal coordinates; h1 (b) Nonna] vectors (c) Element infonnation 3 (d) Nodal restraints
X(J), Y(J), Z(l) , H(J) V3X(1) , V3Y(J), V3Z(J) IN(I , I), IN(l, 2), ... , IN(J, 8) NRL(SJ-4) , NRL(5J-3), ... , NRL(5J)
T,B q(t)
= E B'q(t)
(
TABLE 8.4 Dynamic Load Data for Program DYNASH
,,,,
No. of Lines
Type of Data
1
Dynamic parameters
8.6 PROGRAM DVNASH FOR GENERAL SHELLS 111 Sec . 8. 4 we developed element SHQ8 for the dynamic analysis of ge1w111I shells. Now we present a computer program called DYNASH, which is b11N1 ii 011 lhu~ clement. The easiest way to compose this program is to modify sta111 IIH.·nts in Program DYNASO relating to element H20 (see Sec. 7.8) , from whh h ulcment SHQ8 is derived. But in Program DYNASH, we must use the forn111l11given in Sec. 8.4 to construct the matrices K , M , and Pb(t). Preparation of structural data for Program DYNASH appears in Tabfo H I C'ompuring this data with that in Table 7. 9, we observe thatNEN is omiUcd 1111111 till' list of structural parameters. However, the symbol H is added to the list f111 11 r use where the thickness is constant over the whole shell. We also sec th11t 11( 1t Is included with the nodal coordinates for 11 case where the thickness vuries ov, 1 llll l lt•111l nl. The lint' of normal vectors in tlw f11hll t'onf nins u node 1111111lwr .I 1111d 1
J, J, I, J,
NN NN NE NRN
1hree orthogonal components of the vector Vj at the node [see Eq. (8.4-4)]. In 11ddition, five possible types of nodal restraints are given in line (d). Table 8.4 lists the dynamic load data required for Program DYNASH, which is similar to that in Table 7 .10 for Program DYNASO. However, now we h11ve five possible initial displacements, initial velocities, and nodal loads inM lcad of three per node. As for element H20, we may have a quadratically vnrying line load on an edge of the shell element, with components in _the x, y, 11nd z directions. For this purpose, we need three node numbers and nme force lnlensities (per unit length). Also, the area loads require 24 force intensities (per
~uch stresses should be determined at the sampling points for numerical intc~111 11011.
1
Descriptive title NN,NE,NRN,E,PR, RHO,H
"For sequence of node numbers, see Fig. 8.6(a).
b(t) = {bx, by, b,}
0'
Items on Data Lines
I I
1iroblcm identification Structural parameters
rib (t) /J / dt d71 d(
In this expression, the load vector b(t) is assumed to contain components of 101, (per unit volume) that are uniform through the thickness of the shell. ThuN,
1
No. of Lines
T ype of Data
Because the factors 2 and j effectively become h; and hr /12, the first and sc1:111ul parts of matrix M contain translational and rotational inertias, respectively Equivalent nodal loads due to body forces on element SHQ8 may be fc,111111 using only matrix f 11 , as follows:
Pb(t)
391
1'1 011111111 DYNASH f<>t Clo111111tl llhollu
NNID NNIV
1, D0(51-4) , D0(5J-3) , .. . , D0(51) 1, V0(5J-4) , VO(SJ-3) , ... , V0(5J)
Applied actions (a) Load parameters (b) Nodal loads (c) Line loads (d) Area loads (e) Volume loads
I NLN NEL NEA NEV
NLN,NEL, NEA,NEV J, AS(SJ-4), AS(51-3), ... , AS(5J) J, K, L, BL1 , BL2, .. . , BL9 I , BAI , BA2, ... , BA24 I, BVI , BV2, BV3
1
I
1
Forcing function (u) Jiu netIon p111111111•t111 (b) F11m·tl1111 rn.11111111••
'=
!SOLVE, NTS, DT, DAMPR
Initial conditions (a) Condition parameters (b) Displacements (c) Velocities
Ground accelerations (a) Acceleration parameter (b) Acceleration factors
1
Items on Data Lines
I
Nl10 ~
NNID, NNIV
IGA GAX, GAY ,GAZ NFO K , T(K), FO(K)
fllutoe end Shell s
392
unit area) to describe quadratically var ying components in the x, y, and directions. For element SHQ8 we use n = 2 in the gand 7J directions [see Fig. 8. 6(111 1 to locate points for numerical integration. At each of these four points tho computer evaluates the time-varying stresses, a x', a y' , T.e y', Ty'z', and r ,'x' in lol'11I directions. As for element PBQ8, we needed to use n = 3 each way for tc:1111• in the consistent-mass matrix of element SHQ8 to retain sufficient accuracy .
where the material is reinforced concrete and the units are US . We used this data. in Program DYNASH with DAMPR = 0.05 and solution by ~ubprogram NUM~NT . Figure 8 .8(a) shows a computer plot of the ground acceleration, and the res_ultmg translntions in thex and z directions at node 5 are given in Fig. 8.8(b)_. The ~ax1mum values of these displacements are DJ1 = 0. 82 l l in. and DJ3 = 1.430 m; at tlm~ t . = 7_2 0 ms· Also shown in Fig. 8.8(c) are plots of the flexural stresses in the x and y directions (at ~
100
~
Example 8.3 A quarter of a doubly symmetric cylindrical roof shell appears in Fig. 8. 7. This port11111 of the shell is divided into four SHQ8 elements of constant thickness. Note that the shl•II is symmetric with respect to the x-z and y-z planes. Consequently, nodal restraints 11111•1 be used to prevent translations across those planes and rotations in the planes, 11• indicated at nodes 4 and 14. On the other hand, simple supports at the ends of the ~hl'II prevent translations in the x and z directions, as at node 20. A rigid-body grouml acceleration D8 J(t) occurs in the z direction, and we wish to find the response ol tht structure due to this influence. For this problem the physical parameters are E
393
Program DYNASH for General Shells
Sec. 8.5
= 3.6 x 103 k/in. 2
v
= 0.15
L = 100 in.
R = 3L
p
= 2.25 x
50
C
1000
0
1200
T
(.)
(ms)
< -50 0::
c..,
,!_,
-100
(a)
10- 1 k-s 2/ in. 4
=
h = 3 in.
'"·
OJ3
40° C
(D8 J )max = 115.9 in./s
2
:::; 1
.oo
If)
z
w
Cl 0
T
1
Sym .
/
1.5L~::.L/
s/ 14u 17/ t----=
6
Sym.
G)
/
I I
:/ 0 I
---71 // I
0
o.oo
,-
<
- 1 . 00
0:: I-
( b)
sx .
I
/I // I
IVp
z
20
r.k"
... ._./_ _s____ l ____,s_ _ _2_ 1, /
I
/
_:; 0. 60 -
0-40
w
Cl O. 20
0
z
0::
o.oo
<
I
I
Typ.
W -0.20
z
~ -0.40 (/)
I
',
I
',
I
'-J
/ t ', 1
.~
R'--('''' I ' I ~
~
o· .... (/)
0,60 -
(r l
1,·111111,1, H.H ( 'ylli111t l111I lll'll 111111 (11) p11111111l 11n rk1111h111, (h) dlNpl11n•1m•111s, (,•) tk,111111 ~, 111~1 ~
394 Plates and Shells
Chap. 8
the upper surface) for the integration point near node 1. Their minimum values are SX' ::. - 0.6350 ksi and SY' = -0.1456 ksi at time t = 720 ms. Of course, these stresses must be added to the membrane stresses occurring at the same time.
Sec. 8 _6
395
Element for Axisymmetric Shells z
8.6 ELEMENT FOR AXISVMMETRIC SHELLS 3
As shown by Ahmad et al. [7], it is possible to specialize a ring element with an isoparametric cross section to become an axisymmetric shell element by making one dimension small compared to the other. In this section we demon Strate the procedure by specializing element AXQ8 from Sec. 7.9 to form a shell clement called AXSH3. Figure 8.9(a) shows the axisymmetric solid element AXQ8, for which thl• cross section is an isoparametric quadrilateral with eight nodes. As the first step in the process, we make axes [ and 17 orthogonal and reduce the 1J dimension to the thickness h. Thus, we form the rectangular parent AXSR3 of elem,•111 AXSH3 (before constraints), as shown in Fig. 8.9(b). Next, we may introdut'l' constraints to refer the displacements at each group and pair of nodes to thosl' of u single node on the middle surface, as depicted in Fig. 8.9(c). The noclul displacements indicated at point i in that figure are '
(i
= 1,
i •
I
---+- U ;
2
~l~______i___e
-
-
r
(a)
v',
'Tl
2, 3)
(I) where a a small positive rotation about an axis normal to the t-11 plane. Figurl' 8. I0(a), (b), and (c) show relationships between nodal displacements at an end of clement AXSR3, the middle of AXSR3, and a node of element AXSll.1, respectively. The two types of constraints to be invoked are: 1 is
I. Nodes on the same normal to the middle surface have equal translations in the 1J direction. 2. Normals to the middle surface remain straight (but no longer normul) during deformation.
x', (c)
(b)
s
Figure 8.9 Specialization of axisymmetric solid: (a) e!ement AXQ8; ~b) rect~n gular parent AXSR3 of element AXSH3 before constramts; (c) constramed no a displacements.
1
With these criteria we can relate the six nodal translations in Fig. 8. IO(a) to 1hr lhrce nodal displacements in Fig. 8. lO(c) by the following 6 x 3 constraint
mutri x:
1 0
0
0 0
=
0
()
0
2 0
0
0
h1
0
C,,1
8 imilarly, the four nodal translations in Fig. 8: IO(b) a~e related to the three nodal displacements in Fig. 8. IO(c) by the constramt matnx
( ))
0
"'
()
)
()
which is ol si:11· •I
\, 11 wt· w1•11•
111
h; 2 0 h; 2
(3)
()
11pply Pq , ( >) 111 tht• 1•11d~
1111<1
I\(( , (1) tit
.100
l'h,11111 n11d Uhnlln • 1111,
4
i
1
3-r
i
- ,
------
I
,1~1
r,
2
397
---......
"",\
------
. 1 ..
I
• I
/ 3
h,
.,,~ '
l 1011111111 for Axl11y1tm1011 lo Shull•
1, V
h,
y'
11.e
')
2
_J_
k
t 6
z'
-3
k
5
k
t 4 (a)
x
(b)
(C)
Figure 8 .10 Nodal displacements: (a) end of AXSR3; (b) middle of AXSR3; (c) nocle of AXSH3. (a)
the middle, we could reduce the number of nodal displacements from 16 to 1J I lowever, we will take a more direct approach, which is similar to those in S1•1 • 8.2 and 8 .4 for plates and general shell elements. Figure 8.11 shows element AXSH3, for which the coordinates of any po1111 may be stated as
[r]
=
± [r;] + ±
J
f;
fiTJ~ [ l2; 2 m 2; In this equation the direction cosines for the normal vector z
zi
i= l
l2; =
cos 'Yi
y',ri,v'
t I)
i=I
m 2;
h; 2 ,
T/ -c, .
Vi are
h; .
2 sin Y;
= sin Yi
where y, is the angle between the r axis and the normal at node i. The geon 11111 h interpolation functions in Eq. (4) have the formulas
g
!, = --(] 2
t)
h
= I
-g2
u
= {u , v}
(b )
(h I
11ssu111ing that the loads arc axisymmetric. These di1,placements cnn he exp1·1·ss1•cl In tl•1111s of the nodal displaccmcnls 111, i,1, 1111<1 us fo llowi.;:
,r,,
x',1;,u'
z'
'l'hercf'orc, the th ickness h may vary quadratically in the g direction. Generic displacements at any point in the element are
11111111·•• R.11
(a) Element AXSH3; (b) nodal rotation.
1
308
Plates and Shells
[ U]
=
V
± i=I
j; [Ui]
±
+
fiYJ!!!. 2
i=l
V;
r-
Sec. 8.6
,Y;J
Element for Axisymmetric Shells
399
These derivatives are transformed to global coordinates by the operation
sin ai COS 'Yi
(7)
Geometric details justifying these expressions appear in Fig. 8.1 l(b) . As for lh<' ~cneral shell ~Jement, we arrange the displacement shape functions from Eq. (7) mlo the matnx format
fi
=
[1 0 0
1
h,
""i
- YJ sin 'Yi] hf; YJi cos 'Y;
= [oJ
01 00]/;
fsi
let us define
[00 00 -sin 'Yi]!!!./; cos 'Yi 2
=
-± r::
(H)
0 0
i=l
YJ,
To isolate terms in submatrix fi that multiply fAi
(i = 1, 2, 3)
0 0
-d,s' .ny,] - ei
ai bi
d; cos 'Yi ei cos ,'i
Sill
'Yi
n Vi ai
(15)
where (I})
(16)
Then ( I ti)
und
We consider four types of nonzero strains for element AXSH3. They are
u., v,, u r
€,
( 11)
€,
The _formula in_ Eq. (11) will be used later to determine the consistent nw11N u1ulrix and equivalent nodal loads for element AXSH3. The Jacobian matrix required for this element is
e=
= €9
u,, + v,,
'Yrz
J
= [;,·:
:::]
( I lj
(17)
lJsing the second version of this strain vector, we form the ith part of matrix B If'Om terms in Eq. (15) as
in which
ai 0 0 bi
and so on. Bi=
The inverse of J becomes
(, YJ.,]
[ g_,
YJ. ,
We w_ill need derivatives of the generic displacements with respecl to 1h ut
r
"·'
11, ,,
3
L I I
0
- drf; T/hi. sin y;
b;
ai
d; cos 'Yi - ei sin 'Yi
( I I)
lorn! coordmates, as follows:
u,,,
I! r
1- 1 = J* =
!,,, 0 [ 0 ()
0
-Ji.1;'YJ sin
0
-Ji sin y1 Ji.,'Y] cos y, .fi
CON ,'1
(18)
(i = 1, 2, 3)
Sl111ilar to the general shell element, we isolate terms in submatrix Bi multiplying 1/, which gives
B;
y,
( I I)
-di sin 'Yi ei cos 'Yi
= BAI +
'l7B0;
(19)
r,n,,
(20)
llor the whole matrix, we have
n nA 1lw 111st
10111111111
I
wlll lu•lp IIN to d1•11v1· tht· htlll11nHM 11111td x
101 1•1l 111t•11t
AXSI I I,
400
1•1111111, " '"'
m1011,.
,11 11.0
I 101110111 lur A>
401
Stresses in till· d itl'l'flo11 111 ,,I p1 l1111•d 11x1 i, (rnused by ux isy111111etrlc loud'9 1
Iti I.
er' and the corresponding strains
r,
I
11
I' T, ')•'}
fll'l'
e' = {e.,., e,, , 'Y.,•y,}
II
0
II
(1
+
v)(l - 2v)
I - 2v 2(1 .2)
0
e
0
0
0 I
21, 12
2m1m2
0
(7. 2- J I l
e2 = {l2, m2} = {- m1, / } 1
( ' ~)
As for the general shell element, it is also convenient to have
= T.B
mar, I, n
Bn arc both of size 3
=
21rp
x
+
T]fa) /J /rd( d77
(28)
fyn
fA
+
HHa)/ J / rd(
(29)
We find equivalent nodal loads caused by body forces using only matrix 11 Thus,
Pb(t)
=
f
{,
frr fJb(t) / J / r dO d{ d71
f
fJb(t)/ J /rd{
1
= 41r
I
b (t)
(30)
9, buf flw 1 11
, 11 1 1
,,,,
fin ~
= {b,,
b,}
(31)
I hcsc body forces do not produce any equivalent nodal moments. After solving for time-varying nodal displacements in the vector q(t} , we 1 101 determine stresses for local directions in the element using
u' (t)
r dO dg d71
(
111 this equation the matrices B,; a nd
T]fa?( fA
Nlllts in
(8.4 2HI
J he clement stiffness matrix may now be formulated using matrix D 1 • 'f'huI
+
1111d the components of force per unit volume (constant through the thickness) are
fo_r fhe purpose of calculating stresses in local directions . In this case, wi ll h;~ve only three rows, due to the form of T. in Eq . (24).
I
fTf/J/rdOd{d77
0
.\N before, the first and second parts of M consist of translational and rotational l1111rtias .
'fo cv~luate matrix T. at an integration point, we must find the direction coNI II fo r unit vectors e, and e2 in the directions of x' and y' at the point. This yI 11111 done as follows:
f f frr(B '?EB'/ J /
1,
M
E' = T€e
B'
(27)
111 thi s case the matrices fA and f 8 are both of size 2 x 9 , but the second has only h I ms that are to be multiplied by 71. Integrating Eq. (28) through the thickness
(4
which is of size 3 X 4 . Then the strain relationship becomes
e, = (J,)norm = {l1, m,}
_,
1
= 21rp {, { / fA
11
IT
pcB,:.?E B,:. + i(Bli?E Be] /J /r dg
r r frr
0
V
In or~e~ to relate local strains in the vector e' to global strains in the v1·, 1t we specialize the 6 x 6 strain transformation matrix T. in Eq. (7.2- 28) 1 t
[
f
M = p _ II
0
T =
= 21r
1lw remaining integral must be evaluated numerically, using two integration p11l11ts in the gdirection [5]. The consiste nt mass matrix for element AXSH3 has the form
E
E=
K=
K
r
The stress
E,
1111 ly te rms lo be mullipli(.)d by TJ. Integration of Eq. (26) through the thickness* 111 1hr clement g ives
=
E T. B q (t) = E B 'q (t)
(8.4-36)
l'h1•sc stresses sho uld be calculated at the numerical integration points. In Sec. 7. 9 we expressed nonaxisymmetric loads that are symmetric with 11 ~PL'Cl to the r-z pla ne as the Fourier componen ts: ~To Ni mpllfy lntl~f.1111lio11 thro1111h thl' lhil•k,wss, 11,1111s In r 1111(1 rn11trl x .I contuining 1/ urc 111•11h·l'l1·d ,
402
Plates and Shells m
L
br =
brj
COS
Chap, I
m
j()
I
hz =
j=O
L
bo =
w
=
t
[cos J() O 1 0 0 3
X
{
~.,t;
0 COS
o o o (7.9-12)
b 8j sin J()
0
[u;] V;
W;
1- l
·J
0 0 sin }8
j()
0
u
=
L cl qj j=O
=
(11,
L cj L f;Qu j=O i= I
=
f, = 0 0
0 0
0
-J.,t;sj
0
0
f;,sCj 0
v ,s
3
=I i= l
0
0
w,s
0
w,11
0 0
W,e
j
-f;,s17S;Cj -.,t;s,.cj
0 0 0
jf;77s;Sj
0
-J.,t;sj 0 0 f; ' sS·J 0 0 0 jf;cj
s,.
= sin
y,.
sj Cj
J=
0
772
0 0 0
f;, s17C;Cj j;C;Cj -jj;77C;Sj
0
0
f;, g17Sj
0 0
f;sj jf;77cj
0
U; V;
W;
(36)
h;
2a,.
%/3;
= sin JO = COS j()
(37)
r ,1)
(38)
In this case the inverse of J is seen to be
.,t;
(i = 1, 2, 3)
( II)
h,
r' = J* = [ to,z
OJ
17,·r 77,z 0 0 1
(39)
l )Hing this inverse matrix , we can transform the derivatives in Eq. (36) to global ,•oordinates, as follows:
f/) ~ ; ~ ;:1./i ()
0
[ 0
fr
0
()
U,e
0
0
772 cos y,
()
0 0
rg
ANlwf'orc , lcl us is<>lale lcrms in submalrix f1 that arc multiplied by
1~11
f;,{Cj 0
(7.9- 31)
Also, we put the displacement shape functions into the matrix form 0
(35b)
'l'he required Jacobian matrix is
cos J() 0 0 cos J() [ 0 0
h; -77 sin y,. 2 h;
u,s u,11
C; =COSY;
in which
Cj
0 0
In the coefficient matrix of this expression, the following abbreviations are used
3
m
m
[0
Thus, Eqs. (10) and (11) apply equally well to any case of nonaxisymmetric loads, but matrix f is now of size 3 x 15. For this analysis we must determine the derivatives of the generic displacements in Eq . (32) with respect to local coordinates. They are
V,8
wh.crc t~e angle /3,. is a small rotation about the x' axis [see Fig. 8. ll(a) 1. Abcforc'. 1f_the loads were antisymmetric with respect to the r-z plane, the furn lions sm 18 and cos JO would be interchanged. Equation (32) may be stated more efficiently as
-sin y,. cosy,. 0
= 0 0 0
fBi
v,11
Li77j 1= 1
j
(.l'I
h[-sin Y; cosy,. -
3
+
403
and
b,j cos J()
j=O If the loa_ds were anti.symmetric with respect to the r-z plane, the functions sin jtl un~ cos 18 woul~ be mterchanged. The response of an axisymmetric shell to lhi~ s~nes of hannomc loads consists of a series of hannonic generic displacemenlN l•or element AXSH3 these displacements are expressed as follows:
[u]
Element for Axisymmetric Shells
j=O
m
V
Sec. 8.6
r,. I k 1111
(
U ,r
u,, II, 11
l''\11)
()
11', II /
0
fJ ' ()
.I "'
0
0
~1
., ...
u, , u,'1 II ' II
11 ',11
(40)
404 l 1l11t1111 0 11d Shollo Hue. 8.6
Element for A>
405
Multiplying the terms in this equation produces
u,,
where i Into
a;Cj b;Cj
0 0 - d;S; Cj 0 U, z 0 0 -e;S;Cj 0 U, IJ -JJ;sj 0 0 ! (Jf;'r]s;sA) 0 V,r 0 a;Cj 3 0 d;C;Cj 0 V,z 0 =2 b;Cj 0 e;C;Cj 0 i= I V,o 0 -JJ;si 0 -! (jJ;71c;sA) 0 W ,r 0 0 a;Sj 0 d;si W ,z 0 0 b;Sj 0 e;s1 w , 11 i 0 0 jJ;ci 0 ! }firJcih; where the constants a;, b;, d;, and e; are given by Eqs. (16).
=
Ej
=
'Yrz
U ,z W
'YzlJ
'YrlJ
-U,r IJ
j
,z
+
(H,)1
a:
w 11 ) .
r w
0
I . ; rfic1
I - 2rfi'rJS;cA
b,c1
I . 2r }/, 'f'JC/11
a1c1
0
(d;c1 - e1s1)c1
0
1
b,s1 ("
1
'Yy' z',
(47)
'Yz' .,,.}
which is of size 5 x 15. We write the stiffness matrix for each harmonic loading as
f f, f" = f
(BJ)TEBJ JJlrdOdgd'rJ
1
br
1
+ HB~JTE B~J JJ Jr df
[2(B,11)TE B,;i
where k = 2 for j = 0 and k muss matrix becomes
=
I for j
=
(49)
I, 2, ... , m. Next, the consistent
0 0
I
0
)
Ki=
r
- J,c;
I
Ez• , 'Yx' y' ,
(48)
(I
Q
- r J!,.11
= {Ex•,
1
b; CJ
- ; jfis;
E
llonally, the 6 x 6 strain transformation matrix T. serves to relate E to E , as in l'.q. (7.2-27), except that the second row is deleted. Another useful matrix is
O
0
1
uf
In this case the normal stress
0
r
(45)
lo which the following strains correspond:
11
I
+ 'r]Ba1
(46)
The strain-clisplac~ment relationships shown in the second form of this vccto 1 111 the same as those ~n Eqs. (7 .9-15). Using these relationships and Eq. (4 J ) , w 11111y construct the zth part of matrix B for the }th harmonic response, as follow
a19 0
BAJ
lll'C
+ V,r + V,11 w ,r
(44)
which will be used to find stiffnesses. For the present analysis, nonzero stresses in the directions of primed axes
V,z
-r (u +
= (BA;)i + TJ(Bs;t
B1 =
1
E9
, m. As before, we can decompose (B; )j
11ncl state the total matrix in the form
U,r
E,
= 0, 1, 2, ...
1, 2, 3, and}
(B;)i
For no?axisymmetric loads on element AXSH3, we consider six typt ~ nonzero strains . Thus, Er
=
- 2,)fi'rJC1s/11
-1).
1
1
I .
2r )/,rJ.1·,.11h1
e1.11
(r1, Ji·q!,') 2,. .,,
(50)
which is the same for j = 1, 2, . .. , m (see Sec. 7.9). Then equivalent nodal londs for each harmonic set of body forces are
p,,(t)1 =
J', J', f" f'XcJc1b(t)1 IJlrdOdgd 'rJ k TT
fi
2f'.\'l,(t)1 I .J 1, d £
(/
0 , 1,2, , , , , 11/)
(5 1)
408
Plates and Shells
Sec. 8.7
Program DYAXSH for Axisymmetric Shells
in which only matrix fA is used. Components of force per unit volume (constant lhrough the thickness) consist of b(t)j
= {b,j , b,j, b0J
(52)
TABLE 8.6 Dynamic Load Data for Program DYAXSH Type of Data
Finally, the stresses for each harmonic response may be written as
u'(t)j = E BJ q(t)j
No. of Lines
Items on Data Lines
1
!SOLVE, NTS, DT, DAMPR
1 NNID NNIV
NNID, NNIV J, D0(3J-2), D0(3J-l), D0(3J) J, V0(3J-2), V0(3J-l), VO(JJ)
1
NLN,NEA,NEV J, AS(3J-2), AS(3J- l), AS(3J) I, BAI, BA2, . . . , BA6 I, BVl, BV2
Dynamic parameters
(j = 0, 1, 2, . . . , m)
(5.l)
which are in the directions of local axes.
8. 7 PROGRAM DYAXSH FOR AXISYMMETRIC SHELLS
Let us now consider a computer program named DYAXSH for the dynanik analysis of axisymmetric shells with axisymmetric loads. This program USl'~ clement AXSH3 from the preceding section. It is generated by modifying thl• part of Program DYAXSO dealing with element AXQ8 (see Sec. 7 .10), 011 which element AXSH3 is based. However, in Program DYAXSH we mw,t formulate K, M, and Ph (t) in accordance with the equations developed 111 SL'C. 8.6. In Table 8.5 we see how structural data is to be prepared for Progru111 l>Y I\ XSH. As for general shell data, the symbol H appears in the list ol NII uctural parameters for a case where the thickness is constant over the whok• tihcll. In addition, H(J) is included with the nodal coordinates for a case whl'll' thickness varies over an element. That line of data also contains the compone11t11 Y2R(J) and V2Z(J) of the normal vector Vi for node J [see Eq. (8.6-4)]. Finally, three types of nodal restraints are indicated in line (c). Dynamic load data for Program DYAXSH is displayed in Table 8.6. I ll'll' we have three possible initial displacements, initial velocities, and nodal !0111111 instead of two per node. Area loads for element AXSH3 are similar to those Jor AXQ8, but only the element number I (not the node numbers J, K, and L) lll'l'd be given. Also, the acceleration factors GAX and GAY are replaced by GAZ I111 the uxisymmetric shell.
Initial conditions (a) Condition parameters (b) Displacements (c) Velocities Applied actions (a) Load parameters (b) Nodal loads (c) Area loads (d) Volume loads
NLN NEA NEV
Ground accelerations (a) Acceleration parameter (b) Acceleration factor Forcing function (a) Function parameter (b) Function ordinates
1 1 I NFO
No. of Lines
Prnhlcm identification SI I ucturul parameters i\xlsy111mctric shell data (u) Noclnl coordinates, etc . (b) Element informatio n° (cl Nodnl rcslrnints •• Fo1 Nl'(Jlltl ll l't'
ul
lllltlt• 1111111htll N, Nl'l'
Fl11 t! I 1(11) ,
NFO K, T(K), FO(K)
Example 8.4 The vaporous gas storage tank in Fig. 8.12 has a hemispherical top and a circular z I
Items on Data Lines Descriptive title NN, NE, NRN, E, PR, RHO, II
NN NE NRN
IGA GAZ
With element AXSH3 we use n = 2 in the f direction [see Fig. 8.1 l(a)] to locate points for numerical integration. At each of the~e two p.oint~ the computer evaluates the time-varying stresses
TABLE 8.5 Structural Data for Program DYAXSH Type of Data
407
J, R(J) , Z(J), V2R(J), V2Z(J), ll(IJ I, IN(T, I), IN(l ,2), IN(l ,3) J, NRL(JJ-2), NRL(3J- I), NIU ,( \f J 1111
408
1'1111011 ontl Sholls Sec. 8.7
Prog ram DYAXSH for Axisymmetric Shells
409
z I
We ran the foregoing data with Program DYAXSH using DAMPR = 0.?2 a.nd Subprogram NUMINT for calculating d7n~mic responses. The computer plot m ~1g. 8. 13(a) gives the variation of pressure ms1de the t~~' an.d Ftg. 8.~3(b) s~ows ti.me histories of membrane and flexural stresses in the x direction at the mtegrat10~ fOmt near node 37. The maximum value of the former stress is 205.2 MPa, and the mm1mum value of the latter stress is -308.6 MPa (at the outer surface of the wall). Fu.rther refinement of the finite-element layout near the base would produce a somewhat higher llcxural stress at that location.
j2
t
j3
/
J•-j1
16
®
I
.. a.
3 . 00
I:
0.4H
-+
~ 2.00
:::, (/)
(/)
w
0::::
0.4H
a.
_l_
_J
I. 0 0
< z 0:::: w 1z
- o. oo
2
(b)
4
(a)
cylindrical wall that is fixed at its base. This axisymmetric shell is divided into I H AXSI 13 clements that become progressively shorter toward the base, where high be111ll1111 stresses arc expected. Surface pressure p, from an interior explosion causes dy1111111h response of the shell , and we wish to examine stresses in the vicinity of node 37 d111• 111 this inllucncc. Physical parameters in this example are E = 207 X 106 k:N/m 2
v = 0.30
h = 0.025 m
1
,.:; rd(
a."' 200
"-
(p.,.)mnx = 3 MPa
f r-z J fr•
l
MEMBRANE I:
p = 7.85 Mg/m 3
I")
when: the material is steel and the units are SI. As in Example 7 .6 , we need to find equivalent nodal loads due to press111·1• I' which we assume acts in the direction normal to the middle surface of a shell ek•11w11t (lhl· t / direction). These equivalent loads arc
p1, = 27Tp,,.
8
T Cms l
Figure 8.12 (cont.)
R = H = 2m
6
w O 0
0
8
z
T Cm s l
0::::
< w z
-200 X
1111
(/)
'l'h1s 1'01111ult1 lcw 1>,, is the same as that in Eq . (7. I 0-h) , but in this case there is 110 111'1 d 1111 the pii111l' on f As before, signs for the equiva lent nodul londs urn 11uto11111t11·11 lh d11t1•111il11l·d hy liq. (n) l'or pressure t11ken to be positive in the positive sonsl• of1h1• 1111111111l tliH'l'Iio11 r/.
( b) lt'l1111111 II
II
VIIJ11UUII N )lllN Nllli ll)ll'
11111k:
(11)
Jond ; (h) Nll\'NNCS,
lnt11• nrut Hhnll1
REF
R NC
Ch
9
8
I. M1ndl11.1, R. 1)., "l111lucncc of Rotutwy Mo11111111111I Shl'llt on Flcxurul Mot11u11 Isotropic, Elastic Plates," J . Appl. Mt•ch ., Vol. 7 J, 195 1, pp. 3 1 38. 2. Ahmad, S. , lrons, B. M., and Z icnk.icwicz, 0. C., "Analysis of Thick 1111d lhl Shell Structures by Curved Finite Elements," Int. J. Numer. Methods Elli( ,, Vul No. 4, 1971 , pp. 575- 586. 3. Wea~er, W., Jr. , and Johnston, P R., Finite Elements for Structural A 11 ,,/1 ,/ Prent1ce-HaJ1, Englewood Cliffs, N. J., 1984.
Rigid Bodies Within Flexible Structures
4. Timoshenko, S. _P., and Woinowsky-Krieger, S., Theory of Plates and Shl'!/1 , , 11 ed., McGraw-H11l, New York, 1959.
5. Cook, R. D. , Concepts and Applications ofFinite Element Analysis, 2nd ed. WII, v New York, 1981. ' 6. Yoshida,. D. M.'. and Weaver, W. , Jr. , "Finite-Element Analysis of BeaniH und Plates with Movmg Loads," Int. Assoc. Bridge Struct. Eng. Vol 31 1971 l'I 179-195. ' . ' ' 7. Ahmad, S., Irons, B ..M., a~d Zienk.iewiez, 0. C., "Curved Thick Shell nnd Membrane Elements, with Particular Reference to Axisymmetric Problems," ,.,,,, 2nd Conj. Mat. Methods Struct. Mech., Wright-Patterson Air Force Base Ohio 1968, pp. 539-572. '
9 .1 INTRODUCTION
Occasionally, the analyst encounters a structure containing one or more parts (or bodies) that are very rigid in comparison to the other parts. Such bodies are usually taken to be infinitely rigid [1], and the nodes connecting them to the rest of the structure are constrained to displace in a pattern corresponding to the rigid-body motions. These restrictions on nodal displacements serve to reduce the number of degrees of freedom in a given problem. For example, the joints labeled A, B , and C in the counterweighted plane truss in Fig. 9. l(a) would ordinarily have a total of six degrees of freedom among them. However, the rigid counterweight (shown hatched in the figure) , on which they are located, has only three degrees of freedom. These three displacements are translations in the x and y directions and a rotation in the z sense, as indicated by the arrows labeled 1, 2, and 3 adjacent to the rigid body. Thus, the number of degrees of freedom in the problem is reduced by three due to the presence of the counterweight. Figure 9 .1 (b) shows a second example of a building frame, for which the analytical model is taken to be a rectangular space frame containing floor and roof laminae . Each lamina is assumed to have infinite rigidity in its own plane, but zero rigidity normal to the plane . The x and y translations and the z rotation for any joint on a particular body are dictated by the corresponding rigid-body motions of that lamina. These motions are indicated by three numbered arrows at each framing level. Therefore, if ni is the number of joints at a given level, the number of independent displacements at that level is 3111 I 3 instL·ud of 6111•
411
412
lll11lcl tloclh11 within Flexible Structures
Chip
2
3
413
Rigid Bodies in Framed Structures
In this chapter we examine the effects of including rigid bodies within framed structures and discretized continua. It is assumed that the bodies are connected and supported by linearly elastic, flexible materials and that their displacements are small. We also assume that the bodies are not rigidly connected to each other or to the supports, which would require additional constraints .
t_,
/
Sec. 9.2
A
y
)-,
9.2 RIGID BODIES IN FRAMED STRUCTURES (a)
z 3
1
/
t_2 6
Figure 9.2 shows x, y, and z axes and six indexes p I, p2, ... , p6 for components of actions or displacements at a reference point p on a threedimensional rigid body. The figure also depicts a typical joint (or node) j of a structure connected to the body. If the structure is a space frame, there will be y
f
z
Rig id body
/-5
~y X
(b)
z
)-,
/
X
/ / /Zp, (c)
Figure 9.1 Rigid bodies within flexible continua: (a) counterweighlcd 11 uN•, (b) building with laminae; (c) shell with hub .
/
__________ ___ _y / xP,
Appearing in Fig. 9. I (c) is a third example of a rigid hub at llll' l'l'lllrl 1 shell structure. If the geometry of the shell is general, displacements of no,l 111tn~·lwd lo the hub urc construined lo displace according 10 the iil{11I t,1111 111011011s l11l>l·lcd I through 6 in the n~ure. Othcrwi1-K·, ii' the shell Wl'11.• uxl11v111 111l't11c. lhl· 1111111hl·r of iigid body motions would lw ll'Wl'l • 11
1>'111111·1• 1>.2
Rigid hody consti uint~ .
Sec. 9.2
Rigid Bodies within Flexible Structures
414
~ix indexes j I, j2, ... , j6 for actions or displacements at point}, as indkul 111 thc figure. An offset vector rpj is directed from pointp to point} and has Sllll l'0111~onents XpJ, YpJ, and zpJ· We may calculate the statically equivalent al'flmt ~ ,~ point p due to actions at point j using the concept of translation of axes I >, I I I hus,
II In Ihis equation the symbol Aj denotes a column vector of six action comprn11 nl 111 point}, as follows:
A1 = {A11,
A12, · · · ,
A16}
wh~rc the first three are forces, and the last three are moments. Also, the w11 (1) contains six statically equivalent action components at poi111 I'
A,, 111 Eq.
Aµ = {Aµ1 , Aµ2 , ... , Ap6} l<'inully, the transformation matrix
TpJ
in Eq. (1) has the form
where
Zpj -ypj] 0
Thus, transposition of CpJ merely changes its sign, which is an inherent property of a skew-symmetric matrix. Note that the transformation in Eq. (6) is analogous to that in Eq. (4.5-1) for small rigid-body ground displacements. A similar transformation matrix TpJ can be derived for each of the other five types of framed structures. Table 9.1 summarizes these matrices for beams, plane trusses, plane frames, grids, space trusses, and space frames. Each of the first five matrices in the table can be found by deleting appropriate rows and columns from the sixth. For example, point p for a rigid lamina in a plane truss would have three displacements, which are translations in the x and y directions
1. Beams (x-y plane of bending)
=[
~
Tpj
-zpJ
Pl
Thi s skew-symmetric submatrix contains positive and negative values ot th n>mponcnts of the offset vector rµJ· These components are arranged in a 111111111 lhul pr,c~d~ccs the cros.s product of rpj and the force vector F1 at point}. S11nilurly, the kmematically equivalent displacements at point j 11111y I rnkulntcd from those at point p with the relationship
= T;o,,
1ft
111•11· th~ ~ymbol DP represents a vector of six small displacement co1111>11t1l'III ol lhl' 11gtd body at the reference point p. That is,
'1'111• 11p1•111l111 '1' 1'.1 i11 nq (h) is
Xpj
3. Plane frames (structure in x-y plane)
Tp;=[
b -ypj
~
~] I
Xpj
4. Grids (structure in x-y plane) T p; =
I O [~
b -t
YPi]
5. Space trusses (transposed to save space)
T;;
whl'H' thl.' first lhrcc arc translations , and the last three arc rotations. In 11dd111111 tlt1· Vl'l'tor 1>1 rnnsists of six kincmatically equivalent displacements at point
{D11 , l>p, . .. , D,c,)
b ~]
- ypj
[b ~ ~ -~; t -~] 2
D,, = {D,,1, D,,2, . .. , D,,6}
D,
[L ~]
T·=[
-y~
D1
=
2. Plane trusses (structure in x-y plane)
0
(10)
0
-Xpj
i11 which l, is an identity matrix of order 3 and
CpJ
Xpj
TABLE 9.1 Transformation Matrices Tp; for Framed Structures
:J
Tpj = [:;j
415
Rigid Bodies in Framed Structures
=
0 0
1
YPi
0
-Xpj
6. Space frames
(N T,,,
I
0
0 0
I
()
:,,
0
z,,, ()
1,,1
0 0 0 0 0 0 Y,,1 I 0 0 0
,,,,
()
I
()
()
()
l
416
Rigid Bodies within Flexible Structures
Chap
I
Sec. 9.2
417
Rigid Bodies in Framed Structures
and a rotation in the z sense. But a joint} in the truss has only two displacemcnlN which are translations in the x and y directions. Therefore, we keep only the flr "I second, and sixth rows and the first and second columns of the 6 x 6 matrix ·1;,1 Deletion of the other rows and columns from the sixth matrix results in the 3 x 1 array for plane trusses shown in the table. Note that the 3 x 3 transformal i1111 uiatrix pertaining to grids requires not only deletion of rows and columns , hul also rearrangement as well. The reason for this is that the x- and y-rotation1, nl Ille joint of a grid are taken before the z-translation [see Fig. 6. l(d)]. Now let us consider the task of incorporating rigid bodies into our ann lyt ical models for framed structures. For this purpose, we use a member-orie111, ,I <1pproach to transform actions, stiffnesses, and consistent masses at the end1, 11r 111cmbers to reference points (or working points) on the rigid bodies. Also, ii I usually necessary to transform mass and mass-moment-of-inertia matrices l11r lhe rigid bodies from their mass centers to their working points. Figure 9.3 shows a space frame member i connected to three-dimensionnl ri gid bodies at both ends. Points p and q on the bodies are taken as worhllljl poinls to which information about joints j and k will be referred. Firsl, 1111v lll'lions in vectors Aj and Ak at the ends of the member may be transformed 11110 sluti cally equivalent actions AP and Aq at the working points by the follow11111 /'l'llcrulizcd form of Eq. (1):
AP] [ Aq
[Tp1 0 ] [ A1 ] O Tqk Ak
(I I)
'l'hl' action vectors Ak and Aq are similar to Aj and AP in Eqs. (2) and (3). /\11111 lhl' transformation matrix Tqk is the same type as Tpj given in Eq. (4), b111 1111 points q and k. Equation (11) may be expressed more concisely as
(I ) in which (II) lllld
T; =
[;j
l
~J
( 11)
'l'ht• trunsf'onnation matrix T1 is a combined operator that converts the m:ti1111" Ill AM, (al lhc ends of the member) to the statically equivalent actions in A111 (111 lh working points of the rigid bodies). These actions are all in the dircclirn, ~ of Nl111cl11nd uxcs . Bolh actual and equivalent nodal loads can be trcall'd 111 lhl I lllllllll'I'.
DiHplHl'Cllll'nls at joints j and k wi ll also be expressed in terms ol' lh11111 ti I' 1111d ,1 hy 1111 t·xtcndcd fmm of Eq . (6), as follows:
I I T,~, I u,
l)A
O
o ,, n,. , 'l',/1
I
l>,1
I
I
---
..__
--- ..__
I I
--- --- -- -- --
I I --- --- ..J
Figure 9.3 Space frame member with rigid bodies at both ends.
The displacement vectors in this eq~ation corres~ond to the action vectors in Eqs. ( 13). Equation (15) may be wntten more bnefly as (16) Whl'll'
( I 'I)
418
Rigid Bodies within Flexible Structures
Sec. 9.2
and
T;
= [;
point c, as indicated in Fig. 9.2. If the body is thr~e-di~ensio~al_ and has six degrees of freedom, its mass and mass-moment-of-inertia matnx 1s
;JJ
. In addition, we can transform the member stiffness matrix Ki from joinl• J and k to the reference points p and q. For this purpose, consider lh action-displacement relationships
AM;= K;DM; Substitution of Eq. (16) for DM; gives
AM;= K;T;D8 ;
(20)
Use of this expression in Eq. (12) produces
Asi
= T;K;T;D8 ; = T;K;T;
(" )
The symbol_ Ka! denotes the member stiffness matrix for actions at points p urul f/ due lo unit displacements at those points. Similarly, the consistent-mass matrix M; for the member may be trn1111 lonncd from joints j and k to points p and q. Reasoning as above, but w11h uccclerations instead of displacements, we can derive the fonnula
Mc=
0 0 m f;u 0 0 0 -fyx 0 0 0 0 0 0 -Jzx
MaaDa
+ SaaDa
=
A 8 (t)
( '4)
To form the matrices in this equation, we assemble contributions from individ1111I lllcmbers by the direct stiffness method, as follows:
,,,
S,111
= L KB/
Mao=
L"• Mo;
Ao (t)
"e
= L As (t);
( 1)
This assembly process is similar to that described previously in Sec. 3.5, hut 1h 111otions arc at working points instead of joints. If a rigid body docs nol l'Xllll 111 11 particular joint, all of the offset vectors for members framing into that joinl 111 t11kl·1t to hove zero lengths . Also, the rotational displacements arc 0111illt·d lor 11011t·xislt•111 rigid bodies i11 plane and space trusses. As yet, lht• muss 111111,ix M1111 in Eqs. (24) and (25) is tkvoid ol 11111 l11h11tio1'.s 110111 lht· rigid hodit·s lht·111st·lws, whidt 11111y lw lht• 1110s! 1111p,;111111t 11•1111~ li.11d1 111-t1
0 0 0 -lxy
0 0 0 -fxz
fyy
-fyz
-lzy
fzz
(26)
In this symmetric 6 x 6 array, the symbol m denotes the mass of the body, which is computed from m
=
fvpdV
(27)
The mass moment of inertia lxx is obtained as fxx
= fv p(y~
+ z~) dV
(28)
in which Ye and zc are they- and z-distances of a typical point in the body from the center of mass. The mass product of inertia lxy is
fxy I, Iere the sy~bol Mi_ re.presents the consistent-mass matrix for member i al jn1111• J nnd k, while MBi 1s its counterpart for the reference points p and q. Equations of motion for all of the possible rigid bodies in a framed sll 111 lure may be written as
0 0 0
m 0 0 0 m 0
(l I)
I lence, the matrix relating ABi to D8 ; is
K8 ;
419
Rigid Bodies in Framed Structures
=
L
PXcYc dV
(29)
where x is the x -distance from point c. Other moments and products of inertia in matri~ M have similar definitions. The negative signs on mass products of inertia in Eq~ (26) result from Euler's equations for small motions of rigid bodies 14]. Table 9.2 contains various forms of matrix Mc required for all types of framed structures. The matrix M for a rigid body may be transfonned to a working point p hy an operation si;ilar to that for consistent masses of membe_r i ~iven in Eq. (23). However, only the points p and care involved, as shown m Fig. 9.2. The required congruence multiplication is
Mp = TpcMc TJc
(30)
1n this equation the transformation matrix Tpc is of the sam_e _form as ?1a~ix TpJ discussed earlier, except that c replaces j. Note that for a ng1d body_ m either a plnne truss or a plane frame, the matrix Tpc must be the same (and of size 3 x 3). Also, for a three-dimensional body in either a space truss or a space frame, Tpc 111 11gain the snml' (1111d of si7,l' 6 x 6) . . . As the st·~·o11d 1111•p 111 11HNl'11thling cquutrons of motion , we must add the 11111 ssi•s 1111 d 11111s1, 1111111,rrl! M111 1111·11111 lt11 tht• ri~id bodies to the matrix M1111 in Eq. ()•I) 'l'lti1, piwi. 1111 1111 1111v.1111111l ti 1111111 i,1 M 11 11, 111, lollowi,;
420
Rigid Bodies within Flexible Structures
TABLE 9.2 Mass and Mass-Moment-of-Inertia Matrices Mc for Rigid Bodies 1. Beams (x-y plane of bending)
I~J
Mc= [~
2. Plane trusses and plane frames (structures in x-y plane)
Mc= [ ~ 0
i..l,- - -; ; - - - ~ -- - - ~ - - -- -·+---;;---
-
~z ~] 0 I,,
3. Grids (structure in x-y plane) Mc
=
[
f xx - f yx
-fxy f yy
0
0
OJ 0
j~
...., ~
m
4. Space trusses and space frames
0 m 0 M, -[} 0 0 0
0 0 m 0 0 0
0 0 0
0 0 0
f,x
-fxy
- f yx
! yy
-fzx
-1,y
,,;
g"'
J]
..,c:: -a. "' .s..,"'
-ly,
I,,
...., "b
M1s =Mes+
L
00
Mpk
( II)
.$ -0
'@i
62 ...,.
k=l
wkrc 111, is the number of bodies. After solving the eigenvalue problem for the augmented form of Eq. ( 1 11 wt• cun add modal damping to our analytical model. Initial conditions 111111 g1ound motions may also be included, along with applied actions, if dcsii11I St·t·Iion 9. 3 describes a program named DYRBPF for dynamic analysis ol 111,1111 hodil'S in plane frames, using the member-oriented technique given aboVl' In any case (especially when the mass of the structural framing is 111 h 1wglt·clcd), we could use the mass center c of each rigid body as the wor~ln p11l111 p. With grids, space trusses , and space frames, it may also be convl•11lr11t lo 11.~t· f)l'incipal body axes, for which the mass moment of inertia suh11111111 wtlhin M, is diagonal. If principal mass moments of inertia and the diH•t111111 ol pri11l'ip11I body axes are not known in advance, they may be found by solvl11 1111 l•igl'Hvaluc problem of order 2 or 3 for each rigid body . This method is 1,i111i1111 111111111 lor principal stresses described previously in Sec. 7.2. lf one or 1111111 ut fill' i111plit·d rigid bodies do not exist, we could also eliminate the displ11n•111111t 111 11111ssk ss 11odt•s by stuti,· r<'d11ctio11 , as discussed in Sec. 6.7 .
Q\
... QI
So
~
K1rn111111t, •>.1 1•1µ1111• 11 •I Hli11wH II p1111i11u 11111 pl111111 1111NN wilh lwo dgid 1l'l'l11111;11l111 l11111ii1111• 1·11111111 l liy II pilN11111t11 1111·111lw1 r 111 1111111~ / 1111d A 11,1 11~ dl'Vl'l11p 1•1111l11h111l1111N111 lhl' 1111111h 1111d 1hr 111111111111 l111111d11111p1•tl 1•q11111i1111N 111111111111111111 p111t1l ~ /I 11111l 1/ llt1 ll11, lipid 11111(1
"'c::
'§
.....
422
Rigid Bodies within Flexible Structures
1UC.
9.2
423
Rigid Bodies in Framed Structures
We shall include actions, stiffnesses, and consistent masses from member i, ns w II masses and mass moments of inertia from the bodies. First, we consider actions at points j and k in the column vector
AM; = {Ah Ak} = {A,1, A12, : Aki, Ad Statically equivalent actions at the working points p and q may be obtai111•d II II transformation matrices for plane trusses from Table 9. I , as follows:
TP,
= [
~ ~J
-4L
3L
ABi = T;AM; = {Ap, Aq} = {A11, A,2, -L(4A,1 - 3A12),: Aki, Ak2 , 4L(Ak1 - Ak2)}
K.ii
K; =
[ Ki\i
K,k] Kkk
0
x 4 stiffness matrix for member i, using Eq. ( I , 0.64 Sym.1 0.48 0.36 i - IOL -0.64 -0.48 10.64 EA
3l
O
1
4L
- 4L
2 -8L
64L2
12 0 0 ] mi Mei= 0 12 0 12 [ 0 0 25l 2
(h)
r
1'1nnsformation operators required here are the 3 x 3 arrays
-0.36 ! 0.48 0.36
,:, n l-:L ; :1
0.64 0.48
0.36
- I.l2L
- 0.84L
Sym .
0 Xqcz
i
1.96L 2
------------------1---- 0.64 - 0.48 l . l2L l 0.64 I 0.48
- 0.48
-0.36
0.84L
- 0.64L
- 0.48l
l.l2l 2 J 0.64L 0.48l 0.641.'
5pAI, 1
r~
~] = [
~
0
~]
1
2£
-2l
1
I hey are applied to convert matrices Mc I and 111"rnrding to Eq. (30). Thus,
6 T
_
Mp= Tpc1 Mc, Tpc, -
m1
6
0
[ -12L
Sym.
I
2 0
2
0
I
()
(i)
Mc 2 to the reference points p and q,
0.36
Third , we write the cons istent-mass matrix for member i from Eq. ( I ~ \ Thul is,
M,
-28£2
l0 I 8L
which is found using Eq. (23). . Turning now to the rigid laminae, we take the followmg mass and mass-moment111 -inertia matrices from Table 9. 2 for plane trusses:
Then we transform matrix K; to the working points p and q with Eq. (22), p1111h11 I
EA
(g)
J
- 0.48
= lOL
2
5pAL -8L 6L 50£2 J ·-------- - ----i-1 0 -4L I 2 3 I
Substituting these arrays into Eq. (14), we form the operator T; and use it in h1 1I Thus,
Second, we set up the 4
Sym.
2
M,,
,
T
_
m2
1,,, i M,. 2T ,,,·2 - 3
[
3
0
0
3
61,
61,
~~L]
(j)
321}
')
'1'11111sl11111111111111 nl 11111ld x M, lo llll' wm·~11111 points 11 111ul ,, yi1•lcls
1lw 11 11, 1 o r thl'Nl' 111 rnys 1111n1111·11ts s 11h1111111 Ix M,,1,1 in Eq . (Hl, 1111d the Nl'l' ond is nddcd 10 M,,.,, Thr N Htr11 t111till H th1• 11h11•1 II Vl'N Ht11h•d Ill lh1· '11•11 11111111)\ 111 th1• t'Xlltllpk
424
Rigid Bodies w ithin Flexible Structures
ftoc. 9.4
y
9.3 PROGRAM DYRBPF FOR RIGID BODIES IN PLANE FRAMES
As an example of programming for the member-oriented technique, wr briefly describe Program DYRBPF for dynamic analysis of rigid bodies 111 pl I frames. To simplify the procedure, we take the mass center of each rigul 111 as one of the nodes in the structure (as well as the working point for tht· h111I Within the logic of the program, every member is assumed to have a rigid h, at each end, unless proven otherwise by the input data. This idea providt I key to easily extending the programs for framed structures to incl ud1· 11 bodies. Building upon Program DYNAPF from Sec. 6.6, we can add rigid hrnll to plane frames and create Program DYRBPF. For this purpose, the line c1111t ing nodal coordinates in the structural data (see Table 6.1) must be aug111111I by adding BM(J) and BI(J). These terms represent the body mass m and 1111· 111 moment of inertia /, 2 of the body with respect to its center of mass. Al so , 1t1 II element information we add XCJ(I), YCJ(I) , XCK(I), and YCK(I) , wh11 h note the x and y components of the offset vectors r cJ and r ck at ends j a11d A C course, if there is no body at one end or the other, such offsets are set t•q111I zero. In Program DYRBPF we assemble the stiffness matrix S8 8 , lh1· 111 matrix Mon , and the action vector A 8 (t) by assessing one member al a ti1111 I Eqs. (9.2-25)]. To the second of these matrices we add the mass-inerl111 t 11 for the rigid bodies to form the augmented mass matrix 0 , as given hv I (9.2-31). After the equations for rigid-body motions have been solved , w, find time histories for displacements at the ends of members using Eq. (') ' I Then time-varying member end-actions are calculated from these mc111hr 1 111 placements by premultiplying them with the member stiffness matri x.
425
Rigid Laminae in Multistory Buildings
L-
L
---.1.--
0
0
j2
t
·-i1
4L
j3/ i
M;
- - -- ---x
/ ;:; /
Example 9.2
Figure 9.5
Figure 9. 5 shows a steel plane frame supporting a single rigid body, which is 11 1111111 cube of size Lon each si.1e. We shall analyze the dynamic response of this conllp11111tl to ground acceleration D 8 1(t), caused by an underground blast. If all members of the frame have WlO x 45 cross sections, we can Ninh physical properties as E, = 3 .0
104 k/in. 2
X
p, = 7 .35 X 10- 7 k-s 2 /in. 4
A = 13 .2 in . 2
L = 36 in .
I, = 249. 0 in . 4
Rigid body in a plane frame.
1•11,1. 9.6, the first of which depicts the impulsive x-g~oun.d ac~eleration due to the underground blast. The plot in Fig. 9.6(b) gives the resultmg_t1me history of x_-translat10n 111 node 3 (the center of mass of the rigid body). Also, Fig. 9.6(~) contains y~ots of lli1tc•varying member end-moments AM6 for element.s 2 and 4. Max~mum (or rrumm~m) vnlues for these three types of responses are 0.03943 m., -157 .3 k-m., and 97 .06 k-m. , 1111 of which occur at time t = 52 ms.
where the subscript s implies steel. Al so, the relevant properties of the d gid l1111l y
t>,
2.25 x 10
7
k-s 2/ in .'1 I
Afl, I.'
Ill
{), [ , I
1.050
X
10
2
ks
1
lt,4 RIGID LAMINAE IN MULTISTORY BUILDINGS / i11
2.268 k s 2-in.
for whk h lltt' s11hst·1lpl ,. dt•11otcs rnut·1t•l1•. , Wt• 11111 1111' loi,•1101111• <111111 011 1'10H111111 l)Yltlll'I ; 11Nl111,1 S11hprnp111111 NI IHMC Wllh NM< II >1 1S / 1111d I >AMI'!< () 0 1 l'l11M111 111•111111 p1111h11·1·~ 1111• 1•11111p111l I plnl
M1tllistory bui ldings arc usually constructed in tiers of one, two, or three .stories limi;. I lcm.'l' , llw 1111111l' ti<'t huildil1Rs 15, 61 may be u_sed to descr~be_the •1h yNl'rnrwrs th111 ah111111d 111 l111w· cilk·s. Tlw stl·c l skl'l\.\lons of these lull buil~mgs 1 l!IININI nl ll jllll'l' 11111111·11 111111 IIIIIHI l11•q111•nll y 1111w tht•ir llll' itllll'l'I, lllrttl\At'
Rigid Bodies within Flexible Structures
426
Chlll
Sec. 9.4
427
Rigid Laminae in Multistory Buildings Slab or deck
400
.,
N
y-beam
'- 200
x-beam
C
0
60
20
(.)
<
T
80 (ms )
a:: - 2 00 {!)
I
X
- 400
P,(t)
T---J--.----+----ll'---.1 Y FP
(a) C
·.:; 0 , 0 4 I")
F
~ 0 . 02 0
z
i - - - - --
< o. oo
20
T
z
80 (m a )
X , c - - -- - + I
k---
;: -0. 0 2
t
- - -- -X,p-- - --,.... P,(t)
0
-
- --
--X,; - --
- -- - --1
<
..J (/)
~
a::
Figure 9.7
-o. 0 4
I-
( b)
I )(
ELEM,2
100 C I
~o ~ lO
r
< - I 00
( c)
li'IKurc 9 .6
Rigid body in a plane frame: (a) ground acceleratio n; (b) displut'l'1111111 .
{c) 1'1HI lllOIIIClltS.
111 Ihogonnl ( or l'L'<.: langu lar) pnllcrn . Figure 9.7 shows a lypical floor (111 11 pl1111 of II lit•r building, (.'()Jlsisling o f x hcnms, Y· beams, und a sluh 111 ,It, k ti wr shull l'llllN1dt•r lo tw I igid i11 its own plnnL'. This l11111i1111 is fk x1hh 111 I dtrt'\'111111 11<1111 1111 to ii ~ plt11a• 1111d t·ntdd Ill' 1li s1•11•ti:,,t•d wilh plult' lw111lt11ai I 1111·111 , A 1111,lt-1 l11rl 111111r 1111111111111l y !Hird upp1011d1 IN t11111cl11d1· 11 tr 111111111 v ol lhl' Ninh 111 tlw 1,11,.'I .,r, 111111 111 1· 111 Ii lw11111 Thl''1' hr11111 ~ 11111111• 111111 v rt
Typical floor plan of multistory (tier) building .
rnlumns that are parallel to the z axis (not shown in Fig. 9. 7). An origin of 1cference (or structural) coordinates may be located at any convenient point and llrobably would be taken at the base of the structure. The point labeled Fin Fig. ').7 lies at the intersection of the z axis with the plane of the floor. Also, point ,. represents the center of mass that is tributary to the floor, and point j is a typical joint at this framing level. Because each horizontal lamina is taken to be rigid in its own plane, all points at that level are constrained to displace in a rigid-body pattern. Such ,notions consist of translation in the horizontal plane and rotation about a vertical 11xis. Thus, the rigid-body motions of the lamina in Fig. 9.7 may be described I1y the x and y components of the translation of point F (considered to be attached to the body) and its rotation about the z axis. Moreover, the corresponding displacements at any other points on the lamina, such as joint j or the center of 111uss c, are related to the motions of point F by the concepts of Sec. 9. 2. On the 11t her hand, the x and y rotations and the z translation at a typical joint j remain 11s unconstrained displacements. For simplicity in the analysis , we assume that the geometric layout of the li 11ming is lhe same at ull levels, that there is only one distinct lamina at each ll•vcl, and thut lhL:rt nn• oo shear walls or lateral bracing in the building. Mc.·111bc1s of llw spnl'<' t 1111m• nrc 111kcn to he prisnmt ic and of a lincorly elastic 1111111•1inl Wt• ulNo IIHNIIIIII" 111111 joi11tNIn llw r, 111111• IIH' I i1tiil 1111d thllt displ11L'l'll1Cl1I S 1r l11t i VI' lo ~'111111111 1111 111111111
428
Rigid Bod ies w ithi n Flexible Structures
Because every member in the analytical model is parallel to one ot II structural reference axes, there is no need for rotation-of-axes transformali1111 Figure 9.8(a) shows a beam with its member axis Xm parallel to the x ux1~ I 11ddition, its principal bending axes Ym and Zm are assumed to be parallel lo 11~ y and z. Action and displacement indexes j I, j2, andj3 at the j end of llll' lllh t i denote rotations in the x and y senses and translation in the z direction . 'I h I counterparts kl, k2 , and k3 also appear at the k end of the member. The (1 stiffness matrix Ki for such a beam is the same as that for the grid mcn1h1 1 II Sec. 6.3. Thus,
"' ,.,•- - - - -
I"-
'.zr -
"~.
o
(
'~.s
------=--------'< ·-.
:s VJ
·,;: oj
N
S:l
-~
2
~~
0.
GlxL 2 4ElyL 2
0 0
-6EJ L
lI I
0 0
l O 2 2ElyL i -6ElyL 6ElyL l - I2Ely 0
E ::, 0u ~
12Ely
- ------- - --L- ~- ------ -----2 -GfxL2
C:
Sym .
-in'~
(I
GlxL
E
;,...
'<
0 0
2 a:l
=;;i
12Efy
~
0.
As hcf'orc, the symbol Ix represents the torsion constant, and ly is the 1w11111 1110111ent of the cross-sectional area with respect to the Ym axis. The second type of beam to be considered has its axis parallel to thl' I' 11 I as indicated in Fig. 9.8(b). In this case Ym is chosen to be the member uxlN, 1111 the principal axes x 111 and Zm are parallel to axes x and z. Action and disph1cr111 H Indexes j I through k3 correspond to those for the x-beam. Therefore, tlw h 111ember stiffness matrix for the y-beam is
K, = [~j Kki
Kik Kkk
E oj
..8
e in'
·,;: oj
><
2
.;
E
'<
J
B
0
~
0.
~
Q)
i
!:Q ~
~
4EI.,, L 2
0
I..'
Syni. GlyL 2
..
Q\
I I
,.,• - - - -
I I
0 12Elx . ------------1 -----2ii;i} 0 6ElxL I 4ElxL 2 - GI y L 2 0 I 0 0 - 12£/, 1-6£1,L 6El,L 0 6ElxL
QC>
::,
.!:/1
.... ,,l
GlyL 2
0
12W,
Noh• thul for lhis type of me mber the cross-sectiona l properties h11 v1 11 Ny 111hols lwt·nusc of lhc orientation of the memhcr uxis . T hut is, / 1, is 1111w th 1t11 slo11 rn11s1u11l , whlk / 1 is the st•rnnd 111011wnt ol' the cmss Sl'<:lio,wl 1111·11 wit 11'Nlll'l'I lo till' 1111 uxis
430
Rigid Bodies within Flexible Structures
oc. 9 .4
Figure 9.8(c) depicts a typical column with its member axis z,,, p11111II the z reference axis. Its principal bending axes x,,, and y111 are assu,m·il I parallel to axes x and y. Action and displacement indexes at joints j u11il A numbered in a sequence that expedites the process of transferring co11l11h111 terms from the member stiffness matrix to joint and floor stiffness matrkr~ 11 is , indexes j I, }2, and }3 at the top of the column correspond to joinl dr~pl ments that are not directly associated with the rigid-body motions of tlu above. These displacements are the same types as those discussed previou~I the beams. Similarly, indexes kl, k2, and k3 at the bottom of the colun111 p ti lo joint displacements not directly associated with the rigid-body motio1111111t floor below. However, the indexes j 4, j 5, and j 6 at the top and k 4, k 5, 111111 ul the bottom correspond to displacements that are directly dependt·111 11 rigid-body motions of the floors above and below. They consist of Lru11~l11II in the x and y directions and a rotation in the z sense at both levels j 111111 A We can represent the 12 x 12 stiffness matrix for a column as a p111 1l1 hll urray composed of 3 x 3 submatrices, as follows:
n
K;
=
J_;;_{~K_~L-[Ki, I I K31
I
I
K11 = K 22 =
Kt3
f
_E
=
K,,
!'his equation is similar to Eq. (9. 2-1)' but point F replaces point p. In this case 1he vector Ai contains
Ai
= {Aj4,
(6)
Ajs, Aj6}
[
i
4~
2/.,
0
i
2;; - ly
_ 6E K42 - /} I,
ro
K11
I f
I
(a)
0
0
()
KJ,
lhe x,,,;::t~~:ex;:~sence of the rigid laminae, certain geometric tr~nsforfmationhs . . t . and c to those at pomt F or eac ure required to relate the motions at p~m s J . t . and their lraming level Figure 9.9(a) shows actionsAj4, Ajs, and Ai6 at pom J . . . ·1ents AFI, An, and A F3 at point F. Values of the latter actions are Nlat1cal eqmva n1lculated from the former' as follows: (5) AF = TFjAj
~
4fx
K2, - I [
K,11
For this member the cross-sectional properties are defined as follow.s:hA, = art:· . t t and I x and I y = second moments of area wit respec o / = torsion cons an , 1
Sym.
111 !his matrix subscript 1 denotes action and displacement indexes of lyp, ./2, und }3. Subscript 2 represents indexes of types k I, k2, and k3. S11hN1 111 Nlnnds for indexesj4,j5, andj6. And subscript4 is for indexes k4, k5, 111111 Note that the stiffness matrix in Eq. (3) is also partitioned to scp11111h I 1111t·o11strained displacements at level j from those at level k, as well as I111111 I t·onstrnined displacements at both levels. The submatrices in Eq. (3) 1111
K,i •
431
K32 I K 33
! K42
K11
Rigid Laminae in Multistory Buildings
()
~l
I JW,,
()
0
()
I JFI,
()
(I
()
<;/ I '
(IJ)
li'IMIII ,, 11 ,11 I h 111111 II II ll llll •" l11111111IIIIII N (II) 111111111 11•l11l l1111MhlpN, (t,) d lNpl111•1111111nl 1rl11lh111ijhlp~
Sec. 9.4
Rigid Bodies within Flexible Structures
432
l
which has the same meaning as Eq. (9.2-22), except that the subscript F replaces 8. The results of Eq. (14) in expanded form become
Vector AF is AF= {AFI, AF2, An}
_!_(~~-~---- - -
and the transformation operator TFJ has the form
T,1 - [
K Fi
Dj
_L ,!, ~] = TJ:jDF
Again , this equation is similar to Eq. (9.2-6), but with point F replacing poh p . Here the vector DF contains the terms
TJ:, = [~
I
TFk K42
TFk ~
TT
Fk
(16)
Mee= me O 1 (
0 0
where me is the mass at that level. Furthermore, the mass-inertia matrix with
~
respect to point Fis obtained as tI
0 1
0 0 which is the transpose of the matrix in Eq. (8). Consideri?g no:"' the stiffness matrix for the column in Eq. (3), w1· 11111 [runs.form certa1? of its submatrices to rigid-body coordinates. This opi·r ntlt n.:qu'.rcs a combmed action-displacement transformation that makes usl' 111 ti 11111tnx Sym.l tI
T,.'k 'l'h1· tr:u11,1'0 1111ation matri x 'I; conlains f'our 3 X 3 suh111atric1:s on till' d111~1111 I of whrch thl' first 1111d Nl't·ond 11r1: id1:ntit y mntriccs. The third nnd fo11r11l ul 111atri1·1·s lltl' ol till' typt• giVl'n by Eq . (8), 11nd tht•ir suhs<:tipts dt noh' j11lt11 I th1• 11001 11h11Vl' 1111d joint ~ in tilt• flow lwlow INt't' Fii:. <) H(c)I , 'l'lll' ·,h ~11 nl 1hr
(15)
1 0
t 11
Also, the transformation operator becomes
I
- - - - - -
TFJK33 TJ:1 I ! TFk ~ 3 T TFj I
e
und lhe vector Di is
= {D,4, DJ 5 , DJ6}
l TFJ K 32
Sym.
Note that the transformed stiffness matrix KFi is still partitioned in the manner indicated for K, in Eq. (3) . When the submatrices in Eq. (15) are evaluated, we find that the equalities among them are the same as those given in Eqs. (4), provided that reference points at all levels lie on the z axis. In lieu of more detailed information, we assume that the mass of the building tributary to a given framing level is uniformly distributed over the area of the lamina. In that case the center of mass c will coincide with the centroid of the area, and every floor will have the same radius of gyration re with respect lo point c. Thus, the mass and mass-moment-of-inertia matrix with respect to point c for the lamina at level may be written as
t Ill
DJ
= _ K 21 -+-- K22 - ~-[ TFJ K31 TFk K41
~u~ing next to .displacements, we see in Fig . 9.9(b) the effect at polnl of a n ~1d-body rotation DF 3 at point F This rotation causes a negativl' lran.slation ~nd a positive y-translation at point j. Hence, the kinemuti1 1111 equivalent displacements at j due to rigid-body motions of point F are
l11111 11 lll111111f11111
433
Rigid Laminae in Multistory Buildings
n1l111111111frll111•10111111f11x tH'
(17)
This transformation is of the same type as that in Eq. (9.2-30). The symbol rF In Eq. (17) denotes the radius of gyration of the floor with respect to point F. It Is related to that with respect to point c by (18) which is a familiar expression for translation of axes from elementary dynamics [4).
Horizontal forces P.,(t) and Py(t), applied at a typical framing level, appear In Fig. 9.7. ln general, these forces are eccentric with respect to the reference point F, and the stalically equivalent actions at that point are A,:(t)r
T,..,,A,, (t) r
(19)
In whkh A,,(/l1 • (/',(I) , /\(()\,
(10)
434
Rigid Bodies within Flexible Structures
Sec. 9.5
435
Rigid Bodies in Finite-Element Networks
and
When the forcing influences are horizontal ground accelerations, the resul11111 inertial forces at level are
e
A 8 (t)e = -me{D8 1(t), D82(t)} where D81 (t) and D82(t) are accelerations of ground in the x and y directions . I h forces in Eq. (22) act through the mass center at each story. Therefore, llwy n eccentric with respect to point F by the distances XFc and YFc (see Fig. 9. 7), 1111 these terms replace xFP and YFP in Eq. (21). The transformation relationships discussed above prove useful frn lh purpose of writing equations of motion for a tier building . Our method h handling such equations will be described in Chapter 10 when we study lh modified tridiagonal method for dynamic analysis by substructures.
y
)-,
9.5 RIGID BODIES IN FINITE-ELEMENT NETWORKS I
In this section we shall consider rigid bodies that are embedded wllhll linite-element networks of the types described in Chapters 7 and 8. ln NIH h situation, we cannot avoid having more than one connection of an elenw111 111 particular rigid body. Therefore, the member-oriented approach for 1111111 structures (see Sec. 9.2) will be abandoned in favor of a body-oriented 111,·11,,, Figure 9.10 illustrates an analytical model consisting of rigid 111111111 connected by Q8 elements for plane stress or plane strain. By omitting till' 11111 bodies temporarily and excluding support displacements, we can write equ11111111 of undamped motion for the nodal displacements in the finite-element 1w1w111~ as fo llows:
[DA]
MM MAF] [SM [ Mfoi\ MFF Op + SFA
[DA] = [AA (t)] Ap(t)
SAF] SFF DF
II
In this equation the subscript A refers to nodes attached to rigid bocli1•11, 111 subscript F denotes free nodes (see Fig. 9.10), at which the displact•1111•111 ~ 11 indcpcndcnl. For an arbitrary sequence of numbering nodes , the lcr 11111 11111 l1ihuti11g lo Eq. (I) must be rearranged to put type A first and typcF st.K·orul N11 ll't us hri ng in the rigid bodies and recogni ze that nodes of type A arc <:0111111,1111 lo lllOVl' with the m. Thus, we need lo trunsform actions, stiffnesses, 1111d 11111 I knt 111aSSl'S frrnn nodl'S A to wol'king points on lhc rigid bodies. For lhiHp111p11 Wl' dHll)Nl' lh,· IIIUNH l' l'llll'IS IIS lilt• WOl'kinp poinls. llip1111• 1J Io 111tl1c1111•H rh,· 11111N.~ ,·1·111,•1 , ·1 of hody /.. 1111d 11 1yp11·11 I 111111 wh1•11• l'11·1111•111, 111 1111111l11•d l1uk111·11 /..I ,/.. ' , 1111d /.. I al p111111 , ·1 d1•1111h I
Figure 9.10
Rigid laminae connected by QB elements.
l'igid-body motions, which consist of ~ransl~tions in ~hex an~ y directions and rotation in the z sense. At node J the mdexes J 1 and J2 repr~sent ~nly lranslations in the x and y directions (for a Q8 element). Correspondmg actl~ns nt node j are forces in the x and y directions, and their statical equivalents at pomt ,.k become 11
(2)
which is similar to Eq. (9.2-1), but point ck replaces pointp. Actions in vectors Aj and Ak are
(3) In the latter vector we have forces in the x and y directions and a moment in the sense. The transformation operator in Eq. (2) is
(4) which hus tlw I'
sa111t·
Hq 1111111 11 1 (
1)
ll
frn 111 ll
11N
that for plane trusses in Table 9. I , with k replacing
I ic 1111 I y
Io ti w / 111 11odt• 011 I ht• /.. 111 11 p1d hod y,
11111 I
II II II 11odt·~
Sec. 9.5
Rig id Bodies w ithin Flexib le Structures
436
437
Rigid Bodies in Finite-Element Networks
attached to the body are considered, the equation becomes
DA1l DA2 = [TI1l Th D [~:.:' ~i, "
Ank = TkAAk The expanded form of Eq. (5) is
(12)
When all rigid bodies are taken into account, we have (t\)
Here the vector D8 consists of displacements at mass centers for all of the rigid contains the dependent displacements at all of the attached nodes. bodies, and Of course, T'};A is the transpose of matrix ToA in Eq. (8) . Now let us return to Eq. (1), consisting of the equations of motion for attached and free nodes, devoid of contributions from rigid bodies. To account for the presence of the rigid bodies, we form a transformation operator T , as
DA
AAn1 k where nj is the number of nodes on the body. Each subvector in AAk contains tw ~crms, as in the first of Eqs. (3); and each submatrix in Tk is a 3 x 2 arniy , 11 111 Eq. (4). Considering now all of the rigid bodies in the system, we express the nrt h1n trunsformation relationships by
follows:
An= TnAAA
T =
l II th is equation the vector AA contains actions for all of the attached nodes, 1111 I A,, consists of their statical equivalents at mass centers for all of the r igid hod I" 'l'ht transformation matrix T8 A in Eq. (7) is a large, sparse array contuinln 1nibmatrices of type Tk in diagonal positions, as follows:
TBA=
(13)
DA= TlADB
[T~A
:J
(14)
in which IF is an identity matrix of order equal to the number of free nodal displacements . Th~n we have (15a)
Sym.
T1 0
T2
0
0
Tk
0
0
0
Also,
'"
(15b)
Therefore , premultiplication of Eq. (1) by T and use of the relationships in Eqs.
Tnb
( 15) produces
where 11,, is the number of bodies in the system. Similarly, the kinematically equivalent displacements at node j in 11, '). 10 may be calculated from those at point ck by the formula
DJ = TiDk
whh.: h is of the same form as Eq. (9.2-6). In this case the displacement
Vl'l
M 8 n MnF J [ MFB MFF
[Dn] [Snn SnFJ [Do] _ [Ao(t)J DF + SFB SFi DF - AF(t)
(16)
(II
Terms with the subscript B in this equation refer to motions of the rigid bodies in the system. Block stiffness and mass submatrices in Eq. (16) are
1111
(17a)
llll'
t 111
11 1111 nodes on lhc kth ri gid body arc included , the tru11sf'orrnatio11 in ht'l'Ol lll'S
h(
and (17b)
1U
llccnusc the nrntri x 1'1111 has the form shown in Eq . (8), we can express Eqs. (17) IIIIH l'
1111 whi1 h 1111' 1·,i p11rnh-d 1111111 i11
('Xplil'itly
llH
S • 11111
I
'I'PS IU , 1'l' ft
S1111
S/m
' liS~, 1
( I 811)
438
Sec. 9.6
Rigid Bodies within Flexible Structures
and (18h)
fork = 1, 2, .. . , nb and C = 1, 2, .. , nb, Matrices Snnk. e and SMk, e in Eq., ( J 8a) are submatrices of S88 and SM that contain terms coupling body k wi lh body C, and so on. To complete the equations of rigid-body motion , we need only add 1hr mass-inertia matrices for the bodies to matrix M 88 , as follows: nb
M;n
= Mnn + I
Mck
( 111)
Program DYRBPB for Rigid Bodies in Plate-Bending Continua
Sec. 8.3 (see Tables 8.1 and 8.2) are needed to form action, stiffness, and consistent-mass matrices for all nodes not connected to supports. We also introduce information about the rigid bodies that enables the computer to distinguish attached nodes of type A from free nodes of type F . Then the desired matrices may be generated in the partitioned form shown by Eq. (9.5-1). As mentioned previously, the sequence for numbering displacements at a node of the PBQ8 element is z-translation, x-rotation, and y-rotation. We also use the same sequence of displacements at the mass center ck of the kth rigid body in the analytical model. Therefore, the transformation matrix for corresponding actions at point ck due to unit actions at node j becomes
k= I
where Mck is drawn from Table 9.2. For a typical two-dimensional body k in Fill 9. 10, the matrix Mq is the same as that for plane trusses and plane frames giwn in the table. We shall now discuss briefly other types of rigid bodies in other types 111 discretized continua. If we have three-dimensional rigid bodies embedded in 11 network of H8 or H20 elements, the transformation matrix Tkj becomes the s1111111 ns that for space trusses in Table 9.1, except that p is replaced by k . Also , lh 11u1ss-inertia matrix Mck for a typical rigid body is the same as that for spm lrusscs and space frames in Table 9.2 . Otherwise, the process of settin~ up t•quations of undamped motion remains the same as that described abow 1111 two-dimensional continua, but numbers of actions and displacements art• 111 ncased. Rigid bodies supported by discretized plates in bending are analyzed 111 11 11111nncr analogous to that for grids, except that the theory is body-oril•111111 instead of member-oriented. Also, the sequence of nodal displacements for th plate-bending element is not the same as for the grid element. As shown i11 hj (8.2 I), the z-translation W; is taken before the x- and y-rotations Ox; and IJ,, Three-dimensional rigid bodies connected by networks of general 1,lu II ckmcnts have characteristics similar to their counterparts in space f'r111111 11 1lowcvcr, the rotations in the x, y, and z senses at an attached node are ac11111II dl•pcndcnt components of the independent tangential rotations a; and {31 (s1.•t• I ·11i1 8 .() ).
Finally, rigid rings in axisymmetric solids or shells with axisy1111111'lll lo11ds would be restrained to translate only in the z direction. If loud'I 111 11111111xisymmctric, such rings can also translate and rotate in the r and () dlll'l lions und rotate in the z sense .
9.8 PROGRAM DYRBPB FOR RIGID BODIES IN PLATE-BENDING CONTINUA
'1'11 illusll'lllt• tlw 1111lun· or prog rumming for Ila· hody rn it•nll'd Hppmud1 , w,.,1111 dl Nl'IISN Prn~ tlllll l>YRI\PII 1111' dynun1il' 11111il ys 1~ 1)1 li 1ii d hodi,·N s11pp1111i·d I pl1111•11 Ill lll'111ll11p 1"01 lhl N p11tpUHI', till' Nltlldll111l 1111d d y 1111111ic loud d11111 1111
439
T kj
= [ :~ -xkj
~
~]
(1)
O 1
This array is taken from the third, fourth, and fifth rows and columns of the last matrix in Table 9 .1, with p replaced by k. Similarly, the mass-inertia matrix for rigid body kin a discretized plate is
M,, -
[I ~;~ ~!~l
(2)
which is drawn from the third, fourth , and fifth rows and columns of the last matrix in Table 9.2. Supplementary rigid-body data required for Program DYRBPB appear in Table 9.3, which conveys the number of bodies NB in part (a). Then in part (b) we have NB lines of body-node data containing the body number K, the number of nodes NJ(K) attached to the body, and the body-node numbers JB(K, 1) through JB(K, NJ(K)) . Properties of the rigid bodies are given in part (c), wh~ch indicates NB lines with the body number K, followed by values of the followmg six terms: XC(K) = x-coordinate of point ck YC(K)
= y-coordinate of point ck
BM(K) = mass of body k XXI(K) = mass moment of inertia l:o:k XYI(K)
= mass
YYI(K)
= mass moment of inertia l yyk
product of inertia
lxyk
We also need supplementary dynamic-load data for Program DYRBPB , as shown in Table 9.4. Information contained in this table consists of initial condilions and applied uctions for the rigid bodies. In part (a) of initial conditions we huvc the 1111111lll'r of' bodies with initial displacements NBID and the number of hodics with i1111111I w lul'llit•H NBIV. In part (b) arc NA()) lines with the body 1111111h1·1 K 1111d lh11·,· po10,1hll' 11111i11l displ11l•1•111,•11ls 1)1\0( 'K 2), DB0(1K I ). und
Rigid Bodies within Flexible Structures
440
Chap
TABLE 9.3 Rigid-Body Data for Program DYRBPB• Type of Data Rigid-body data (u) Number of bodies (b) Body nodes (c) Body properties
No. of Lines
Items on Data Lines
I NB NB
NB K, NJ(K), JB(K, I), JB(K, 2), ... , JB(K, NJ(K)) K, XC(K), YC(K), BM(K), XXl(K) , XYl(K), YY((KI
"Supplements (and follows) structural data in Table 8.1.
TABLE 9.4 Rigid-Body Dynamic-Load Data for Program DYRBPB" No. of Lines
Items on Data Lines
lnitiul conditions (u) Condition parameters (b) Displacements (c) Velocities
I NBID NBIV
NBID, NBIV K, DB0(3K-2), DB0(3K-1), DBO(.IK I K, VB0(3K-2), VB0(3K-l), VBO( IKJ
Applied actions (u) Load parameter (b) Rigid-body loads
I NLB
NLB K, AB(3K-2), AB(3K-1), AB(3K)
Type of Data
11
Supple ments (and precedes) dynamic-load data in Table 8.2.
DII0(3 K). Also, in part (c) we see NBIV lines containing Kand three poN~lhl
initial velocities VB0(3K-2), VB0(3K-1), and VB0(3K). Because thesl' i11ltlAI rnnditions are given for rigid bodies, their effects on attached nodes nuo,t h t·o111putcd within the program and are not included as dynamic-load data 1111 th Ntrncturc . Actions applied directly to the rigid bodies are listed as the second lypr or d11tu in Table 9.4. The only parameter required in part (a) is the number of 1111111 d hod ii.ls NLB . In part (b) the data for rigid-body loads consist of the body 1H11nh K untl applied actions AB(3K- 2), AB(3K- l), and AB(3K). This forcl.l und th lwo nwmcnt components are assumed to act at the mass center ck, as drn " th 111L11tiul l'orcc in the z direction caused by ground acceleration. Of course, 11\'I 11111 11t 1111y other points on a body could also be handled, but they would requi11· dnl Im lot'11tio11s as well as magnitudes. As in Program DYNAPB, the rotational displacements at free nod1111 11 t•li11il11atud by Guyan reduction . After solving the equations of motion 1111 rt.aid hodil·s und f'rcl.l nodes, we can find time histories of displacements for 1111111 h I 11odt·s with Eq . (9.5 13). Then the displacements in vectors 0 11 and D, 11111y I 11Nl·d to obtain ti111u vmying stresses in the finite clements. 1,:,ui111pfo
'> ..,
l•l1•t11r 'I 11 d1•ph IN hull ol II " y11111111tdr Hlit1rl1111•, \'IIII NiH(lllU ol II q11111h11 111 11 -1111 , 111111111 dlH~ H11pp,111rd hy II ph111• 111 lw11dt111l 1111 pl111t· ih d1vid1·d 111111 two I'll
442
Rigid Bodies with in Flexible Structures
elements that are fixed along edge 3-5-8 and symmetric with respect to a plane co111111n ing edge 8- 10-13. A step force of magnitude P, = 0.1 kN acts in the z direction 111 th mass center of the first rigid body, which is labeled point c 1 in the figure. Phy~1111I properties of the plate are
E = 103 x 106 kN/m 2
L
v
= 0 .34
= 0.05 m
p
~
E
= l = 0.05 m
0.1s
w a o. 0 z
= 8 .66 Mg/m 3
°<
h = 0.002 m
h1
L
=5 =
[ 2(1
I xy =
16) m,-Rf(! 1r 2 - 91r
0.01 m
.,
~
:c
hf] =3112X 10 -8 Mg·m 2 .
"' -10
w a 0 z
= -8.913 X 10- 9 Mg·m 2
0 .2 0
o. 15 fl Q
0
0 , 10
0,05 0 00
0
Q..
+12
We used this data in Program DYRBPB with DAMPR = 0.05 and direct 1111111 r icul integration by Subprogram NUMINT. For this purpose, half the load was applll·,I In l'nch of two cases to find symmetric and antisymmetric responses. For the sy1111111 t1I l'Use, restraints against x -rotations are required at nodes 10 and 13. However, 1111 th unlisymmctric case, we need restraints against z-translations and y-rotations 111 1h11 nodes. Figure 9. l2(a) shows computer plots of the resulting z-translations at poi111, 11n 1 lhl' rig id body for the symmetric and antisymmetric cases. In part (b) of the fig1111·, 111.~o sec the z-translations of node I for the two cases. Finally, in part (c) 1111• 11111 hi stories o f the flexural stress SX at the integration point near node 3. Of course, lhl IW 1·111 vcs in each of the plots must be added to obtain total values for the responsck 111v II
-
0 . 05
-, 0. 00
a:: <
-2 0
w z
whe re the symbols R, , h 1 , and m, denote the radius, thickness, and mass of the qu111 t disk .
ID
10
a
1rRf m, = ph,- = 1.700 X 10- 4 Mg 4 16) I.u =IYY =m, R, -2 4 -97T
0.20
E
and both the plate and the solid are made of brass. The rigid body is attached to 110,I I , 4 , 6, 9, and 11, and its properties are as follows:
R,
443
References
Chap. 9.
X -30 U)
( C)
Figure 9.12
REFERENCES W W Jr "Dynamics of Elastically-Connected Rigid Bodies," in Develeaver, . , ., . d w A Shaw Pergamon opments in Theoretical and Applied Mechanics, Vo1. 3 , e · · · , Press, New York, 1967, PP· 543-56Z. of Framed Structures, 2nd. ed. , . 2. Weaver, W ., Jr., and Gere , J . M ., Matrix Analysis
!.
Van Nostrand Reinhold, New York, 1980. . 3. Weaver , W . , Jr. , and Johnston , p · R · , Finite Elements for Structural Analysis, Prentice-Hall, Englewood Cliffs, N.J. , 1984. . . 4 . Beer, F. P., and Johnston, E . R. , Jr. , Vector Mechanics for Engineers: Dynamics, 4th ed., McGraw-Hill, New York , 1984. . . . Analysis 5. Weaver, W. , Jr., and Ne1son , M . F . , "Three-Dimensional 6 385-404of Tier Bu!ldings " ASCE J . Struct. Div., Vol. 92, No. ST6 , 196 ' pp. . . . . 6. Wea,ver, W . , Jr., Ne lson , M . ~-, and Manning , ~ - A. , "Dynam1ct4~~~:~7:_u1ld.1ngs, . " A<'( Ml'C"h. Dtv. Vol. 94, No. EM6, 1968, PP· o} ' /• " , t , F,"' , ("i •
"''""l'I' '),
ll lt1111d hmly 111 11 pl1111•· (11) hrnly fl1111Nllllh1IIN, (h) 11nd11l 111111Nl111 11111N, 11) 111·~111111 ~l11·on
(cont.)
'
10
Sec. 10.2
445
Guyan Reduction Methods y
I
Substructure Methods
10.1 INTRODUCTION
When the number of degrees of freedom in a structure becomes very lar!,w, w 1u;cd to divide the analytical model into substructures. Figure 10.1 ill ust1111 such a case, which is a computer plot of a large radio telescope antenna desi~n I as u space truss. This type of structure consists of a reflector and a s11pp11rl structure. We can take advantage of the facts that the reflector has two pl1111 or symmetry and the support has one such plane. Therefore, we need 111111ly only a quarter of the former and half of the latter, using appropriate restrni111- 1111 tht· planes of symmetry. Also, the more complicated reflector can be dividl•d 1111 substructures and analyzed by one of the methods in this chapter. For analysis by substructures, we must distinguish techniques thul 111 1rn ituhlc for statics from those more conducive to dynamics. In static 111111ly~I 1111 nodal displacements for a substructure can be eliminated from equl lth1111111 t•q1111tions during a frontal reduction procedure [1, 21. However, in dy1111111I 111111lysis, we need to retain a certain number of scattered degrees of frccdrn11 with low stiffnesses and high inertial actions, for which approximate cq11nl11111~ 1 111011011 cnn be written. Three approaches that work well for dynamks 1111 th 1i1di11gonul 11wlhod (with modificutions), tht• parnllcl elimination llll·lhcul , 1111 I lht• t·o1111HH1t·111 111mk ll't'hnique . Euch ol' these suhslnic1111·c 111c1hods will I d('Sl'iilwd in till' l'IISlllll~ Sl'l'liOIIS or thi s d111pl1•1·.
Figure 10.1
Radio telescope antenna structure (pointing to zenith).
10.2 GUYAN REDUCTION METHODS Tridiagonal Method for Substructures in Series
Pigurc 10.2 shows n two-dimensional discretized continuum divi?ed into s1thstruclurt•s 1h111111l· l'lllllll't'lcd in series. Suhstructurc numbers appear in ~oxes lwlnw till' llg111t• 111,11• 1111• My111hol (1 dl'IHlll'S II lypil'ul s11hstruct11re, 1111d 11, 1s lh(:
446
Substructure Methods
B,
Chap, 11)
Sec. 10.2
Guyan Reduction Methods
447
B,
(4)
Now we will apply Guyan reduction (see Sec. 6. 7) to decrease the number of degrees of freedom in the t'th substructure. For this purpose, let the displacements of type Ae be dependent upon those of types Be and Be+i· Thus, (5a) Figure 10.2 Substructures in series.
in which
number of substructures. For each substructure we have interior nodes of lyp, A (or Ae) an~ i~sulating boundary nodes of types Be and Be+ 1 on each side. Tiu sy~bol Be md1cates boundary nodes common to substructures 1 and / while Be+1 represents those joining substructures and + 1. Let us form 11 11 column vector of nodal displacements as
e
e-
e
(5b) Equations (5b) are of the same form as Eq. (6. 7-7b). The acceleration relationship similar to Eq. (5a) is
..
DAe
D
=
{on
I'
DA p Da2, DA2, .
, DAn,' Dsn,+}
( 11
.. = TAeBeDBe +
..
TAeB e+1 Dne+,
(5c)
To reduce the equations of motion in Eq. (3) to a smaller set, we construct the transformation matrix
With this sequence, the stiffness matrix for all nodal degrees of frccd11111 h1..•comes
(6) Sa,s 1 Sn,A, SA1B1
SA1A1
SA,82
S82A1
SB2B2
S82A2
SA2B2
SA2A2
S=
0
Then we have ( 'I
0
SAnsA"s
SA"sBns+I
Sn11s+ IA11s
Sn,,s+1B,1sf
(7a) and
I
m~t1:ix Sis a tridiagonal array of submatrices, this approach is rdr 11111 lo ns. lhc tridiagonal method. Without writing it, we can also observe tlllll 11 11 t·1111s1stcnt mass matrix has the same tridiagonal form. Suhstructurc .e contributes terms to the equations of motion as follow
MeDe + SeDe 111 l'Xpaudcd form, thi s equation is
Mu,n, M, ,11, 0
M11 1A1 M,1,1, M11 111 11
= Ae(t)
(7b)
De=
lll•1..·11usc
Substitution of Eqs. (7) into Eq. (3) and premultiplication of the result by TJ produces
( 11 (8) In this equation the reduced matrices are
St
'I'~ Sr'I}
S ' "' 11 rll r
Ist,,
111,
(9a)
448
Substructure Methods
Chap. 10 Sec. 10.2
A i (t) =
TI A e(t) = [
J
~;e ABe+1
Guyan Reduction Methods
449
(9h)
Example 10.1
(9tl
The unrestrained beam in Fig. 10.3 is divided into two substructures, each consist!ng of two equal prismatic flexural elements. Let us determine the coefficient matrices s;B and M!8 by the tridiagonal method, as required in Eq. (11).
Modified submatrices in Eqs . (9) have the definitions s;eBe = SBeBe
+ SB eAe 'I:i eBe
MrieBe = M BeBe + T Jen eMA eBe
( 1011)
+ MB eAeTAeBe + T JeneMA eAeTAeBe (!Oh)
A;e = Ane + T JeBe A At
(!Ot)
11nd so on. Finally, the reduced equations of motion for all of the boundary nodes tu~, the form (I l l
Substructures contribute to the matrices in Eq. (11) by the usual direct assenihl v pr•occss. The reduced stiffness matrix is still a tridiagonal array of submatrit·,,11 ns follows:
Figure 10.3 Beam divided into substructures.
For substructure 1, the stiffness matrix in Eqs. (3) and (4) becomes
6 3L 2L2 i -------t--- 6 - 3L i 12
0 ( I I)
L2
0
0
0
0
0 Subrnatrices of S!n having single asterisks receive contributions from onl y 11111 .~ubstructu re, but those with double asterisks have two contributing 1rnh Nltuctures. The assembled mass matrix M% 8 in Eq. (11) has the same tridia111111,1l u11·11 11gcmcnt of submatrices as the assembled stiffness matrix in Eq. (12) . S1111 ll 11rl y, the assembled load vector is
A*()lJ t - {A* BI' A** 8 2 , A** 83 ,
· · · ,
A*** Dn,, An,,,+, }
2
I I
- 6
-3L
l 3L
L
1
6
I I
! - 3L
2
2L2
Terms in this array are drawn from Eq. (3 .4-24) . Similarly , the consistent-mass matrix for substructure I is 156
22L
4L2 i
--- - - - -- - -l- - -- -
l 312 . l O 8L i - - - - - - - - - ~ - - - - - - - - -~-- - 0 0 l 54 13L i 156 0 0 - 13L -3L l - 22L
( l II
wli il'h aga in has single and double contributions. For convenience in co111p11h 1 pr ogrnmming, we assemble the reduced matrices while proceeding fro111 11111 1Hil11,t1uctun..: lo the next, thereby increasing the number of type B di splrn:t·1111•11t lk cuusc the assembled stiffness and mass matrices both have It idl11v11111il form~. vibn11 ionnl and dynamic analyses can be more efficient 131than 101 111111 1 111t thods wlwrt• the rn111t iccs arc filled. The l'o llowing example dl'111<11111l11111 Hppl lt-a1lo11 ol lhis npproul' h to II lwum 1ypt· ol st111t·t111t·.
(a)
l O 4L - --- - --T------- 1--3L
54
13L
- 13L
- 3L2
(b)
2
2
J
4L2
in which the terms arc obtained from Eq. (3 .4-26). Next, we use Eqs. (Sb) to fo rm TA tne and T AcD e+ i for substructure 1, as follows: 'I\
1/11
1 S ,1 1fl I S ,1 1/1 1
I
"'·
I
21
2/,
L
l
/,
(c)
450
Substructure Methods
TA 1B2 = -S;iA I SA 1B 2 =
IO
-L
3
Sec. 10.2
.------~------------/ 78
]
I
-L
1
With these submatrices, the transformation operator in Eq. (6) may be written as
T I --
[
M* _ pAL BB 105
lo O I B, _9___~-L~___Q_. O
J _
In 1
0
TA1B1
TA1B2
0
In 2
l
= 4L
L2
2L
l 2L -L 1 3
,:-3
1
- L2 -L
A
- -- -- -i---- - - - - -
0
0
0
J
'
0 / 0
4L
(i)
B
EI 3L 4L2 I ---------f---- Sym.]:' 2L3 -3 -3L I 3
SI - TI S1 Tl = -
1
2L2
3L
-3L
4L2
J
0 0
\ 27
13L
i- 13L _ -6L 2 _ -
(j)
78 \ 22L __ 8L 2 \
B,
Modified Tridiagonal Method
( fl
I
/
Sym.
1,
8L
which may now be used in Eq . ( 11).
3 T
i
2
Then the reduced stiffness matrix becomes
*-
22L
0 0
O
4L
[I]
2
r------ --,----------1 I 27 13L \ 156 I \ 2 2 II______ -13L -6L 16L II] II ____ TI _ _O____ __ _ I I I
4L
451
Guyan Reduction Methods
and
2
__!_[ 2 L 4L
Chap.
2
Now we consider a modification of the tridiagonal method for substructures in series that abandons the idea of insulating boundary nodes. Instead, there are three definitions for types of displacements, as follows: (1) Subscript Ae implies dependent displacements in substructure C; (2) subscript Be represents dependent displacements at boundary nodes joining substructures C and C+ 1; and (3) subscript Ce denotes independent displacements associated with substructures 1 through C, including boundary nodes common to C and C+ 1. The dependent displacements of types A and Bare to be eliminated, but the independent displacements of type C will be retained. Note that the number of type C displacements increases for each consecutive substructure. Also, they may be scattered within the substructures as well as at boundary nodes. At each stage in the reduction process, the partially-formed equations of motion are symbolically the same as in Eq. (3). Thus,
MeDe + SeDe = Ae(t)
(14)
However, the expanded form is now f."or substructure
2, we can proceed in a similar manner and find that
s;
=
st
Ml= Mi'
< \,11sl•quc11!1y, assembling the reduced stiffness and mass matrices f O b 111111 J yields ' r m su st ruct1111 • I
------------
,- 3
I/ 3L s,~"
m 2/,'
OJ
4L2
1-1/,3
I 2// 3L
()
()
()
()
fl,
I
I
0
8/~-i---
I/
,I '
6
Sym.
fJ I
[II
IJ 2
,I/ '
II I
fl
/11
(15)
Men
Mee
De e
ScA
Let the displacements of types Ae be dependent on those of types Be and Ce, as follows: (16a) where
I/ I/
:::J[;:J [t +
(hi
11)
TA1llr Also,
Wl'
= s;;,1 ,St1 r1J1
Ti1,cr
=-
s;~A eSAtCt
(16b)
r
( I Cw)
huvl' llw lll'l'l'il'l'Hlion rclnlionship If
ll1,
U
II
'l\1111>11,
j
'I\,, 11),
452
Substructure Methods
Chap. 10
Then the transformation operator becomes
Tc=[~: T~cJ 0
le
( If I c
With this matrix , the displacement and acceleration transformations are
( l Hnl und De=
[2:J = .. De e
Te
[~BJ De e
C, the mass matrix is diagonal, as in the next example. For such a cas~, transformation of the associated eigenvalue problem to standard, symmetnc form should involve factoring the mass matrix instead of the stiffness matrix. Also, an efficient overlay technique for stiffness terms is demonstrated in the following example (see also Sec. 10.3). Example 10.2 Figure 10.4 shows a rectangular plane frame divided into _two substructures, each of which consists of a beam and the two columns below. For this structure we shall find the reduced stiffness matrix in Eq. (21) by the modified tridiagonal method , using the following assumptions . Tributary masses having values of m and 2m are lumped at the two framing levels, as indicated in the figure. Flexural rigidities of beams 1 and 4 are equal to 2E/ and 4E/, while those for columns 2, 3, 5, and 6 are equal to El. Members are all prismatic , and axial strains are to be neglected. y
( IHhl
Substituting Eqs. (18) into Eq. (14) and premultiplication of the equation hy Tl yidds
Me*
[Dn] .. + Sc* [Dn] = Ae*(t) c c De
De
st = TISe'fe = [s!B s:c]
Ste e
Sen
TIMe'fe = [M! M%c] McB Mtc e At(t) = TIAe(t) = [A!] Mi
=
8
T7
( IV)
QJ
'l'hc l'educed matrices in this equation have the definitions
L
CD ®
0
-- -Da,
L z
* Dc .. + Sfc Dc = A7,(t)J,,, [M,c
/4
- oc,
Da,
L
[I]
Modified submatrices in Eqs. (20) are the same as those given by Eqs . ( IOI t'Xt'llpt that subscript Ce replaces Be+ 1• Wl~en proceeding from one substructure to the next in a series, displ111 1 1111•11(s of type Be become a subset within those of type Ae+1. Also, for lht· In 1 suhslrncture, displacements of type B,,, are usually omitted. Thus, tht• 1111111 1•q1111tions of' motion become l 'II which invol ves displacement s of' type C for the whole structure . 111 the nwdilkd tridiu1,onal muthod , cli111i11ution (or release) of du,pli111 1111111111 ol hotli 1y1ws I\ 1111d II l'llllHl'S s7, 1111tl M1,· in Hq . (21) 10 ht· IIJlc d I lt1Wl'Vl'I , ii l1111qwtl lllllNSt'H 1111' II Mt•tl ill ('lllljlllll'llo11 wllh lll'l't' lt•mliot1N 111 l\' 11
©
3
/
-oci
/
---
(.ltlhl
2
DA,
(.!OnI
Ac e
453
Guyan Reduction Methods
Sec. 10.2
®
® / 5
/ I
X
/ 6
2L
.1
Figure 10.4 Plane frame divided into substructures .
Fm suhNhul'1 1111• I , w1• Sl'I up the 6 x 6 stil'f1wss 111ntrix in Eqs. {14) und ( 15), as
lollows:
4114
li111J1 1r1u.h1111 M11thrnh1 1
,110.
10.2
(11 1y1111
llnd111.. tlot1 Molhoda
455
,I/
S,
WI
-u
Syn1. A1
I. '
It '
J;'
()
,I '
0
I;'
()
21.1
3L
3L -3L
I
3/~
3L
1
- 3L
- 3L A,
In !his instance, (hem arc no displacements of type B2 because the bases of columns are
llxcd. Also, displacements of type C2 are the same as those of type C1 in the first ijilbstructurc. Terms in matrix Si from Eq. (o) may be added to those in matrix S2 from I !q. (p) to produce
II,
I
II
12
I
-3L ! - 12
B1
116L2
(' ,
12
A
S2
C1
=
* + S2 =
2£1 15£3
--
SI
3IL2
116L2 i
(q)
-36L--36il --126-
wl11l'h 1s J)tlt l_i(ioncd according to displacements of types A 1, B 1, and C 1 • 'fhl'll I 'q~ 1II lo form T A e11e and TA ece for substructure I. Hence,
9L !' - 126
9L
306
IIH' IINl'd
TA
-
5 A1A1 -1
-
5 A- 1A 11
-
101 -
T A 1C:1
-
S 1[-4 -~J A1 8 1
S
A 1C 1
=
=
'J'his superposition of terms represents an overlay technique that will prove useful for 1·omputer programming in Sec. 10.3. In preparation for elimination of type A2 displacements, we determine TA 2c 2as
l
15
3[-1 11]
SL - 1
II
TAC 2 -_ _ 8, A_21A 2 s'A 2 C 2
I JNin,-: lhcs1.: suhmatrices, we compose the transformation operator in Eq. (I '/) 11111 111
-4L L
/ -9
9
I I
-9
9
O / 0 ISL O
0
L - 4L
15L O
0
I --------~------0 0
0 0
/ ISL / 0
2
0.2449
Ti
B1
36L / 126
36L
- 36L
-36L
!-
126
0 ISL C ,
I,
I
I;;:: ()
6L
Sym.l :' 126
"·
A,
L[
~}~~- ~:S)~.!_~-L
0
0
-L
(s)
J
2
=
S*2
=
rs T = L3 2
2
2
El [
14.45 - 17.39
-17.39] 40.65
(t)
Parallel Elimination Method
A more general manner of substructuring does not produce coefficient matrices composed of submatrices in a tridiagonal pattern. For example, the two-dimensional discretized continuum in Fig. 10.5 is divided into four substructures with interior nodes of type Ae and insulating boundary nodes of type Be having no special arrangement. In this approach the meaning of subscript Be is that it includes independent nodal displacements for substructures 1 through As in the tridiagonal method, the technique is to eliminate dependent displacements of type A 1 from each substructure and to retain a growing number of independent displarcmcnts of type Be. For each st1•p in the reduction process , substructure contributes the following tc11111, In 1111' l'q1111tions of mo tion:
e.
11 I
0 (' ,
l/,
= _ _!_
I
2
()
[TA2C2]
0.061221
This matrix contains actions of type C due to unit displacements of type C, with those of types A and B eliminated (or released).
hy Hq . (20H) . 1'1ol'l'l'di11g l'rnm su~structurc I to substructure 2 (sec Fig. 10.4), Wl' 11 ch 1111 iltNpl11c1·m1•111s_ ol lype lJ in the former to become type A in the htllur. Adcl111 1111 I 11111111h1111011Nfrn111 suhslruc1url' 2 to the stiffness matrix arc:
WI
(r)
Now we can determine the final reduced stiffness matrix to be
IIN HIVl' II
Si
=
l c2
S* [ cc
1-~l~--~~~l----
EI 'f'/' S I T1 = l 5L3
1 [0.2449 0.06122] L 0.2449 0.06122
Then the abbreviated transformation matrix T2 becomes
'l'h111t•fon: , the reduced stiffness matrix Si is
St
-
A1
---------j-------
1 ISL
_
-
2
()
17
e
I
I
Sec. 10.3
456
Substructure Methods
457
Modified Tridiagonal Method for Multistory Buildings
Chnp
and
..
De=
B,
= [1?.AJ Dn e
TeDnc
(26b)
Substitution of Eqs. (26) into Eq. (22) and premultiplication of the result by TJ gives (27) A,
A,
The reduced matrices in this equation are symbolically the same as in Eqs. (10), but displacements of type B are different. As for the tridiagonal and modified tridiagonal methods, we assemble stiffness, mass , and load matrices in a direct fashion while proceeding from one substructure to the next. We could also devise a modified parallel elimination approach that abandons the notion of insulating boundary nodes and introduces retained displacements of type C. In that case, the displacements of types A and B would both become "slaves" to the "master" displacements of type C.
B,
B,
A3
A.
10.3 MODIFIED TRIDIAGONAL METHOD FOR MULTISTORY BUILDINGS
Figure 10.S Substructures in parallel.
This. cx~ressio n again contains the same symbols as Eq. (3) but its ex 1 Vlll'H10 0 IS
'
I 11111
MAB] [~A] + [SM SAnJ [DA] _ [AA] [MM MnA Mnn Dn e SnA S88 D8 e - A8 e
1
( \I
which is diffcrentfrom Eq . (4). ThedependenceoftypeA d' l type JJ,. is written as e isp acement" upon
DAc = TA elJ eD11r
I'
I ll'rl' .the n~atri ~ TA.i•!J,e has the s~me formula as the first of Eqs. (Sh), hut 111t•11111ng of IJ1 1s different. T he required transformation operator is
T1 =
f
II
lh
~:'l.
Thii. hu.~ lh.t· uppt'lll'!lllt'l' of Eq . (6.7 9), but it pl)rlltin s to only onl) s uhs 111111111 Now thl' d11q1l11t•t•111t•11t 111111 lll't't•ll'rn1io11 tn111sl'o1111Hlions urt·
n,
1) I 1
u,,
I t
'I', I >11 I
( 1t111)
We now apply the modified tridiagonal method described in Sec. 10.2 to twoand three-dimensional multistory building frames. Most planar building frames can be handled in the high-speed core storage of a large-capacity digital computer without dividing them into substructures. However, we wish to use the same technique for both plane and space frames in order to take advantage of inherent similarities while explaining these two applications. For a multistory rectangular plane frame, we take_the analytical model illustrated in Fig. 10.6. The frame is assumed to have linearly elastic prismatic members, rigid joints , and fixed bases. We also assume that there are no shear walls, diagonal braces, or setbacks in the building. Each substructure in the figure consists of the beams at the framing level A (above) and the columns between levels A and B (below). Displacements of type A appear at the upper level , and those of type B are at the lower level. From the joint displacements labeled in Fig. 10.6, we see that axial strains in the beams are to be omitted; whereas, those in columns are to be retained. The reason for keeping the latter strains is that their influences are known to be significant in analyses of tall buildings r4, 5) . Therefore, at each framing level there is only one translation l),,.r in the x direction and its corresponding lateral force AFt· Thus, the subscripts in the fi gure match those for the modified tridiagonal method in Sec. 10.2, l)X<.:Cpl that F rnplaces C. Nole that the numerical subscripts o n displacements of types A n11d I/ (lcf'I to liphl ) nrc I, 2, . . . , 211,, whcrn 11,. is the number of l'OIIIIIIIIS , AINO, till' Nlli>M'ripl ~ llll lhmll' 111 typl' F (lllp lo botto m) Hl'l' I , 2, .. .•
e
458
Substructure Methods
t
Sec. 10.3
Modified Tridiagonal Method for Multistory Buildings
t
Ym
/
/
t
t
/
459
/
Level A
-/),
n 1~ (j) t -~- '-------=------"7 ----
k2/ k
i2/ i y
Xm
/ (a)
B __ loA, _
Leve l A
t
t0A3
DA2nc-•
~~~....C....~~~~~-4~~~--;
_,,
·, Ym
z
/
/
DA4
DA2nc
t
Level 8
~
0
83
t
DB2nc- 1
Level A
_,,
. "-.
- ~ -
t
.
--13
j2/
/
/ 084
I
11
082n c
t
t /
(j)
_,,
/ Level 8
/
---x
~!_
y
}--,
/
Figure 10.6 Analytical model for Program DYMSPF.
n,, where ns is the number of stories. Finally, this analytical model has tr ih11t111
masses m 1, m2 , •• • , mn, lumped at the framing levels, which includt· 11111 tributions from beams and columns. As for a tier building, every member in Fig. I0.6 is parallel to om· 111 th structural axes; so no rotation-of-axes transformations arc rcquirl•d . 1•11:1 111 10.7(a) depicts a beam with its member uxis I',,, pnrallel to tht· 1 11xiN 111111 It principal bending axis z,,, purnlkl lo lht· ., uxis . Aclio1111nd displ11c1•1111;111 1111h 11
z
- k 3 - -- - Xm
/ k2/t
-~
Zm
(b)
Figure 10.7
(a) Beam parallel to x axis; (b) column parallel toy axis.
· 1 ·2 k L and k2 imply that the 4 x 4 member stiffness matrix is the same as .I 'J ' ' that in Eq . (3.4-24) for a flexural element. . . .
On tht• otht·r h1111d , Fig . I0.7(b) shows a column with its member axis 1m purnllcl to 1lw v 11xt'l 1111tl ti s prinripnl bending axis z,,, ugain par~llel to the z axis. Action urnl dt ~plm i· 11 w 11 1111d1•x1•s 1111•111ls.J 1111d k ml' 1111111lwn·d 111 a s1•q11c11ct thut
480
i,11h11t1111 turn Mnll10d1
line. 10.3
IN l'<'.IHhll'.i;l' fo~· l1'1111Sll' lll11p 11111111111 ~llll111•~Nl'S lo joint 1111d Horn 11tl 111ut11ct·s. I hus, indcxcs1 I 1111111 'Ill lhl' lop ol lhc column arc lhl' .~1111 11• 11 t Im II lw11111 ul lcvcl A , whllL· A I 1111d A.> 111 lht• bottom coincidc with tho~ h hl•11n1 at level B. However, lhl' indl·xcs j I al the top and k3 ul lfll' hull crn IL'spo'.'~. to the x-~ranslations or tloors al levels A and B. Wu muy w, 11 I '1 "< <> s~1flness matnx for a column as a partitioned array consistinp ol , s11h111utnces, as follows:
r,
0
4L
-r, K,-L3 0 _ El ,
Ir I O 4L2 I --------~--------t----0 6L I O 6L I 12 0 -6L I O -6L I -12 0 2L2
.i I
}2 A
JI
Sym.
i
2 _J __ _ ________
}2
kl
1
kI
k2
j3
B
A
k2
II
II
j3
12 k3
k3
F
whid1 .is a rearranged version of Eq. (6.2-1). Note that subscripts in Eq , ( t) l11l..1·11 111 the sequence A , B, and F Wh~n stiffn~ss~s for beams and columns in substructure arc m.~r 111 hl lhl· resultrng matrix 1s
e
Sn- e Bl•11111s cont~ibute ter~s only to submatrix SM, but columns contribull' 11, nll till' su~n~utnces. Matnx Se augments stiffnesses from previous subst 111r111, 11 prcl~n1111~ 1ry lo reduction . From the modified tridiagonal method , Ntrlln ll'lh1L't1011 formulas for each substructure are
= Sm1 + s,IA TAB = Sou
- SIJA S,J SAIi
s;,. = s,JF + s,1ATAF = s,JF -
s:,.
Sn:
I1
s,1As;;-J sA,..
I \I
+ s,.i\ TM' = Sn: - Sri\ SA) SA/o'
n
Whl'll'
T,1/1
- SA) SAIJ
TA/o'
=
- SA) SA/o'
(~
A~ llw .~L·rics elimination proceeds from one substructure to the nt·xt, thi· 1111111 1 nl 111ws 111.1d <.'ol1111111s of _types A and B remains equal to 211 ThL•H•lorr , 11•d1·'.111~·. d1spl11L'l''.1,'L'nls of .type 11 i.11 subs1r'.1t·turc (' to hL'L'mnc tyrw A 111 ~111 Hlllll lt11l ( I I . I his 1l'qtlll'l'S pl11c1ng 11111(11L'l'Nof' typl' JI 11110 posillonic nl IVI , \ 11111•1 hJs, ( \) llll' itpplll'd IO l'lll'h SllhStllll'lllll' f'lll'II lll'W l 'Olllllhllllllll ll frt1I ,
1
1
461
the next substrucllltl' 11re added to the residual arrays from the preceding substructure. On the other hand, the number of rows and columns of type F Increases from 2 for the first substructure tons for the last substructure. For each 11cw story (except the last), we pick up one more lateral displacement, so that matrices s;F and SJF keep expanding in size. This overlay technique requires 1·omputer core storage for only one substructure and n, framing levels. The forward elimination procedure is completed at the lowest story, where l'Olumn bases are assumed to be fixed. At this stage, we have thens equations of undamped motion (5) In which MFF is a diagonal matrix of lumped masses and AF(t) is the vector of !literal forces. Damped or undamped story displacements may then be found using either the normal-mode method or direct numerical integration from Chapter 4 or 5. After time histories of story displacements DFhave been determined, other Items of interest may be calculated in a backward-substitution procedure. Starting at the lowest level and working upward , we compute joint displacements DA from (6)
For this purpose, it is necessary to have saved back-substitution matrices TAB and 'l~F for each story during the forward elimination process. When applying Eq. (6) recursively, we redefine displacements DA for any particular level f to l above. Of course, the vector D0 become displacements D 8 for the level Is null for the lowest substructure. Time histories of member end-actions for beams may be obtained by placing appropriate terms from DA into a 4 x I member displacement vector l)Mi· Premultiplication of this vector by the 4 x 4 beam stiffness matrix K; yields: (7)
e-
Sym .J
S1t1
Mudl!lnd I I lrlln11111111I Molhod lor Mulllstory Buildings
in which the end-action vector AM; contains a shearing force and a bending moment at each end. For columns, time histories of end-actions are found in a similar manner, but vector DM; contains two terms each from DA, Do, and DF. Also, the 6 x 6 column stiffness matrix K; in Eq. (I) is used in Eq. (7) to produce six member end-actions . Of course, these time-varying end-actions for hcams and columns could be added to any static actions existing at time t = 0. Turning now to three-dimensional multistory tier buildings, we must acrnunt for the rigid lamina existing at each framing level. Figure 10.8 shows ~ubstructure consisting of a rigid lamina, the beams at level A , and the l'Olumns below . Recall that stiffness matrices for x-beams, y-beams, and zl olumns were giwn in Eqs. (9.4-1), (9.4-2), and (9.4-15) and that the last is trnnsl'ormed lo I lit id hndy rnordinates . In addition, formulas for the transformed 11111ss inert in 1111d ~1111 y 11111d 11111trices appear in Eqs. (9 .4 17) and (9 .4- 19) .
e,
462
Substructure Methods /
Sec. 10.4
I
/
I I I/ /
/ / /
1-------t I
I I Level A
x-beam /
--------- z-column
/ /
/ /
Level 8
/
1?.
10.4 PROGRAMS DYMSPF AND DYMSTB
/ / /
- ----,
In this expression , TT is the transpose of the matrix in Eq. (~.4-13~,. and DF,. contains story displacements from levels A and B i~ the last s~x P?s1t1ons. In retrospect, it is interesting to note that the axial const~a~nts in .the ~eai:ris of a multistory plane frame are analogous to those due to th~ ng~d laminae in tier buildings. Recall that each framing level of the pl~ne f~ame 1.n F1? .. 6 has only one translation in the x direction. This implies infinite axial ng1d1t1es for the beams and we can visualize a one-dimensional rigid body at each framing level. As for' the laminae in tier buildings, these unseen constraints in plane frames serve to reduce the number of degrees of freedom in the analytical model.
/
/
/
463
Programs DYMSPF and DYMSTB
/ 1----
In this section we discuss Programs DYMSPF and DYMSTB for dynamic analysis of multistory plane frames and tier buildings. These pro.gram~ use auxiliary storage as well as core storage to analyze two- and three-d1mens1onal . building frames by the modified tridiagonal method: Starting with the multistory plane frame, we give the outline for Program DYMSPF, as follows:
Outline of Program DYMSPF z
Figure 10.8 Typical story framing in tier building.
For tier buildings , assemblage of stiffnesses and the forward elimio11111111 proceed the same as with plane frames. However, the number of rows 11ml columns of types A and B is 3nc because each joint has three unconstrn111r1J displacements. Also, the number of rows and columns of type Fis 3n., dur 1t1 the fact that each lamina has three rigid-body displacements. The overlay 11·1 h nique also works the same as for plane frames, and computer core storugl' 111 I only contain information about one substructure and 3ns rows and colu11111~ Ill matrices of type F Calculation of story displacements in tier buildings and the buck WIii I substitution process follows the sequence given for plane frames. But to oh111111 column end-actions, we must transform story displacements from the rel1•11•m point F to joint j at the top and joint k at the bottom. Therefore, Wl' haw
l. Read and write structural data a. Structural parameters b. Bay widths and story heights 2. Generate story mass and stiffness matrices (by substructures) a. Clear stiffness and mass matrices b. Read and write story mass and member information c. Augment residual stiffness matrices with S1 d. Calculate reduced stiffness matrices by Eqs. (10.3-3) e. Place story information into auxiliary storage f. Shift matrices of type B into locations of type A g. Repeat steps b through f for each story (top-to-bottom) . 3 . Determine frequencies and mode shapes a. Convert eigenvalue problem to standard, symmetnc form b. Calculate eigenvalues and eigenvectors c. Write natural frequencies (cycles per second) d. Transform, normalize, and write modal vectors e . Normalize modal vectors with respect to mass matrix 4 . Read and write dynamic load data n. Dynn111ic.: parnmctcrs .. b. l11itinl conditio11s (stmy displnl'l' lllcnts nnd vcloc1t1cs) c , Appli,•d 1111111m1 ( 1 loil'l'~ 111 lrn11ii11l' ll•vi-l M )
464
Substructure Methods
d. Ground acceleration (in x direction) e. Forcing function (piecewise-linear) 5. Calculate story displacements a. Set up modal damping matrix b. Cal_culate story displacements for each time step c. Wnte and/or plot story displacement-time histories d. Find and write maximum/minimum story displacements 6. Determine. member end-actions (by substructures) a. Retrieve story information from auxiliary storage b. Calculate joint displacements DA c. Cal_culate member end-actions for each time step d. Wnte and/or plot member end-action time histories e. Fi~d and write maximum/minimum member end-actions f. Shift elements of DA into D8 g. Repeat steps a through f for each story (bottom-to-top) ·1:ublc JO . I shows preparation of structural data for multistory plane frnr11 Str uctu~al parameters i~ the table include the number of bays NB and the minih of sto:';~ NS·. For _d1.men~ions of. bay widths and story heights, Prn~IIIIII DYMSJ .1 requ1r~s mm1mal mformation. Each line of that data gives the nurnh N~JM of sequential _occurrences, followed by the repeated dimension of tlw hll w1d1h BW (left-to-right) or the story height SH (top-to-bottom). Each of thr N blocks of substructure data contains the superimposed story mass SMA (t,o p lo-bottom) and member information (left-to-right). For the latter wt· 11r J i-:1.vc only the number NUM of repetitions, followed by the cross-sectionnl 111 of th~ member AX or A Y and its second moment of area ZI with respct·t to th
, ,,, axis.
TABLE IO.I Structural Data for Multistory Plane Frames No. of Lines
Type of Data l'mb 1cm identification St,uc·turnl purumctcrs D11nc•n,,ionN (11) Buy widths (b ) Story heights
I
I
llt'IIII I~
('nh111111N
n
"N'I hl111~~
NUM, BW NUM, SIi
ft
I
SMASS
•
• --
• A~ 11•q11i11•d
Descriptive title NB, NS,E, RII ()
I
Suh, tI Ul' llll C d11tu11 (11) Stoiy lllllSN (h) Ml•111lw1 info,mution )
Items on Data Li
,~
NlJM, AX, /I NlJM, AY, t'I
Sec. 10.4
465
Programs DYMSPF and DYMSTB
Preparation of dynamic load data for multistory plane frames appears in Table 10.2. The dynamic parameters are the same as before, and the initial conditions involve NS story translations and velocities at time t = 0. Under applied actions we see the load parameter IAF, indicating whether story loads exist or not. As shown in Fig. 10.6, each story load is a force An in the x direction at framing level The ground-acceleration and forcing-function data in the table have been discussed before in conjunction with other types of structures.
e.
TABLE 10.2 Dynamic Load Data for Multistory Plane Frames Type of Data Dynamic parameters Initial conditions (a) Condition parameters (b) Story displacements (c) Story velocities Applied actions (a) Load parameter (b) Story loads Ground accelerations (a) Acceleration parameter (b) Acceleration factor Forcing function (a) Function parameter (b) Function ordinates
No. of Lines
Items on Data Lines
I
!SOLVE, NTS, DT, DAMPR
1
IND, INV 00(1), 00(2), ... , DO(NS) VO(l), V0(2), .. . , VO(NS)
a a
a
IAF AF(!), AF(2), ... , AF(NS)
1 I
!GA GAX
I
1 NFO
NFO K, T(K), FO(K)
• As required .
Example 10.3 Figure 10.9 shows a two-bay, ten-story plane building frame having a rectangular layout. Beams in this structure are all steel rolled sections of size W 21 x 55, but the steel columns vary, as follows: (Cl) W 8 x 31, (C2) W 10 x qO, (C3) W 12 x 85, and (C4) W 12 X 106. The length Lis 144 in., and the mass superimposed at each framing level is 0.06 k-s2 /in. For this frame we have two dynamic loading conditions. The first is an atmospheric blast that causes the triangular force An in Fig. 10. lO(a) at each level , except t~~ top (where it is half as much). The second consists of rigid-body ground acceleration D81 in the x direction that has the sawtooth shape in Fig. 10. IO(b). We analyzed the frame for these two load ing conditions using Program DYMSPF, with a damping ratio of 0.10. Resulting time histories of the translation D1,, at the top level are given in Fig. 10. lO(c) . I .oud case (11) l' llll\t'\ 11 mnximum response of 4 .090 in . at time t = 0.480 s, while the 11111xi n111111 1·x1111~H111 lrn lontl 1·nsc (h) is 4 \'i7 111. nt tinw t 2 . 16 s. After those IIIIIXIIIIII, 1hr 1np1111\1·, il1111llll\h h1•1·11111w ot d11111p111p
i;111>atr uutur u Muthocl11
41111
467
Soc. 10.4
1'10111111111 l>YMSPf' and DYMSTB
AF, (k)
o~I\ 5
- - - - - ' - - - t (s) 0.04
0.2
0.4
0.6
0.8
(a)
0.1
Du, (g l o -----+-----1-----~------l--2.0
t (sl
-0.1
C2 10@ l
(b)
4 3 2
t (s) 011 (in.)
0
- 1
ly
- - ~I
-2
Dy,
7,7,:--
X
~~
2L
'}l
Fl1t11n• I0. > Ten 1
~~
Nl ory cxumplc
1hr Program l)YMS1'J1
-3 4 (c)
Figure 1().10
Results for ten-story example: (a) atmospheric blast; (b) ground
m;cclcrntion ; (c) 1r11nslutinn 1),. 1 due to (a) und (h).
488
Sec. 10.5
8ul111tr ucturn M11thncl1
469
Component-Mode Method
The outline of Progrn111 l>YMS I B lrn tin buildmgs 1s s11111l111 111 1h11 Program DYMSPF for plane frames. Ilowcvcr, structural <.lulu in p111 t I 1111 11 I building is given in the sequence \ direction, y direction, and negat1v,· ,II tion. Also, for a typical floor plan, we must provide Boolean dutu ti,, 11111111 joints within a rectangular x-y pattern. That is, an existing joint is i11d1111t ,I h a one in an integer matrix, while a nonexisting joint is indicated hy 11 1, 111 It part 2 of the outline, the program calculates the location of the ccn11·1 111 111 (point c) for each story. Also , the mass-inertia matrix is generated with ll ~I to point F, as shown in Eq. (9 .4-17). In part 4, the dynamic load d11t 11 1, 1111 extensive than for a plane frame. Three types of initial displaL·1·111r111~ 1111 velocities are possible for each lamina. In addition, we handle both , nn y-forces applied at each level, as well as x- and y-components of gm1111d 111 I erations. Therefore, the data must also define two independent pict·cw1,,· 1111 forcing functions (for the x and y directions). If shear cores, bracing, and setbacks are included in the analy11rnl 11111 I of a tier building [6], we need more data than that described ahll\T WII substructures, it is also possible to analyze soil-structure interaction I I I 1111d I calculate inelastic responses [8] of tier buildings.
10.5 COMPONENT-MODE METHOD
The original ideas for the component-mode method are attributed to I 11111 y I I 1lowever, Craig and Bampton [10] also made useful improvements. In thr htlt work, generalized displacements in a substructure consist of a limitL·d 1111111 of vibrational mode shapes and a finite number of nodal displacement, 111 Ill luting boundaries. The component-mode theory produces mass (or dy1111111I coupling between the modal and nodal accelerations in a typical suhsl111d\l Figure 10.11 shows a substructure C that is arbitrarily located within discretized continuum. Nodes of type A are indicated at interior position, . whll those of type Bare at boundary locations. Undamped equations of mot11111 l111 lh substructure may be written as
Me De + Se De
= Ae(t)
(I
whil'h is the same as Eq. (10.2-14). However, the expanded form is
1111w
] [DA] [SAA SABJ [DA] [AA I [MM MAB MBB Ds ( + SsA s,JB = A,, M11A
D11
I
With boundary nodes restrained and zero loads, we can set up 1·1~l·nv1llue problem
e for component-mode method.
2, ... , m" where m, is the number of retained modes. By normalizing the
modal vectors with respect to the mass submatrix MAAl, we can state the relationship (4)
In this expression, the symbol DNe represents a vector of m, normal coordinates, and «l>Ne is the normalized modal matrix. For nonzero boundary displacements (occurring statically), we have
(5)
I
111111
Figure 10.11 Substructure
where "'Iv~ lh
TAB e = -S;leSABt
(6)
Now let us define a transformation matrix Te as 111 wl11l'11 (1)11 is the ungulur frt•qut'ncy of motk I und <1>11 is tlw n111,·,p1111,ll11 111111k sl1111w. Frn this ll11111l·d vihrnt101111l 111111lys1s, tht• 111od11I indl'Xl'' 1111· 1
T( =
[N T,111] 0
I,, ,
(7)
470 Substructure Methods
Sec. 10.6
This operator relates the displacements in vector De to generalized displacemt•nl 111 a vector De, as follows:
De =
[~:l T{~:l =
= TeDe
(Hu) 8
De= Te.De
(Hh)
.Substitution of Eqs (8) int0 Eq (1) d · · · yields · · an premult1phcat10n of the result by 'I J
= Ae(t)
1\1NB] [~NJ + Moo Do e
[ro;,,O
_o
S88
J[DN] = [AN] D e A e 8
(10)
8
= ct>iMM cf>N = I m, MNB = ct>iMAA TAB + ct>XMAB = MIN
MMN
T1 MAB + MBA TAB + T1 MAA TAB
( I I ft I ( 11 h)
(lld
Also , the stiffness submatrices become
SNN
= «1>isM cf>N = ro m,2
sNo = ct>XsAA TAo + «1>isAB = o = siN Son = Sno + T Xn SAB und the applied-action subvectors are
( I '11) ( I h)
( I l)
AN= «1>rAA Ao
(I In)
= Ao + TXoAA
= L Ae(t)
(=I
(J 5)
(=I
where n, is the number of substructures . Of course, the substructure matrices in Eqs. (15) must be expanded with zeros to become conformable for addition. Now the equations of motion in Eq. (14) can be solved for the displacements in vector D by either the normal-mode method or direct numerical integration. Then displacements in vector DA e for each substructure are found using Eq. (5) as a back-substitution expression. In the next section we shall apply the component-mode method to trusses, for which some additional complications arise.
M I) I S llw ll~~l·111hll·d 11111ldn·~
In earlier chapters of this book, we considered only axial deformations for dynamic analyses of trusses. However, inertial and body forces also occur along the members in transverse directions, which cause flexural deformations as well. These influences are most significant in trusses composed of only a few members. We have found that the best approach for analyzing such structures is the component-mode method [ 11, 12]. By this technique, a member with both axial and flexural deformations constitutes a substructure. Treating the member as a simply supported beam, we include a limited number of its exact vibrational mode shapes as flexural displacement shape functions. Figure 10.12 shows a prismatic plane truss member i with local (primed) and global (unprimed) axes. We shall handle such a member as a specialized finite element acting as a substructure within a plane truss . Flexural displacement shape functions in member i are taken to be the vibrational mode shapes of a simply supported beam [13]. These mode shapes will be superimposed on linear displacement shape functions due to translations at joints (or nodes) j and k. Thus, the generic displacements u' and v' in the x' and y' directions may be written in terms of modal and nodal displacements as
u: = f Df
(1)
or
ur
.M. '.
t•q1111111111
n, -
A(t)
Me
( I lh)
:;• El(l, ( l )I"!, the symbol Im, den~tes an identity matrix of order m,. Also, .., in
-
ns -
I
10.6 COMPONENT-MODE METHOD FOR TRUSSES
The mass submatrices in this equation (without the subscript e) are
Moo = Moo +
=
471
(1))
In expanded form, this equation is
[ ~,, MoN
-
M
t• J
~:;~~ !:t !~:on~;,;~~:o~e~;ti~:::i~s DNe in the first part and D e in the scrnnd
MeDe + SeDe
Component-Mode Method for Trusses
I) Jill
A(t}
( Ill
(2) In this equation the symbols DA; and D8; represent the displacement vectors
1>At -=
DA, 0A2
I
D,11
D111
• • •
I
Ds,I Ds2
(3)
DJo 1)/,,1
I
472
Substructure Methods
Sec. 10.6
473
Component-Mode Method for Trusses y'
I D~, D~, __... j
v'
tr-----+.::...._....:.....;_ t- _ _0_ _ _ _ t u'
-
k
- -- -
x
,
~
~ -- - -- - L~-----~ (a)
1~
V
(e)
(b)
1
(f) (c)
Figure 10.12 Plane truss member with local and global axes.
'l'l-1111s in vector DA; are the amplitudes of a limited number m, of retained 11111111111 111odc shapes for a simple beam, as depicted in Figs. 10.13(b)-(d). In add1 111 111 the terms in vector D 8; are the four translations D 81 through D 84 at joints 1111111 k, shown in Figs. 10.12 and 10.13(a). Matrix fin Eq. (1) contains displacement shape functions, and its 111 I s11h111ntrix is
(d)
Figure 10.13
= = [
f A
0 0 0 sin 1r{ sin 21r{ sin 31r{
.. ·] ...
Displacement shape functions for plane truss member.
in which the linear differential operator d; is
111 which the di mcnsionless coordinate is g = x ' IL. Appearing in the scco11d 111w 111 this s11hmulrix are the natural mode shapes for vibrations in a simply .,Ill' po, tL·d prismatic beam. To keep them dimensionless, these mode shapcs 1111· 11111 1111111111lii'.l'd with respect lo the mass of the member. On the other h1111d, th Nl'l 'OIHI sul>lllatrix in f has the form
0 I
~
g 0
OJ g
II,
cl, I'
111 ,
II,, I ,
1
(7)
L 2 ag2 ;
The first submatrix in B; has the resulting form t ~,
'lht•sl' li11l'11 r shupl· !'unctions result froin unit displuccments of' 1>;11 th ro11ph /J 111
IINl11dic111t•d i11 Figs. 10. Uk) nnd (!') . S1111i11 diNplllL'L'llll' llt n•l11tio11ships f'tu· lhi s L'lt•111t•11t 11111y lw st11tl•d nil
.o
y' a2
0 0 0 I 7T2 [ 8 Ai = d;f" = Y L2 sin 1rf 4 sin 21rf 9 sin 31r(
.. .l
(8)
und the second is I'll/
cl, r,,
,
I
I
o
o
I
llllllll,
('))
474
Substructure Methods
Chap. 10
Sec. 10.6
475
Component-Mode Method for Trusses
The stiffness matrix Sf in local directions for member i has the known frn 11 1
L
s: =
=
BrE B; dV
[soM snB oJ
2 ,
(tm
_
_
Iv ,,r,r, dY 1
M,m -
pAL 0
6 [
i
~
In this equation the stress-strain matrix E is (I f l
where 12 is an identity matrix of order 2. The first submatrix in
-sMi --
[~
f BT
7T4EJ, Ai E BA; dV = 2L3
v
_ (pAL -2-w"',
2 )
-
Sf
becontl'H
Sy1
16
0
_
;
SBBi -
r BB;E Ba; dV = -L )v 1
[
(II)
1
0
Mf = ( pf f dV = Jv
Mf
[M~ M~J MBA
2
(17) 1 0
2 ;
=
[A~J
(18)
Aa ;
A,; -
=
[by'bx']
(19)
has the form
i
~Tb'(t) dL -
l
f];b'(t) dL =
1[:~f::]~,
dL
and the second is
An;
=
l [g ~b;i::1
( Ill
Mos ;
is
thy'
dL
For local (or member) axes , the undamped equations of motion for small displacements of the plane truss member are
As before, the3mbol I,,,, represents an identity matrix of order submatrices in Mf are
-
f
~
v p'I' f 11 r/V - pAL [ 7T'
() '
1111d
(21)
i
Mf Df + Sf Df = Af (t)
T\it'ill/I
(20)
0 0 ;
which is the same as matrix K' in Eq. (3.5-25) . . Next, the consistent mass matrix Mf (in local directions) for memhl't 1 111 obtamed from 1
Af (t)
Sym.]
1 0 0 -1 0
r r1 b'Ct) dL
Ji
b'(t) The first subvector in
0
The first submatrix in
=
(I 'I
EA
0
Here the vector b' (t) contains body forces (per unit length) in the x' and y' directions, as follows:
and the second submatrix is found to be
,
Sym.l
which is the same as matrix M' in Eq. (3.5-32). Also, equivalent modal and nodal loads in local directions are calculated as
A; Ct)
81
2
I
0
l
1/2
()
1/2
I/ l
()
1/J
111,. < >11 111
By rotation of axes for the parts of type B, Eq. (22) may be transformed to global (or structural) axes to become
M;D; (If,)
(22)
+ S;D; = A;(t)
(23)
The displacemcnls D/ and.accelerations fi; in Eq. (22)_are related to the corresponding vectors 1)1 ond 1)1 in Eq . (23) t,y the expressions
n;
ft,1>1
f>;
ft,n,
l111h1•1111 11·l11ti1111 -.hlpH tlu· 1011111111111111,mM 111111Hl11111111lHlll 11111111 ,i 111
(24)
476
Substructure Methods
Chap, 10
( 1~)
Note from the form of matrix R; that only the nodal displacements D transformed fr~m structural to member directions. (Displacements D 81 th;.:11;":, 'f?n4 are shown m the x and y directions in Fig. 10.12.) Submatn·x R'811n . · E q. ( ~,) lS
Rn;=
[R
OJ
0 R;
( '" I
where
= [ _ Cx
Cy] Cx i
Cy
(.! II
Dire~tio~ cosines in the rotation matrix R; were defined in E s (3 5-24) 11 substitut~ng Eqs. (24) into Eq. (22) and premultiplying the J~t~er by R} · w~ convert it to Eq. (23). In that equation we have the following matrix procl:H 1~ -
S;
M
l
A;(t)
l
( '')I
l
( 1111
c; =
EA L
c2y
CxCy
[ -c;
c2X
-cxCy
s:]
( 111
-::12] Cx/3 • . •
A
- R" TA'
Ill -
/Ji
(I - f)(c,.bx· - cyby,) I + c,by•) • ' dl 1. g(c,bx· - Cy by,) [ {(cyb, • + C\ bv,) . I
-1
01 -
I
(I - g)(cyb,,
( I\)
or
Noll· tlu.11 submutrix. M11111 M f1111 is invariant with rotution axes I Afl~r tlw rotat,011 nxcs lntnsf'ornrntion f'or cuch nw111h<.1r is ~·0111plrt I I I(' l'ljlllll IOIIS of Ill()( ion for ( ht• whok• H( 1111.'ltll'l' 11u1y lw USSl'lllhlt•:, (O t11il11l1'.'
or
Mi>
I
Sh
i=I
A(t)
=
m
L A;(t)
(35)
i=I
where m is the number of members. The equations of motion for the structure can now be solved by the normal-mode method or by direct numerical integration. Either approach may be preceded by a vibrational analysis, from which a damping matrix can be established for the whole structure, as before. Extension of the theory for plane trusses to the analysis of space trusses is straightforward and appears in Ref. 12. The primary change to recognize in three dimensions is that each member has two principal planes of bending, defined as the x '-y' plane and the x '-z' plane. But the flexural mode shapes in each plane are still the sine functions sin 7Tg, sin 27Tg, sin 37T{, and so on. Definition of a principal plane in space may be aided by using a third point p in addition to points j and k, as explained in Sec. 6.4.
( 'HI
= RTMiR'. = RT A; (t) l
m
s = LS;
10.7 PROGRAMS COMOPT AND COMOST
'T- ' = R; Sf R;
The results of these operations are as follows:
Son,= R.];,Snn,R.n,
477
Programs COMOPT and COMOST
in which
R; =
R;
Sec. 10.7
A(()
( 11)
Now we briefly discuss Programs COMOPT and COMOST for dynamic analyses of plane and space trusses by the component-mode method of the preceding section. These programs calculate responses to initial conditions and piecewiselinear forcing functions that may be either applied actions or translational ground accelerations. Using the normal-mode method, we determine axial forces and bending moments in the members, as well as time histories of modal and nodal responses. Example 10.4 Figure 10.14 shows a plane truss having only two members, with a step force P applied in the x direction at the quarter point of member 1. Both members are prismatic and have the same values of p, E, A, and /,. Although realistic sizes were used, we take the dimensionless parameter AL2 / I, to be 250 for this problem. Dimensionless frequencies for the truss solution without flexure are w i = 1. 000 and wi' = 3.023, which have been normalized by dividing them by the first. Table 10.3 contains twelve such dimensionless frequencies for the structure when five vibrational modes are included for each member. Also, Fig. 10.15 shows the corresponding mode shapes of the truss for the first four modes. If members 2 and 1 are taken separately as simply supported beams, their fundamental frequencies become only slightly more than those for modes 1 and 2 in Table 1. Figure l O.16 gives responses of the dimensionless displacement Di, plotted against the dimensionless time r *. Herc the dimensionless displacement is obtained by dividing the dynamic value by the static value for the same load . On the other hand, the dimensionless time , ,.,. is time r divided by the period of the first mode for the truss solution without lk•x11n•. The curves in H11, 10. l
478
Substructure Methods
Chap. 10
Sec. 10.7
479
Programs COMOPT and COMOST
y
\ \ \ \ \ \
\ \ \ 3L
2
(a)
(b)
(c)
(d)
----x
~I.-3L-.I Figure 10.14
Plane truss with two members.
TABLE 10.3 Dimensionless Frequencies for Two-Member Truss Mode
I 2
w lf<
Mode
w lf<
0 .204
7
J
0.117 0.7%
8
3.034 3. 163
11
9
I , I 8()
1.628
~
I .MH J II I
10 11 l :l
~. 146 t ~·, ,
h
H )ti/
Figure IO.IS Mode shapes for two-member plane truss: (a) mode 1; (b) mode 2; (c) mode 3; (d) mode 4.
400
•;11lin11111ill1111 MulhocfN
Sue. 10, /
1110111111111 C:O MOI' I 1111d COMOS1'
y
2.5
I
4
D;t-~-t
,
2.0
p
T l
I
3
5
L
1.5
6--------....:o----...,__,..;u-----s
'
1.0
I
I
1.0
Figure 10.16 Displacement time histories for D
Figure 10.17
I
Plane truss with np panels.
I
;,.o
1.5
6
'/.
I
0.5
0
4
R
t. 1.6
flexure included in the members. The first three flexural modes in membor I ll'lld 111 dominate the response because the load is applied directly to that member. Addi118 1111,,h 4 and 5 to each member does not improve the accuracy of the response very 111111 Ii Example 10.5
1.5
1.4
1.3
Figure I0. I7 illustrates a truss with an indefinite number of members, having u Nh•p 11111 1 P applied in they direction at the center of member 1. By increasing the numbL·r ol 1111111 I frn111 I to 11,,, we can study the effect of this parameter upon the dynamic rcsp1111~1 / I of joint I in they direction. All members of this truss have the same values for 11, I . I nnd /,; and we take the dimensionless parameter AL2 / I, to be 2250. Por this example the truss solutions without flexure will be compml·d uv.11111-1 solul ions with one flexural mode included for each member. In Fig. I0. I 8 w1• pl111 1111 rulio R of the maximum response of D 2 for the model with flexure to th at wilho111 lln 1111 versus the number of panels n,,. The ratio R approaches unity as the nurnlwr ol p11111 1incrcascs. Thus, the flexural deformations in the members have little cffcl'I 011 1hr 1111111 rusponscs when the number of mem bers becomes large.
1.2
0.9
l•'l uun• J0. 18
Displacement ratio versus number of panels.
481
482
!h1h11t1uut11111 M11tlu1d1
R
R NC
S
I. 11011s, 13 . M. , "A Fronlnl Solul11111 1'1111{111111," /111 . J . Numl!t. Method.,· ll'IIH • V11I No. I, 1970, pp. 5-32.
i. llinton, E., and Owen, D. R. J., F/11/tl' Weme11t Programming, Act1dl•111h 1111 London, 1977 . \ , Wcuvcr, W. , Jr., and Yoshida, D. M., "The Eigenvalue Problem for B1111d1•d 1\11111 I ccs," J . Comp. Struct., Vol. 1, No. 4, 1971, pp. 651-664.
Notation
'1 . Weaver, W., Jr. , and Nelson, M. F., "Three-DimensionaJ Analysis of Ti1•1 ll11tltl lngs," ASCE J. Struct. Div., Vol. 92, No. ST6, 1966, pp. 385-404. ~. Weuver, W., Jr. , Nelson, M. F., and Manning, T. A., "Dynamics of Tier ll11lldi111-' ASCJ~· J . Eng. Mech. Div., Vol. 94, No. EM6, 1968, pp. 1455- 1474. (1 . Wcuver, W. , Jr., Brandow, G. E., and Manning, T. A., "Tier Buildings with ~lu 11 Cores, Bracing, and Setbacks," J. Comp. Struct., Vol. I , Nos. 1/2, 1'1/1. I'll 57 84. 7. Wcnvcr, W. , Jr., Brandow , G. E ., and Hoeg, K., "Three-Dimensioruil ~1111 Slructure Response to Earthquakes," Bull. Seismal. Soc . Am., Vol. 63, No. I, l 1J/ I pp. 1041- 1056. H. Wcuver, W. , Jr., and Bockholt, J. L., "Inelastic Dynamic Analysis of'Tit11 ll111hl lugs," .I. Comp. Struct., Vol. 4 , No. 3, 1974, pp. 627-645. 1
1 ll11rly , W., C. , "Dynamic Analysis of Structural Systems Using Co111p11111111 Modes," A/AA J., Vol. 3, No. 4, 1965, pp. 678-685.
10, ( 'ndg, R. R., Jr. , and Bampton, M. C. C., "Coupling of Substructures for Dy1111111h Annlysis," A/AA J., Vol. 6, No. 7, 1968, pp. 1313-1319. 11 , Wl·uvcr, W. , Jr. , and Loh, C. L., "Dynamics of Trusses by Componc111 Mn1h Mclhod," ASCE J. Struct. Eng., Vol. 111, No. 12, 1985, pp. 2526- 2575. 12. Loh , C . L., "Dynamics of Trusses by Component-Mode Method," Ph .Jl ,/11 .1·1•rratio11, Department of Civil Engineering, Stanford University, May I 98'1 11 , Timoshenko, S. P., Young, D. H., and Weaver, W., Jr., Vibration Proli/1•1111 111 l:11t,1i11t•t•rl11g, 4th cd., Wiley, New York, 1974.
1. MATRICES AND VECTORS
Symbol 0 A B C
D E F G H I
J K L
M
Definition Null matrix Action vector (also coefficient matrix) Strain-displacement matrix Strain-stress matrix (also damping matrix and constraint matrix) Displacement vector Stress-strain matrix Flexibility matrix Constraint matrix Characteristic matrix (also Householder matrix) Identity matrix Jacobian matrix Element stiffness matrix Lower triangular matrix Mass matrix (also concentrated moments) Concentrated force vector
483
484
Notation
Symbol Q
R
s T
u V
X y
z A
b d
e f
j k p q u
Definition Factor in QR algorithm Rotation matrix (also factor in QR algorithm) Stiffness matrix Transformation matrix (also tridiagonal matrix) Upper triangular matrix Eigenvector matrix Vector of unknowns Vector of unknowns Vector of unknowns Spectral matrix Eigenvector matrix Body force vector for element Linear differential operator for strain displacement relationships Unit vector Interpolation function matrix Unit vector Unit vector Unit vector Nodal load vector for element Nodal displacement vector for elc11w111 Generic displacement vector for elc111rnt
2. SUBSCRIPTS FOR MATRICES AND VECTORS
Symbol A 8 F /, M
N I' N II
Definition Nodal displacements eliminated (<1/.w1 attached) Nodal displacements retained (al.1·0 body and boundary) Free (also floor or fram ing leve l) Lumped Member Nornrn l rnordinntes J>ri11l'ip11I rnordin111t•s
485
Notation
Symbol b
d e
f g j k
C m n p q I'
s X
y z
Definition Body Damped Element Forced Ground Index Index Index Index Number Number Working (or reference) point Working (or reference) point Radial direction Structure x direction y direction z direction
3. SIMPLE VARIABLES
Symbol A
B C D
E G I J L M
p
R 'f'
l{~•Nll llllll'd
(}
l111tlul
\'
Definition Area Constant Constant Displacement Young's modulus of elasticity Shearing modulus of elasticity Moment of inertia (second moment of area) Polar moment of inertia Length Moment Force Radius Period Strnin l'llorgy l 1111r11t 111l l'lll'I HY
486
Notation
Symbol
w
X a b C
d
e
f h j k
e m n p
q r
s t
u V
w X
y z
Definition Work Generalized action Constant (also acceleration) Constant Constant (also damping constant) Constant (also displacement) Base of natural logarithm Interpolation function (also frequency , cycles/ sec) Thickness Index for ... (also v=T) Index for . . . Index for ... (also spring constanl) Index for ... (also length) Number of ... (also mass) Number of degrees of freedom (also damping parameter) Action at element node Displacement of element node Radius (also cylindrical coordinate) Segment length Time Translation in x direction Translation in y direction Translation in z direction Cartesian coordinate Cartesian coordinate Cartesian coordinate (also cylindric11l coordinate)
487
Notation
Symbol
Definition
a
Rotation or angle (also Hilber constant) Magnification factor (also Newmark constant) Shearing strain (also damping ratio and Newmark constant) Increment Normal strain Dimensionless coordinate Dimensionless coordinate Rotation or angle (also Wilson constant) Direction cosine (also eigenvalue) Frequency coefficient Poisson's ratio Dimensionless coordinate 3.1416 .... Mass density Normal stress Shearing stress Curvature Twist (dOx/ dx) Angular frequency (also angular velocity)
/3 )'
5 E
( rJ ()
A µ, V
t 'TT
p
a T
!/; w
5. PROGRAM NOTATION
Symbol AO( ) AOP AB( ) AE( ) AF( )
4 . GREEK LETTERS
Symbol ~
Definition
...,.
Increment Summation
FU1K·lio11 or ,nod<.· A11~11l111 I 11•q111•111 y
u
ALPHA AM( ) AN( )
AR( ) AS(
AX(
)
Definition Initial accelerations of nodes Initial acceleration of moving load Actions applied to bodies Actions at element nodes Actions at free nodes (also actions at floors or framing levels) Hilber parameter a Actions at ends of members Actions in normal coordinates Support reactions Actions at structural nodes Cross Sl'l'lionnl nn·as A,
488
Symbol BETA BI( ) BL1 , BL2,. BM( ) BS1,BS2, .. . BV1,BV2, . . . BW CME( , ) CMS ( , ) CV( ) CX,CY, CZ
DO( ) DAMPR
DBO( ) DE( DF(
) )
DM( ) DN( ) DR( ) DS( ) DT E E l ,E2, . EL( )
F( ) FO( ) G GAMMA G AX, GAY, G AZ II
l,J, K, L IA(' IAF
489
Notutltm
Ocfi11ltlo11
Nl•w111111•k puramclcr f3 Body i11crtit1s Jntcnsilics of line loads Body masses Intensities of surface loads Intensities of volume loads Bay width Consistent mass matrix for clc111t'11I Consistent mass matrix for strnt·l111, Characteristic values (eigenvalul'N ) Direction cosines c.,, cy, and c, Initial displacements of nodes Damping ratio Initial displacements of bodies Displacements of element nodes Displacements of free nodes (also displacements at floors o r 1'1·1111111111 levels) Displacements at ends of membt•rN Displacements in normal coorcli,11111·11 Displacements of restraints Displacements of structural nodes Duration of time step /:l.t Elasticity modulus Elasticity constants Element lengths Frequencies (cps) Function ordinates Shearing modulus Newmark parameter y Ground acceleration faclors for 1, I' , and z directions Thickness Indexes Indicalor for imposing axial co11s1111l111 ~ lndicalo r for actions al lloors
Symbol ID( ) IBO( ) !G A IML IN( ), JN( IPL IPS IR,IC IRO !SOLVE
IWR JB( ' ) JNO( ) LN NB NBID NBIV NC NDF NE NEL NEN NEO NES NEV NFO NJ NLB NLN NLS NMODES NN NNA
)
Definition Displacement indexes Element numbers for output of stresses Indicator for ground accelerations Indicator for moving load Indexes for nodes of elements Indicator for plotting Indicator for plane stress or plane strain Row and column indexes Indicator for eliminating rotations Indicator for method of solution Indicator for writing Body-node numbers Node numbers for output of displacements Loading number Number of bodies (also number of bays) Number of bodies with initial displacements Number of bodies with initial velocities Number of columns Number of degrees of freedom Number of elements Number of elements with line loads Number of element nodes Number of elements for output Number of elements with surface loads Number of elements with volume loads Number of function ordinates Number of joints (or nodes) on a body Number of loaded bodies Number of loaded nodes Number of loading systems Number of modes Number of nodes Number of nodes of type A
490
Symbol NNI> NNF NNII > NNIV NNO NNR NRL( NRN NS N'l'S NUM OM EGA( p
)
Pl II (
PR )
R(
)
I{( '(
RI 10 SH( SIi SS( SX ,SY, . T( ) TIME
VO( ) VOP VHO( ) VN(
)
VS(
)
X( ),Y( ), Z( ) XC'( ),YC( ), ZC( ) XC'.I( ),YCJ( ), ZCJ( ) XC'K( ),Y CK( X I( ) YI(
). ZI(
)
), ZCK(
)
l)()llnlllon N11111lw1 of nodal displucc11K•111s N111uhcr of nodes of typu F Number of nodes with initial di spl acements Number of nodes with initial wlrn llh Number of nodes for output Number of nodal restraints Nodal restraint list Number of restrained nodes Number of stories Number of time steps Number of repetitions Angular frequencies w Moving load Eigenvectors ct> (mode shapes) Poisson's ratio Rotation matrix Radii of gyration of rigid bodies with respect to centers of mass Mass density p Element stiffness matrix Story height Structural stiffness matrix Stresses Times Time Initial velocities of nodes Initial velocity of moving load Initial velocities of bodies Velocities in normal coordinates Velocities of structural nodes Nodal coordinates Coordinates of point c Components of offset vectors Components of offset vectors Torsion c:onslanls /, of cross scc:lions Sl•t·wul 111011wnts of arcu /1, 1111d I of l ' I OHM ~1·1·1io11s
General References
TEXTBOOKS ON STRUCTURAL DYNAMICS (CHRONOLOGICAL ORDER)
1. Rogers, G. L. , Dynamics of Framed Structures, Wiley, New York, 1959. 2. Norris, C. H. , et al. , Structural Design for Dynamic loads, McGraw-Hill, New York, 1959. 3. Biggs, 1. M. ,Introduction to Structural Dynamics, McGraw-Hill, New York , 1964. 4 . Hurty, W. C., and Rubinstein , M. F., Dynamics of Structures, Prentice-Hall, Englewood Cliffs, N. J., 1964.
5. Lin, Y. K., Probabilistic Theory ofStructural Dynamics, McGraw-Hill, New York, 1967.
6. Przemieniecki, J. S., Theory of Matrix Structural Analysis, McGraw-Hill, New York, 1968.
7. Rubinstein, M. F., Structural Systems-Statics, Dynamics, and Stability, Prentice· Hall, Englewood Cliffs, N. J., 1970. 8. Fryba, L., Vibration of Solids and Structures under Moving loads, Noordhoff, Groningen , The Netherlands, 1972. 9. Fertis, D. G. , Dynamics and Vibrations of Structures, Wiley, New York, 1973 . 10. Clough, R. W. , and Penzien, J. , Dynamics ofStructures, McGraw-Hill, New York , 1975. 11. Belytschko, T., Osias , J. R ., and Marca!, P. V., Finite Element Analysis of Transient Nonlinear S1r1U'/11ml ll<•ltGvior , ASME, AMD , Vol. 14, 1975.
12. Bathe, K . J,,
1111d WIINon, E. L ., Numerical Methods in Finite Element Analysis, Prcnlicc-llnll , I111plnv1111cl C'lilTN, N. J., 1976 .
491
492
{1111111r11l llnftunm 111 13. Blevins, R. I)., Floll' l11rl11, ,•rl I l/1111tlr111.,·, V1111 Nostrnnd lfrlnliold , NI'\\ , 1 1977. 14. Simu, E., and Scanlan, R. II ., Wind HJ)i•ct.1· 011 Strt1ct11r,•s, Wiley, New Y111 ~, I •J 15. Meirovitch , L., Computational Methods in Structural Dy11a111ic·.1·, Sljlliufl Noordhoff, Alphen aan den Rijn, The Netherlands, 1980. 16. Craig, R. R ., Structural Dynamics, Wiley, New York, 198 1. 17. Paz, M. , Structural Dynamics, 2nd ed., Van Nostrand Reinhold , New Ylll ~. l'Jk
TEXTBOOKS ON VIBRATIONS (CHRONOLOGICAL ORDER)
1. Rayleigh, J. W. S., The Theory of Sound, Dover, New York, 1945. 2. Den Hartog, J.P., Mechanical Vibrations, 4th ed. , McGraw-Hill, New YOI~, l•J~h 3. Myklestad , N. 0., Fundamentals of Vibration Analysis, McGraw-Hill, Nm• , 111h 1956. 4 . Jacobsen, L. S. , and Ayre, R. S., Engineering Vibrations, McGraw-lltll , N York, 1958. 5. Bishop, R . E. D. , and Johnson, D. C., The Mechanics of Vibration, C11111l11hl11 University Press, London, 1960. 6. Tong, K. N., Theory of Mechanical Vibration, Wiley, New York, 1960 . 7. Church, A.H. , Mechanical Vibrations, 2nd ed. , Wiley, New York, 19(,l 8. Crandall , S. H., and Mark, W. D., Random Vibration in Mechanical S1•1t, 111 Academic Press, New York, 1963. 9. Bishop, R. E. D., Gladwell, G. M. L. , and Michaelson, S., The Matri, A1111/\ 11 of Vibra,ion , Cambridge University Press, London, 1965. 10. Chen, Y., Vibrations: Theoretical Methods, Addison-Wesley, Reading, ~111 ~ 1966. 11. Anderson, R. A., Fundamentals of Vibrations, Macmillan, New York, I% I 12. Vernon, J.B ., Linear Vibration Theory, Wiley, New York, 1967. 11. Vierck, R. K., Vibration Analysis, International Textbook, Scranton, Pu. , J 11(, I 14 . llaberman, C. M. , Vibration Analysis, Charles E . Merrill, Columbus, Ohru , l!JriN 15. Thomson, W. T., Theory of Vibration with Applications, Prentice-Hall , Englrn 11, ut Cliffs, N. J., 1972. l
17. Meirovich, L. , Elements of Vibration Analysis, McGraw-Hi ll , New York, Jilt, IH. Newland, D. E., An Introduction to Random Vibrations and Spectral 1\11,1/\ 111 Longmans, London, 1975. I9. Tse , F S., Morse, I. E., and Hinkle, R. T. , Mechanical Vibrations 'J'h,·1111• Applications , 2nd ed., Allyn and Bacon, Boston, 1978.
,111
I
493
TEXTBOOKS ON FINITE ELEMENTS (CHRONOLOGICAL ORDER)
1. Przemieniecki, J. S. , Theory of Matrix Structural Analysis, McGraw-Hill, New York, 1968. 2. Desai, c. s., and Abel, J. F., Introduction to the Finite Element Method, Van Nostrand Reinhold, New York, 1972. 3. Oden, J. T. , Finite Elements of Nonlinear Continua, McGraw-Hill, New York, 1972. 4. Martin, H. C. , and Carey, G . F., Introduction to Finite Element Analysis, McGrawHiU, New York, 1973. 5. Norrie, D. H. , and de Vries, G., The Finite Element Method, Academic Press, New York, 1973. 6. Strang, G. , and Fix, G. J., An Analysis of the Finite Element Method , Prentice-Hall, Englewood Cliffs, N . J., 1973. . 7. Gallagher, R. H., Finite Element Analysis Fundamentals, Prentice-Hall, Englewood Cliffs, N. J. , 1975. 8. Bathe, K. J., and Wilson, E. L., Numerical Methods in Finite Element Analysis, Prentice-Hall, Englewood Cliffs, N. J. , 1976. 9. Hinton, E. , and Owen, D.R. J. , Finite Element Programming, Academic Press, London, I 977. 10. Desai, C. S . , Elementary Finite Element Method, Prentice-Hall, Englewood Cliffs, N. J., 1979. 11. Cheung, Y. K., and Yeo, M. F. , A Practicallntroduction to Finite Element Analysis, Pitman, London, 1979. 12. Hinton, E., and Owen, D.R. J. , An Introduction to Finite Element Computations, Pineridge Press, Swansea, Wales (United Kingdom), 1979. 13. Owen, D. R. J. , and Hinton, E., Finite Elements in Plasticity, Pineridge Press, Swansea Wales (United Kingdom), 1980. 14. Cook, R.' D., Concepts and Applications ofFinite Element Analysis, 2nd ed. , Wiley, New York, 1981. 15. Becker, E. B., et al., Finite Elements (five volumes) , Prentice-Hall, Englewood Cliffs, N. J., 1981-1984. 16. Bathe, K. J., Finite Element Procedures in .Engineering Analysis, Prentice-Hall, Englewood Cliffs, N. J. , 1982. . 17. Huebner, K. H. , The Finite Element Method for Engineers, 2nd ed., Wiley, New York, 1983. 18. Weaver, w., Jr., and Johnston, P. R. , Finite Elements for Structural Analysis, Prentice-Hall, Englewood Cliffs, N. J., 1984. 19. Segerlind , L. J., Applied Finite Element Analysis, 2nd ed., Wiley, New ~ork, ~985. 20. Zienkiewicz, O. C . , The Finite Element Method, 4th ed., McGraw-Hill , Maidenhead, Berk shill', Fnj.\lnnd, 1987.
494
General Referenc11
TEXTBOOKS ON MATRIX ANALYSIS OF STRUCTURES (CHRONOLOGICAL ORDER)
1. Laursen, H. I., Matrix Analysis of Structures, McGraw-Hill, New York, 1966. 2. Martin, H. C., Introduction to Matrix Methods of Structural Analysis, McGruw Hill , New York, 1966. 3. Rubinstein, M. F, Matrix Computer Analysis of Structures, Prentice-Hall, 1•11 glewood Cliffs, N. J. , 1966. 4. Hall, A. S., and Woodhead, R. W., Frame Analysis, 2nd ed. , Wiley, New York 1967. 5. Willems, N. , and Lucas , W. M. , Jr., Matrix Analysis for Structural Engin1'1'11 Prentice-Hall, Englewood Cliffs , N. J. , 1968. 6. Beaufait, F. W. , Rowan, W. H. , Jr., Hoadley, P. G., and Hackett, R. M. , Co11111111,, Methods of Structural Analysis, Prentice-Hall, Englewood Cliffs, N . J ., 1970 7. Rubinstein, M. F. , Structural Systems-Statics, Dynamics, and Stability, Prc11ll1 1 Hall, Englewood Cliffs, N. J., 1970. 8. Wang, C. K. , Matrix Methods of Structural Analysis, 2nd ed., International '1'1 book, Scranton, Pa. , 1970. 9. Meek, J. L., Matrix Structural Analysis, McGraw-Hill, New York, 1971. 10. Kardestuncer, H., Elementary Matrix Analysis of Structures, McGraw-Hill , Nr w York, 1974. 11. Vanderbilt, M. D. , Matrix Structural Analysis, Quantum, New York, 1974 , 12. McGuire, W., and Gallagher, R.H., Matrix Structural Analysis, Wiley, New Y111k 1979.
13 . Weaver, W., Jr., and Gere, J. M ., Matrix Analysis of Framed Structures, 2nd Id Van Nostrand Reinhold, New York, 1980. 14. Meyers, V. J., Matrix Analysis of Structures, Harper and Row, New York , 1'1"1 15. Holzer, S. H., Computer Analysis of Structures, Elsevier, New York, 19H~
Appendix A Systems of Units and Material Properties
A.1 SYSTEMS OF UNITS
The two most commonly used systems of units are the International System (SJ units) and the United States Customary (US units). The first of these is called an absolute system because the fundamental quantity of mass is independent of where it is measured . On the other hand, the US system has force as a fundamental quantity. It is referred to as a gravitational system because the unit of force is defined as the weight of a certain mass, which varies with location on Earth. In the SI system, the three fundamental units required for structural dynamics are mass (kilogram), length (meter) , and time (second). Corresponding lo mass is a derived force called a newton, which is defined as the force needed lo accelerate one kilogram by the amount one meter per second squared. Thus , we have 1 N = l kg · m/s 2 which is based on Newton's second law that force = mass x acceleration. In the US system, we use force (pound), length (foot) , and time (second). (Note that the unit of time is the same for both systems.) Corresponding to force IN u derived mass, which carries the name slug. This quantity is defined as the 11111ss that will be n<.·cclcrulcd one foot per second squared when subjected to a lorcc of one po1111d I ll'11t•t·,
lhlll t'Olllt'N I 111111 lltt 111111111111 1111111N
lcH('l'/ncn•krnlion ,
496
Systems of Units and Material Properties
Table A. I presents conversion factors for calculating quantities in SI unit ~ from those in US units. The factors are given to four significant figures, whid1 usually exceeds the accuracy of the numbers to be converted. Note that stress i~ defined in SI units as the pascal. That is, 1 Pa= l N /m 2 TABLE A.I
Conversion of US Units to SI Units
Quantity
US Units
Length Force Moment Stress Mass
inch (in.) kilopound (kip or k) kip-inch (k-in.) kip/inch2 (k/in. 2 or ksi) kip-sec2/inch (k-s2/ in.)
x Factor 2.540 4.448 1.130 6.895 1.751
X
10- 2
10- 1 103 X 102 X X
Sec. A.2
Material Properties
(GN/m 2 or GPa) , acceleration ii in meters per second squared (m/s2), and so on. [Note that the force kilonewton corresponds to the mass megagram (Mg) .] Also, in US units we give force Pin kilopounds (kips or k), length Lin inches (in.), modulus E in kips per square inch (k/in. 2 or ksi), acceleration ii in inches per second squared (in./s2 ) , and so on.
A.2 MATERIAL PROPERTIES = SI Units
meter (m) kilonewton (kN) kilonewton · meter (k.N · ru) kilopascal (k.Pa) megagram (Mg)
To analyze solids and structures composed of various materials, we need to know certain physical properties. For structural dynamics, the essential material properties are modulus of elasticity E, Poisson' s ratio v, and mass density p. Table A.3 gives these properties in both US and SI units for some commonly used materials. Note that the shearing modulus G is not listed in the table because it can be derived from E and v.
For any numerical problem in structural mechanics , we must use a co11.111 tent system of units . By this we mean that all structural and load parameters 11111~t be expressed in the same units within each system. Some examples of consisll'III units for force, length, and time appear in Table A.2. For instance, in SI( I ) wr must express an applied force Pin newtons (N), a length Lin millimeters (111111 1, the modulus of elasticity E in newtons per square millimeter (N/mm 2). 1111 acceleration ii in millimeters per second squared (mm/s2 ), and so on. TABLE A.2 System
SI
us
49:
Consistent Systems of Units
Force
Length
Time
(I) (2) (3)
newton kilonewton meganewton
millimeter meter kilometer
second second second
(!) (2) (3)
pound kilopound megapound
foot inch yard
second second second
When programming structural dynamics for a digital computer , it ,~ , pecially important that the system of units for input data be consistent.
TABLE A.3
Properties of Materials•
Modulus of Elasticity E Material k/in. 2 Aluminum Brass Bronze Cast iron Concrete Magnesium Nickel Steel Titanium Tungsten
1.0 1.5 1.5 1.4 3.6 6.5 3.0 3.0 1.7 5.5
X X X X X X X X X X
104 104 104 104 103 103 104 104 104 104
GPa
69 103 103 97 25 45 207 207 117 379
Poisson's Ratio 11 0.33 0.34 0.34 0.25 0. 15 0.35 0.31 0.30 0.33 0.20
Mass Density p k-s2 /in. 4 2.45 8. 10 7.80 6.90 2.25 1.71 8.25 7.35 4.20 1.80
X X X X X X X X X X
I0- 7 I0- 7 I0- 7 10- 7 10- 7 10- 7 10- 7 10- 7 10- 7 10- 6
Mg/m3 2.62 8.66 8.34 7.37 2.40 1.83 8.82 7.85 4.49 19.2
"Numbers in this table are taken from J. M. Gere and S. P. Timoshenko, Mechanics of Materials, 2nd ed. , Brooks/Cole, Monterey,. Calif., 1984.
Sec. B.1
499
Inverse Iteration
on the left-hand side, we find that the reverse iteration converges to the fundamental mode. To simplify notation, let
X; = ;
(4)
Then Eq. (3) beco mes
(5)
Appendix B Eigenvalues and Eigenvectors
which will be used in the iteration procedure. To begin inverse iteration , we first assume an approximate shape of the fundamental mode. The usual arbitrary choice for this starting vector is a column of ones. Thus, (X 1) 1 = {l , I , I , ... , 1}
(6)
Substituting this vector into the last of Eqs. (4), we calculate (B 1) 1 as (B 1) 1
= M(X 1) 1
(7)
Then solve the simultaneous algebraic equations in Eq. (5) to obtain (Y1) 1 8 .1 INVERSE ITERATION
All eigenvalue solution routines are iterative because we seek the roots ot tho <'11<1racteristic equation (see Sec . 3.6) , which is a polynomial of order n, 1111 mogencous action equations of motion provide the form of the eigenvul11 problem to be solved. Thus, we have S ; =
WTM ;
(i
=
1, 2, . . . , n)
(I)
s = UTD u
()
In this lype o f factorizatio n, D is a diagonal matrix; and U is upper trinn itlllM with vulues of unity in diagonal positions. Substituting Eq . (2) and I/ A 11110 Eq . ( I ) produces
,vi
Ur I) U A11
(8)
This expression is only an approximation, unless the estimated mode shape satisfies Eq. (5) exactly. Also, a first approximation to the eigenvalue A, may be found by dividing any term in vector (Y1) 1 by the corresponding term in (X1), . That is,
(A) = I I
(}j),
(1 :s j :s n)
(XJ,
(9)
If (};) 1 is chosen to be the largest (positive or negative) term in vector (Yi) ,, normalization with respect to that value gives us the second trial vector:
which is a slightly rearranged version of Eq. (3.6-4). Direct (or for11•111tl) ifl'f'otion of Eq . (1) would involve substitution of trial vectors for , 011 lh ll'ft hand side and evaluation of WT on the right-hand side. This techniqul' 11 I converges to the dominant (largest) eigenvalue w?, and the corresponding l' ijll'll wt·lor <1>11 • J lowever, to extract the smallest angular frequency first , we muNI 1111 " '"''I',\'£' (or inverse ) iteration. For this second approach, let A; = I / "'i 1111~ fnclor the stiffness matrix S by the modified Cholesky method [2 , 3], as folloWM
= (A 1).(X 1) 1
1
(X,h = (A,), (Y1)1
(10)
in which the normalization constant is (11. 1) 1 itself. This procedure is repeated until the eigenvalue A1 and its associated eigenvector X, are determined to some specified accuracy. In the kth iteration, the recurrence equations for the steps described above are:
I. Calculate vector (B 1)k as
(11 )
M <1>1
ll y 11si11H 11 iul Vl'l'lors for <1>1 011 the ri ghl hand side of Eq
( 12)
600
Eigenvalues and Eigenvectors
App. B
Sec. B.1
501
Inverse Iteration
3. Find the new trial vector (X1)k+ 1 to be
2 .72 M = m [
(I J) where (A 1h is the largest term in vector (Y1)t. To check convergence of thl' eigenvalue, we use the expression
which nd is the number of significant digits of accuracy desired. Bathe [4] recommends calculation of (wr)k = 1/(A1h from the Raylcixh
D =
and
t//10tie11t (
2) _ WI k -
(XThS(X1h (XT)k M(X1h
TM i
= 0
(i = 2, 3, ... , n)
( I~ )
( It!)
Adding this modal constraint condition to the eigenvalue equations [see Eq . (III Wl' cnn red uce their order from n ton - l . For the reduced equations , the Sl'l 11ml 1111llk becomes dominant and also may be calculated by inverse iteration I hi dt•l1111ion iteration sequence is repeated to extract as many modes as dl'"tt ~d 1lowl'Vl'I', to retain accuracy in each iteration, we must M -orthogonali tl' rn, h tll'W v1.·c tor with those found previously. This may be accomplished 11111111 <,'111111 Schmidt orthogonalization, as shown in Ref. 4. E111111111lt• H. I N11w w1· sh11II apply inverse iteratio n to extract the fundamental mode from lh1• 1•l1t II vu ltH· pt 11hll·m I'm the plane truss in Fig. 3. I I (a) (sec Sec. 3.5). For thi s truss 1h1· NI ti In 111111t Ix wns found to he
0.36
w ltr1r ,1
/ •/\ / /
Al1.o,
0. 1(1
0.481 0.48
0.48
I .M
tlt1• 1•011Mis h•111 11111 ~11 11111t1i x
IN
t ,1
0
0
(b)
0
0.2°647
~
0
0
1.000
s[l.3
_[l U =
whkh produces much faster convergence. Unfortunately, improving the rail' 111 l'11t1vergence of the eigenvalue has no effect on the rate of convergence of 1111 l'l~l·t1vcctor. However, improvement of eigenvector convergence can be altat11l·d wtlh spectral shifting, as described later in this appendix. After the fundamental mode has been determined, it is usually elimi1111h·d I111111 the eigenvalue equations by the process of deflation [l , 4]. For this ll'l'h 11iq11l'. we express the orthogonality of the eigenvectors <1> 1 and i with n·~pr, I lo lhl' mass matrix M, as follows:
3.28
for which m = pAL/6. In accordance with Eq. (2), we factor matrix S to obtain
( 14) 111
1
60
- 0.2647 1
0
0
]
-0.3529] 1.333
0
(c)
(d)
I
As the starting eigenvector, assume
(X 1) 1 = {I , 1, 1}
(e)
In the first iteration, we evaluate (B 1) 1 from the last of Eqs. (4), as follows:
(B1)1 = M(X1)1 = m{3 .72, 4.28, 3.28}
(f)
Then solve for (Y1) 1 from Eq. (5):
m (Y, ), = - {8.000, 23.12, -2 .426}
s
(g)
Normalize this vector with respect to its largest term to find the new trial vector 1
(X,)z = - ( ) (Y1 ) 1 = {0.3460, 1.000, -0.1049} A1 I
(h)
In this iteration the first approximation to the eigenvalue is
m (A 1) I = 23.12-
(1')
s
which is the largest term in vector (Y1)1. Results of successive iterations are listed in Table B.l. We see that convergence (to fo ur significant digits) for the reciprocal A1 of the eigenvalue wf occurs in six cycles of iteration. However, the eigenvector X1 requires another cycle to attain the same nccuracy. Their final values are m 2 1 s w, = - = 0.04501 A, = 22.22(j) A, m s 1111d X, {0.2314, 1.000, 0.2472} (k)
wl11l'11
1111• 1111• ~111111· 11~ 1lt11~1· 111111111
hy Pro~rnm Vllll'T 111 Ex11111pll' l •I
Sec. B.1
503
Inverse Iteration
Spectral Shifting
r-
~ ~
8:
"0"<1" NON
c:;;....;c:;; I
To improve the rate of convergence of eigenvectors, we use the technique known as spectral shifting. For this purpose, let the constant ai be a number close (but not equal) to WT, which is the eigenvalue to be calculated. Then subtract aiM ct>i from both sides of Eq. (1), and divide by ai to obtain
wr -
N N
c--i
(17)
S* A; «l>i = M «l>i
N
where "
~
'D
8:
N
;;; 8~
NON
o....; o I
°' ~8~ NON 0 ....;c:;;
N N
s* = S
c--i
V)
8:
I
A;=---
wT-
wr.
N N
c--i
N
s* = u*To* u*
..
~
"""
8:
"""
<"l <"l N
§
Also, the eigenvalue
0 V)
""" N
o....; o I
<"l
N
!"i
2 W·
N
~
~
8:
r°' § °' N
'D "
N
c:;; ....;c:;; I
N
Q.)
u
8: ~
8: A
C 8: --'--
~§s°'
o....; o I
-- .. 9 ~
-
1 =-*+a· Ai I
(20)
which need only be computed once (after convergence). Spectral shifting may be used to aid the following tasks:
r-
N
!"i N
1. Extraction of modes near an expected resonance (ai--" n), where !l is the angular frequency of a periodic forcing function. 2. Extraction of modes corresponding to repeated roots (ai--" w;,) of multiplicity m, including rigid-body modes (ai--" O) 3. Extraction of sequential modes by using (ai--" wT) after deflation
0 N ~
(19)
wr becomes I
<"l
(18)
ai
By this manipulation , the dominance of A; in Eq. ( 17) is greater than that of Ai in Eq. (3), because in the second of Eqs. (18) we see that A;--" oo as ai--" After the spectral shift, inverse iteration proceeds as before, except that the factorization of s* is expressed as
'D
~ ~
- aiM
N
"""
V)
!"i N
If the stiffness matrix Sin Eq. (1) is semidefinite, one or more rigid-body modes exist. In that case, a small (but finite) spectral shift away from zero makes s* in Eq. (19) positive-definite. Then by inverse interation we can extract the first rigid-body mode. Elimination of that mode from the equations by deflation allows iteration for the second rigid-body mode, and so on. A similar procedure is required when calculating nonzero repeated eigenvalues.
~
..,.; N
A
~ "'IE:
-
Example B.2 Let us use a sp~ctrnl shif1 to make the second mode dominant for the plane truss in Example n . 1. Thi• tollowl ng shift
(f)
no4
Elgonvaluoe and Eigenvectors
App
I
Sec. B.2
w1ll ,.11ll11w, cxtrm:tion ) or the second eigenvalue and eigenvector w1'thout dcflat·1011. I·111111 I i11st o , 1·,qs. ( 18, 111• the modified stiffness matrix becomes -0.72
0.3808
S"' - S - a2M = s -0.72 [ - 0.48 1111l•tmi1.11tion
or S"' as in Eq. D*
-0.48 0.48
0.48
J
(1111
0. 3808
0
0
- 2.182
0
0
[
-oLJ
111111
TABLE B.2 Iteration of Second Mode
Cycle
1
2
3
4
5
(X2h
(X2)1
(X2h
(X2h
(X2)•
(X2)s
I
Vector
1 1
0.8904 -0.1400 1.000
0.8677 -0. 1701 1.000
0.8673 -0.1715 1.000
0.8673 - 0. 1715 1.000
0.4592
(19) yields
s
=
- 0.8208
(11)
505
Transformation Methods
.!... (Ai)t m
-92.68
-84.2 1
-83.27
-83.25
-83.25
B.2 TRANSFORMATION METHODS
_ [l U* = 0
- 1.891 I
0
- 1. 261] 0.1~59
Io J
0
As lw lon:, we ussumc the starting eigenvector (X2)1 = {I, I , J}
q,1
tlw first iteration, the vector (B2 ) 1 is computed to be
111
Jql
(82), = M(X2), = m{3.72, 4.28, 3.28}
AV=VA
wll1d1 rnnt uins the same values as Eq. (f). Solution of Eq. (17) for vector (Y2 ), p, rnhll m (Y2)1 = ;{- 82.52, 12.97, -92.68}
(I)
N111 11111 li z11lion or this vector with respect to its largest term gives I
(X2h Ill
= (A;), (Y2)1 = {0.8904,
When most or all of the eigenvalues and eigenvectors are desired, transformation methods [5] prove to be more efficient than inverse iteration. In this section we describe the Jacobi, Givens, and Householder transformation procedures, as well as QR iteration. To confirm ideas, a numerical example follows the discussion of each approach. As a preliminary matter, we assume that the expanded eigenvalue problem has been converted to the standard, symmetric form:
- 0.1400, 1.000}
as described in Secs . 3.6 and 4 .2. In Eq. (1) the symbol V denotes an orthogonal modal matrix of n eigenvectors that are normalized to unit lengths and listed column-wise. Also , the eigenvalues ,\ 1 , ,\2, • . . , ,\11 appear in diagonal positions of the spectral matrix A. The basic process in all transformation methods is to diagonalize matrix A, as follows: VTA V = A
wh1l'11 (At), = - 92.68~
UI
.!'
l,u11
l'uhk 13.2 ~h.ows .the resu lts of successive iterations. In this case conver~Mlt (hi At and the second eigenvector X 2 occurs in fi w I y 1I
Nignilk-111_11 d1g_11s) for both l-111111 vnlm·s for llu s mode arc
At
- 83.25
111 ,\'
2
W2 =
I 7*'
+ a2 = 0.3480 •
"2
1111d
X, wh11
h 1111· 11i 11111 1•1111111 lo thos1• 1
j0.8<17.l, - 0. 17 15, 1,000} 111 Ex11111pk•
l.4 ,
1·
Ill
(11)
( l)
(2)
When this is accomplished, we have found not only the spectral matrix A but also the normalized modal matrix V. Jacobi Method
Sinee its development in the nineteenth century [6] , the Jacobi method has enjoyed extensive usage by mathematicians, scientists, and engineers throughout the world. The essential idea in this approach is to zero a selected offdiagonal term A,,., = Aqp of the coefficient matrix A in Eq. (1) . This is accomplished by opl·rnting upon that matrix with a generalized rotation matrix R.kin
step k, us follow~·
1508
I lunrwnl11n1 1111d t l11n11vncto11
The form of the
II
X
II
Soc. B.:.>
rotutiou 11111111 11 Is
1rorrnformotlon Methods
and Eq. (8) becomes
1
(10) which is an approximation of Eq. (2). The Jacobi method may be applied selectively to annihilate the largest off-diagonal term in matrix A. However, this approach requires searching for that term before it can be put to zero. More commonly, we operate in sweeps, systematically annihilating all of the off-diagonal terms by rows or columns. Each sweep of this kind requires approximately 2n 3 multiplications. Equations (6) give the formulas to be used for altering either the upper or lower triangular part of matrix A. During a particular sweep, terms that are zeroed do not necessarily remain zeros. However, the square root of the mean of the squares (RMS) for offdiagonal terms is reduced after each annihilation (and even more dramatically after each sweep).
sin fh-------- · row I'
-sin ()k
cos 8c - - ----- · row q
I I I I I I
I I I I I I
col. p
col. q
In this rather sparse array, we define the terms (Rpp)k (Rpq )k
= (Rqqh = cos fJk = - (Rqp h = sin fJk t= p
(Ruh = 1
(i
(Rij h
(i or j
=0
(;")
Example B.3
1,111
With one sweep of the Jacobi method, determine approximately the spectral matrix A and the modal matrix V associated with the 3 X 3 array
or q)
(~d
t= p or q)
(111)
= (APP cos 2 fJ + 2Apq cos () sin fJ + Aqq sin 2 fJ)k (Aqq)k+ , = (App sin2 () - 2Apq cos () sin () + Aqq cos2 fJ)k (Aµq)k+, = [(Aqq - App) cos ()sin()+ Apq(cos2 fJ - sin2 fJ)]k = (Aqp h+ I (App)k+r
(A,1,)H r = (A;p cos () + A;q sin fJ)k = (Ap;)k+,
+ Aiq cos
(Aiq)k+1 = (-AiP sin fJ
8)k
= (Aq)k+ ,
(fut)
[
tan 28k
Io~ I -s 'Tr/ 4.
=
After nr rotations, Eq. (3) yields T R,,, ... R2R,A 1 R"T"" 1 R2 A
....
A
..•
"'T R,, "" A r
Tlwrcforc, the matrix of orthonormal eigenvectors is
V -= ""T "T Kr R 2 ••
• ""' K I' 11,
0
-0.48 1.44
0 ] -0.48
- 0.48
1.92
(a)
In the first transformation of matrix A, we shall make the term A,2 = A21 equal to zero. Therefore, p = 1 and q = 2, so Eq. (7) becomes
((1h)
tan
281
=
2A12 = (2)( - 0.48) = I 2 A11 - A22 0. 64 - 1.44 ·
(b)
From this expression we determine the angle 8, and its sine and cosine as ( (1\:)
(<111)
81 = 0.4380
sin 81 = 0.4242
cos 81 = 0.9056
(c)
Then the first generalized rotation matrix in Eq. (4) takes the form
((10)
'l'lw unglc 8k is chosen so that the term (Aµq)k+ , becomes zero. Thus, fro111 I 11 (C,l') we have
( 2Ae9 ) Aµp - Aqq k
0.64 -0.48
A =
'l'ht! multiplication in Eq. (3) alters only terms in rows and columns p and , 111 1 mutri x A , which become
whl•rc
507
It =
0.9056 0.4242 - 0.4242 0.9056 [
0
0~ ] (d)
0
and the first transformation in Eq. (3) produces
0.4152 ,
, T
0
A2 = R , A1R1 =
[
- 0.2036
0 1.665 -0.4347
- 0.2036] - 0.4347
(e)
1.920
As a conNl'< t111,m·t•, lht• fll'sl rotation has reduced the RMS of off-diagonal terms in matrix A from OAH 111 II I 11l•I h11 1111· N1·111ud 1111us twm11ti o n, WL' ,unl..1• A,, • A II t·q1111l 10 11•111, so 1h11t
608
Eigenva lues and Eigenvectors tan
" _ 2 u2 -
2A13 A11 - A33
---
App,
a
Sec. B.2
0.8927
(2)(-0.2036) 0.4 152 - 1.92 = 0 ·2706
v
(1)
= RTRI RI =
Thus, we have sin (Ji= 0.1317
82 = 0.1321
[
-0.0~727 [
- 0.05727 1.665
O ] - 0.4309
-0.4309 1.947 for which the RMS of off-diagonal terms is 0.3074. The third transformation involves zero1·ng A 23 -- A 32, wh'ch · 1 gives 2A23 tan 283 = = (2)(-0.4309) _ A22 - A33 1.665 - 1. 947 - 3.o53 I knee,
(h
= 0.6272
sin 83 = 0.5868 Sil the third rotation matrix becomes
R,- G
-0.5868
tan Ok = (Ap-i ,q)
cos 83 = 0.8097
I I
( j)
0.5~68] 0.8097
( I)
-0.04637 1.353
0.03361
0
0.03361] 0
0 .30881
0 1.353
O 0
2.259
l
0
0
2 2 'ii)
nddllhlll, IIJlJll llXi lllillt' l'i)tl'II Vl'l't\ll ,~ Ill\' l'llk 11) 11(1•d IIN
X
X
0
0
0
0
X
X
X
X
X
X
0 0 0
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
0
X
X
X
X
X
(12)
(Ill)
. le ent <~ 11e sweep, the diagonal terms in matri x A4 represent ,m,d 111110 11 the eigenvalues of the original matrix . That is, g
111
(1::)2
An-2 =
l11111t1011s lllr
r
Rotation in the 2-4 plane yields
(k)
wlt1•11• tlw RMS of off-diagonal terms is O 0405 1 Note that each off d' t I• . . · · - 1agona I te1111 IN 1111w n 111st 1111 order ol magnitude smaller than at the beginning of th At ti . . I r I . . e sweep .
A
(11) k
and so on. After we clear the first row (and column), matrix A has the form
0.3881 - 0.04637 [
Ap- 1,p
where Ok :s 7r/2 . For the first row of matrix A, rotation in the 2-3 plane gives
Suhstituting It into Eq. (3) results in A,1 = RJ A3 RT =
0 .8026
(ll
0 0.8097
0.5817
(o)
Instead of annihilating the p, q term (as in the Jacobi method), Givens [7] proposed annihilating the p - 1, q term instead. Then a zeroed term will remain zero during a forward sweep. For this purpose, we set Eq. (6e) equal to zero to obtain
Using R2 in Eq. (3) yields
A3 = R2 A2 RI =
0.1317
0.1523] -0.5767
(h)
- 0.1317
0.388 1
-0.4 135 0.7005
Givens Method
0.9913 0
R2 =
[
0.4205
Modal vectors appearing column-wise in this matrix are automatically normalized to have unit lengths.
cos 82 = 0 .9913
and the second rotation matrix is ,
509
Transform ation Methods
1111
T his technique produces a tridiagonal matrix T in a finite number of steps, consisting of n, = (n - 2)(n - 1)/2 rotations. The entire sweep requires approximately 4n 3 /3 multiplications. While the Givens method does not lead directly to the spectral matrix, we consider it a useful preliminary to the Householder method, which is described next.
Example U.4 By tlu1
~---------111111111111111111111111111111 610
I l11u11vnl110111111d l:lgo11vooto1H
A=
I
0.48
1,0()
- 0.48 U<,
1.64
0
- 1.00
0
1.36
0
0.48
0
Sec. B.2
5
oq
Substituting this matrix into Eq. (3) yields
(p I
1.64 3 with a rotation in the 2-3 plane. For this p11IJH1~1. We start by annihilating the term A, Eq. (11) results in 3 -1 = tan - I A1 - = tan -
01
1.00 -0.48
A12
Then we have
Transformation Methods
= 1.123
A
A
1.360 I.109
-
A3 = R2A2RI = [
-0.4463
0
-l.109 1.412 -0.4 463
0
0
- 0.1953
0
1.538
-o.iL,]
which is now in tridiagonal form.
1.689
(q)
Householder Method
sin 01 = 0.9015 and the first rotation matrix R. 1 becomes
cos 01 = 0.4327
(1)
0
r i 0
A
R, -
0 0.4327 0.9015 -0.9015 0.4327 0
0
Using this operator in Eq. (3) gives
A Householder transformation operator [8] causes all of the terms in a vector to become zero except the first, which becomes the length of the vector itself. The operator has the form
f]
A
AT
A,-R,A,R, -
i
[ -1.109 13~
0
-o. 1092
l.412 -0.1092
l.588
0.4327
04~27]
(II
0.2077
0.2077
1.640 In this case only one transformation is required to clear the first row (and column) 011f~hh of triangular part. The term A 14 = A4 1 was zero initially and remained unaffct·t,•cl h\ thethe operation above.
82
= tan
-
= tan
-I
A23
0.4327 - 0.1092
Then fhc second rotation matrix
R.2
= pT = p - 1
fl
s2
= L aJ
(16a)
J=l
0.2447
0.9696
0.9696
0.2447
0
e=--
l
. Vc1c
(16c)
In Eq. (16b) the sign of s is taken to be the same as that of a 1 , so that there is no possibility of getting zero for the first term in vector c. The unit vector e in Eq . (13) is the result of normalizing vector c to unit length, as indicated in Eq. (16c).
is
0
(16b)
C
(Ill
cos 82 = 0.2447
0
, O}, wheres
(15) We form the matrix P by first creating e in Eq. (13) from the vector a , as follows:
= - I.324
0
(14)
Pa=b
c = {a1 ± s, a2, a3, . .. , a,.}
From thi s we get
sin 82 = - 0.9696
(13)
Hence, P is equal to its own inverse. Consider a vector a to be converted to b = {±s, 0, 0, . is the length of a. That is,
The second step in this example is to annihilate element A by a rotation l11 tl11 3-4 plane. Thus, Eq. (11) produces 24 - I A24
- 2e eT
in which e is a col umn vector of unit length. The matrix Pis both symmetric and orthogonal, so that p
-1.109
=I
P
Now let us express the product Pa in Eq. (15) in terms of the vector c. Using Eqs. (13) and (16c), we obtain
fi
(\\ I
c
CT)
Pa = \1 - 2cTc a = a Rcsull s ol flll'
l1111!'l'
2c(c Ta) (cTc)
(17)
pr·oducls (in pi11·e11Jlll'st•s) from llw st•rnnd form of Hq . ( 17)
512
Eigenvalues and Eigenvectors
App. II
Sec. B.2
Transformation Methods
513
are Recurrence equations for the kth step are the same as those in Eqs. (19) and (21), except that k replaced 1 and the lower limit on j is k + 1. After n - 2 Householder transformations, we evolve the tridiagonal matrix
n
era= (a 1 ± s)a 1
+ 2 aJ = s 2 ± a 1s j =2
c Tc
= (a1 ±
s)
II
2
+
2 aJ = s 2 ± 2a 1s + s 2 = 2(s
2
j =2
± a 1s)
(22) where the Householder matrix is
Substitution of these expressions into Eq. ( 17) produces 2
=a
p a
s) 2(s ± a 1s)
_ 2c(s ±
a1
2
= b
( 18)
as desired.
In a similar manner, the last n - 2 terms in the first column of th1• coefficient matrix A may be zeroed . With this objective, we let
2 AJ.1
= {O, A21
±
S1, A 31, A41, . . . , A,.1}
( 191,)
( 19\')
VcTCi
=
From the same matrix A used in Example B .4, we shall create a tridiagonal matrix T, using Householder transformations. To begin, let us annihilate the first row and column outside the tridiagonal region, as shown by Eqs. (19) and (21).
I - 2e 1 ef
L" AJ,
= (-0 .48)2 + (-1.00)2 = 1.230
1
SI
= ±1.109
C1
= {0, A21 ± s1, A 31,
e,
=
(z)
C1
C1
\/c"fCi =
l.
878
A41}
X
X
X
X
X
X
X
X
X
X
X
X
X
0 0
P,A1
X
0
X
X
X
X
(2())
X
X
- 0.~327 - 0.9015
0
0
A2 = P1A1PT
=
s, 0
0
0
0
X
X
X
X
X
X
X
X
X
X
()
,\'
,\'
X
X
(a')
~
(b')
°
-0.9~ 15 0 ] 0.4327 0
0
1.109
0
1.412
0.1092
0.1092
1.588
(c')
l
1
. -0.4327 0.2077
s,
0
A2 = Pi A1 p I =
[1300 1.109
X
Also, the last n - 2 terms in the first row of matrix A may be zeroed as wi•ll by X
T
{0, -1.589, -1.000, 0}
0
P1 -_ I - 2e, e1T -
S1
=
= {O, -0. 8464, -0.5326, O}
-r~
( 19d )
Then
(y)
j =2
( 1911)
e, = -C1- -
P,
Example B.5
sf =
"
j=2
C1
(23) 3
Equations (22) and (23) require approximately 2n /3 multiplications.
= {±s, 0, 0, ... , O}
ST=
= Pf PI, .. PI-2
ff
=a_ c
-04~27 0.2077
(d')
1.640
Matrix A2 is the same as that found before in Eq. (t) by the Givens method, except for the signs on several off-diagonal terms. Similarly, for the second row and column we have II
s~ =
L
A].2
= (0.1092) 2 + (-0.4327)2
=
0.1992
(e')
j- 3
S2
= ± 0.4463
C2
= {O,
(.'I l t• ,
(f')
0, A 32 ±
S2,
Ad = {O, 0, 0. 5555 , - 0.4327}
C2
CJ
Vt· !t'J
0.704 f•I
{O, 0 , 0 /HKI/ ,
0 .61 tl~f
(g') (h')
514
Eigenvalues and Eigenvectors
P2 = I
-
0 I 2e2eI = [ 0 0
0
0
0
A,~ P, A,P, ~
0
1.109 1.412
~
0 -0.4463
-0.4463
1.538
0
0.1953
= Pf PI=
[i
- 0.9015
oiLl
(J')
1.689
0
0
0.9696
T k+I
-08~411
( (' ' )
0.4196
(28)
(29)
= QITk
Rk
(30)
This formula gives us the means for determining both of the desired matrices, as will be explained next. At the operational level for this method, we wish to annihilate a lowertriangular term (1j;h of the tridiagonal matrix Ti,, where j = i + l. For this purpose, we set equal to zero the product of row j of a Jacobi rotation matrix and column i of matrix Ti,, as follows:
+
-sin O;(T;;)k
cos 8;(1j;)k = 0
(31 )
From this expression, we have
Assume that Householder transformations have converted Eq. (I) to
tan
TW=WA
(.?•I)
. (} '
(1j;h C;
sm · = - -
or
(32)
-1!.
cos (} . = I
(T;;)k
-C;-
(33)
where
V=HW
(34)
the form
Then the Jacobi rotation matrix (R;)k premultiplies matrix Tk (for i = I, 2, ... , n - l) to produce
T1 = Q1R1 which is known as Givens factorization. The symbol Q1 in Eq. (26) dcnoll'II ,111 o~thogonal matrix obtained by n - 1 Jacobi rotations, and R 1 is an upp1•1 triangular array. The QR algorithm l9] derives its name from the factors in h1
(26).
'
Q'/' and
(T;;T,)
Consequently,
(}~ul
. Prcniultiplication of Eq. (26) by gives
e; =
k
where
= T1 into
QITkQk = RkQk-A
where n1 is the number of factorizations. The recurrence algorithm expressed by Eq. (28) indicates that we must generate Rk and Qk and multiply them in the sequence shown. From that equation we see that the definition for Rk is
0.2447
QR Algorithm
Now let us factor T
=
W = Q1Q2,, · Q,!f
(k 'l
H
515
which iterates to the spectral matrix A. The modal matrix Vis calculated from Eq. (25b), in which
where
0.2206 - 0.1059
Transformation Methods
(I')
Again, matrix A3 is the s~e as t~a.t in Eq. (x), except for the signs on off-diagonal tc 1 IIIN Thus, we see that the desired tndiagonal matrix is
0 -0.4327
Sec. 8.2
than T1. Recursively, we have
oL]
0 - 0.2447
0.9696 0.2447
['1.109 360
T
App, II
postmultiplication by
Q 11~1·11 1
(35) After the first operation in this sequence, matrix Tk is no longer tridiagonal; and after n - 1 operations it becomes the upper triangular matrix Rk in Eq. (28) . Comparing Eq. (35) with Eq. (30), we see that (36a)
( I/)
l'lw 1ww 1111111 ix 'I'.! is unotlwr 11 idingo1111l 111 rny huving s11mllt•1off di11g111111I
11·1111 111
or
( Hih)
516
Eigenvalues and Eigenvectors
App. B
App. B
References
Therefore, matrix Qk is defined as the product of the transposes of n - I Jacobi rotation matrices.
Example B.6 Now we shall apply one cycle of QR iteration to the tridiagonal matrix T obtained by the Householder method in Example B.5. For the first rotation (in the 2-1 plane), Eqs. (34) and (33) give
C1
= YTf1 +
.
T:i.1
sm 81 = cos 81
n1 = Y(l .360)2 + (1.109)2= 1.755
1.109 1.755
- = - - = 0.6320 C1
Ti.1
=- = C1
1.360 -1.755
TABLE B.3 First Cycle of QR Iteration
(R2)1
(m')
[!
l
0.7749
0 0
•
T32
(J
sm 2 = C
2
T-i.2
COS 82 = -
C2
T52
= Y(0.3935)
0 0
0 I
00 01 0 1 0 0
(p')
2 + (-0.4463) 2 = 0.5950
- 0.4463
0.3935 0.5950
= 0.6613
=
.
Vr ~J + 43
srn 03 = -
C3
7;,
T~3 = Y(0.7578) 2 + (0.1953)2 = o.7826
0. 1953 0 .7826
= - - = 0 .2495
0.7590
cos 0, - Ci == 0. 7826
(r')
R1
= 0.9684 (I{ift,((,)1'1'1
1.109 0 0
0 0
[0.7749 0.6320 0 0
(t ' )
(11 1 )
0 -0.4463 1.538 0.1953
oL] 1.689
(R1)1T1 1.752 0.3935 -0.4463 0
[Lr
-0.2821 - 0.3459 1.538 0.1953
J,,] 1.689
(R2R1)1T1
ot]
1.752 0.5950 0 0
[17550 0 0
0.9684
= (RTRIRD1
R1
0ll83] [1755 0
- 0.4180 - 0.4591 0.5152 0.5629 - 0. 1450 -0.7501 0.6404 -0.1650 0 0.2495 0.9684
-00.1291 '~"] 1.689
= (R3R2R1)1T1
1.752 0.5950 0 0
0 0
-0.282 1 -1.383 0.7578 0.1953
-0.2821 -1.383 0.7826 0
Qi = (RfRillr)1
-01~65] 0.5466 1.604
l
(x')
is listed in the lower left-hand position. Using Eq. (27), we now compute a second tridiagonal matrix, as follows:
(N')
T2
=
Ri Qi =
[
2.468 0 .3760 o
0. 3760 1.342 - 0.5870
0
.0
0 .0000 0.0000 -0.5870 0.0000 0.6375 0.4001 0.4001
(y')
1.553
Note that computed zeros must appear in the upper triangular part of matrix T2 to satisfy the symmetry guaranteed by the congruence transformation in Eq. (27) . We also see that the RMS of off-diagonal terms is reduced from 0 .6993 (for matrix T1) to 0.4645 (for matrix T 2). Spectral shifting can be used in successive cycles to accelerate the rate of convergence, as explained in Sec. B. l.
(v' l
REFERENCES
(w ' l
l . Timoshonko , S. P., Young, D. 11. , and Weaver, W., Jr., Vibra1ion Problems in l~11J(l,11•1•rl11}( , 4th L'd., Wiley, New Yor·k, J1)74, l. W1•11v,,1, W , Ji , 1111d Ci1•11•, .I . M , Motr/1 /\1111!1w/,1· 1!/' l•111111NIS1r111•111r1•,1·, 2nd 1•d., Vn11 Nrn1t111111l lfr l11h11ld , N1•w Yrnk, l11HO
'l'lw third rot11tlo11 mulrix (R,) 1 ulso uppeurs in Tublc B.3, ulong with the prodlll'I
l1111ddil in11 1 th,• 11111t11 ~
~] ~]
(q')
'l'uhle 8 .3 contains the second Jacobi rotation matrix (R.2)1 below the first, and the l'l'Ntrll of multipl ying it with (R1)1T1 is shown below the latter matrix. The third rotation (in the 4-3 plane) involves c3
[!
0 0 0.9684 -0.2495
Qi
I
= 0. 5950 = -0.7501
=- -
0 -0.7501 0.6613 0
["ffi
1.109 1.412 - 0.4463 0
(R3),
Tnhle B.3 shows this matrix in the upper left-hand position. After multiplying it with mu1ri x T = T1 (in the upper right-hand position) , we list the product (R. 1) 1 T1 in tla• positi on below T1. For the second rotation (in the 3-2 plane) , we have
C2 = YTi2 +
0 0.6613 0.7501 0
(0 1)
0.7749 0.6320
-0 .6320
T1
0.6341 0 0.7776 0 0 1 0 0
-0.6341 0 0
Then the first 4 x 4 Jacobi rotation matrix becomes
• (R1)1 =
(R1)1
[ o.m,
(n')
= 0.7749
517
618
I l111111vnh1111 nnd I l11111w1u:tor •
At>t> I
3. Weaver, W. , Jr. , and Johnston, I'. R., Vi11it1• J:'l,•1111•111.v for S1m, ·111m/ A111,/n11 1 Prentice-Hall , Englewood Cliffs, N.J ., 1984. 4. Bathe, K. 1. , Finite Element Procedures i11 Engineering Analysis, P1·c11t1n• 111111 Englewood Cliffs, N .J. , 1982. 5. Wilkinson, J. H., The Algebraic Eigenvalue Problem, Oxford Univcrntly l'll'HN, London, 1965. 6. Jacobi, C. G. J. , "Uber ein leichtes Verfahren die in der Thcoiie d1•1 Sh, ularstorungen vorkommenden Gleichungen numerisch aufzuloscn," Crl'l/1•' 1 I , Vol. 30, 1846 , pp. 51-94. 7. Givens, W. , "Numerical Computations of the Characteristic Values of a Rl·11I !·i\ 111 metric Matrix," Report No. ORNL-1574 , Oak Ridge National Laborut01 y, I >11t. Ridge, Tenn. , 1954. 8. Martin, R. S., Reinsch, C., and Wilkinson, J. H., "Householder's Ttid111pn1111I ization of a Symmetric Matrix ," Numer. Math ., Vol. 11, 1968 , pp. 18 1 l
Appendix C Flowchart for Program DYNAPT
1. SUBPROGRAM SDATPT FOR SUBPROGRAM VIBPT
a. Problem Identification Descriptive title
Descriptive title
b. Structural Parameters
He;iding: STRUCTURAL PARAMETERS Subhead: NN NE NRN E RHO
NN. NE. NRN, E. RIIO
NN . NE. NRN, E. RHO
Read and write struc1ural parameters.
1120
I l11woh1111
1111
1110111 11111 IJYNAI' I
App. C
I lowol11111 Im Pt Ol)l'Ot'II OYNAPr 521
<·. Nollul ('oordlr1111t•,
Heading; NODAL COORDI NATES Subhead: NODE X Y
r
I I
J. X(J), Y(J)
J I
Read and wri1c 11od11 I coordinates.
J, X(JJ, Y(J)
e. Nodal Restraints
I
L_
CONTINUE NND = 2*NN
d. Element Information
Calculate number of
nodal displacement, possible.
r1
I
Heading: ELEMENT INFORMATION Subhead: ELEM. J K AX EL CX CY
I
I I
NRL(J) = 0 3
L_
r----1
I
I I I I I
+
I I I I
I, JN(f), KN(!), AX{!)
XCL = X(KN(JJ) - X(JN(l)) YCL = Y(KN{I)) - Y(JN(J)J EL(!) = DSQRT(XCL *XCL + YCL*YCL) CX(J) = XCUEL(I); CY(!) = YCUEL(l)
I, JN(l), KN{!), AX(!). EL{!), CX(I). CY{!)
L ____ _
Read infonnmion ltu Element I. Compute x and y ~·01111~, ne111s of clement lc11ptl1 XCL and YCL. tlw Jt,1111111 I I and lhc direction l'Cl\1111·~ CX and CY. Write infon11111io11 lrn Elemc n1 I.
2
CONTINUE
Heading: NODAL RESTRA INTS Subhead: NODE NR! NR2
Clem· nodal rcs1111i111
list.
I l11w,1h 11 lnr 1 1 11111111111 I VNAl'l
22
,-1
I h1w1.h 11t lo1 1'1ot:J1 mi OYNAPT
App
t I I I I
4 J
I K. NRL(2*K-1). NRL(2*K) I I K. NRL(2*K-1), NRL(2*K) I I 4 I L-CONTINUE
Rend 11ml W1lh 11111111 ,cstruinl I M
L __ _
I I I
NNR = NNR+NRL(J)
,-
I
f---
I I I I L_
Cal ulah.: 1111111111•1 111 degrees ol l1,·1·1h1111 NI •I
I
I
J = I. NDF
I
K = J. NDF
SS(J.K) = 0.0 CONTINUE
b. Calculate Element Stiffness Matrix
6 J = I, NND
NI = NJ+ NRL(J) ID(J) = J-NI
TOP
Write NDF 11nd NNII
Initialize N l 11111111
1 I I
Otheiwise, put rD(J) among supp rt re ·train ts.
2. SUDPROGRAM STIFPT FOR SUBPROGRAM VIBPT a. Clear Structure Stiffness Matrix
Calculalc 111111111< 1 111 nodal rc�11ul111, NNII
I'. Displacement Indexes
r--
UE
END
L--
NUMBER OF DEGREES OF FREEDOM (NDF) NUMBER OF NODAL RESTRAINTS (NNR)
6
99
5
NDF = NND - NNR
ID(JJ = NDF+NI
RETUR
lni1ializc 1111111h 1 111 11101 I rc,1rui111, hi 11•111
,-1
NRL(JJ > 0
523
By clcf'uull, p111 lllt l1 lllllll11g 1'11:1· 1h�11li11 I 1111 111
r-----
5
1
I I I I
•I I I
SCE = AX(l)*E/EL(I) SCEXX = SCE*CX(l)*CX([) SCEXY = SCE*CX(l)*CY(I) SCEYY = SCE*CY(l)*CY{I) SE(l.1) = SCEXX; SE(l.2) = SCEXY SE(l ,3) = - SCEXX: SE( 1.4) = -SCEXY SE(2,2) = SCEYY; SE(2.3) = -SCEXY SE(2,4) = - SCEYY; SE(3,3) = SCEXX SE(3,4) = SCEXY; SE(4.4) = SCEYY
Compute stiffness constants and ti II upper triangular part of SE, as shown by Eq. (3.5-26).
f lowi:hnrt for l'rournrn l>YNAl'I
624
App C App .C
t
525
f luwohurl for Progrom DYNAPT
c. Calculate Element Displucemcnt Indexes e. Fill Lower Triangle of Structure Stiffness Matrix IDE( l) IDE(3)
= ID(2*JN(l) - l); IDE(2) = ID(2•JN(I )) = ID(2*KN(I) - I); IDE(4) = ID(2*KN(l))
Eq,. (J . 5 30)
d. Transfer to Structure Stiffness Matrix JMI = J -1
r---
r
I Check for row restrnint
r--
I I
Check for column rc,11,11111
I I I
+
.. I
= IDE(J ):
IC
= IDE(K)
I
L __
SS(J,K) = SS(K,J)
ITEM
= IR; IR =
I
2
I
SS(IR,IC)
I IL __
L ____
L _____
IC; IC
= ITEM
CONTINUE
7 CONTINUE
Detenn ine row and col umn indexes for structur ~· stiffness matrix.
= SS(IR, IC)+SE(J, K )
3
If not. interchange row ,111d column indexes . Transfer a term in mat1 1x SE to matrix SS, u, i111 plied by the fi rst of Eq~
Place an upper triangular tenn into a lower triangular position.
6
RETURN
Check whether row intk·~ i, less than or equal to coh111111 index.
+
I I I I I I I I I
IR
I I I
I
I I L
END
3. SUBPROGRAM CMASPT FOR SUBPROGRAM VIBPT This subprogram is similar to STIFPT. but the consistent mass matrix is generated instead of the stiffness matrix. 4. SUBPROGRAM STASYM FOR SUBPROGRAM VIB a. Decompose Stiffness Matrix and Copy Mass Matrix to CMU
(3.5- 14).
CONTINUE
r-
I
J
I , NDF
1
4
CONTINUlo
I
CONTINUE
I L
5
r U(J,K) = SS(J,K)
n ansfer stiffness matrix to U.
626
Flowchart for Program DYNAPT
App
App.C
Flowchart for Program DYNAPT
527
r--DECOMP(U, NDF, & 3)
1
Dccompo~c Mil Im•" 111,1111 inro the factorl'd lt11111 I 11 as given by Eq. (.I " •11
I
r-
1
r---
2 J
=
5
L .-----''----, CMU(J.K) = SS(J,K)
STIFFNESS MATRIX DECOMPOSED
Copy stiffness matrix to CMU.
I. NDF
1
,-
6
l
1
STIFFNESS MATRIX /\ND MASS MATRIX NOT POSITIVE-DEFIN ITE
2
L
CMU(J.K)
= CMS(J,K)
Copy mas~
1111111 ix 111 t
Ml 1 STOP
c. Calculate
u-T 7
b. Decompose Mass Matrix and Copy Stiffness Matrix to CMU INVERU(U, NDF)
3
,-
4
J
=
Call Subprogram INVERU to obtain the inverse transpose of U.
I. NDF
1
,I
l
L
d. Transform to Standa rd, Symmetric Form (I)
Prcmultiply by
u-T
4
U(J,K) = CMS(J,K)
lhtnsfcr mu,~
11111111\ 111
I'
r----1
I
I = NDF-11+1
Set up decreasing index I.
I
DECOMP(U, NDr, & 6)
Dcco111pm1• 11111" 11111111,
11110
Slll+N l·.SS M/\TRIX NOT 1•os 1l"IVI, 1) 1'.l •IN I J'I :, M/\SS M/\TRI X Dl :C'OMPOSED
the iill'lorl'd
1111111
I 11
• ,---I I
'
I I I I I t I
I
9
J
JI
I , NDF
NDF JI f I Tl•.MP 00
Set up decreasing index J. 1111<.l ini1iali1c TEMP to zero.
I lowt hnr I for 1'1ou1nm llYNAl 1 1
I
I I I
•I I I I I I
I
!
•I ,--I
I I I TEMP = I I I IL ___ I
App C:
X K
T EMP
I
l ,J
+
I
U(J.K)•CMU(I.K)
P1 c111u lt1ply C'MlJ hy II 1 indicmcd 111 tht.· 111 , 1 ,,1 I 11
I
(3.6 11 ).
1lowr.hurt fur 1'109run1 DYNAPT
828
IL __ _ _
•IL _____ _
8
I
RETURN
•I
CMU(l,J) = TEMP
•I IL ____ I I ______ L
(2) Postmultiply by
u-
END
Put result back into ('~ ll I
5. SUBPROGRAM EIGEN2 FOR SUBPROGRAM VIB
9
Calculate eigenvalues and eigenvectors to double precision.
10
1
6. SUBPROGRAM TRAVEC FOR SUBPROGRAM VIII a. Calculate Angular Frequencies
r----I I
13 II = l,NDF TOL I = NDF - 11
+I
= 0.000001
Set up decreasing i ndex I
1 , 'l'I
IOl1•1tl lll'I' JIii /l' III
~ l)ll.'IIVllilll'\
I
I I I
r-- --
I
12 J = l ,NDF
I I
•I I I
lf mass matrix wa, de composed, go to 2.
I I
TEMP = 0.0
Initialize TEMP to zero
r--1
L
CV(I) = 1.0/ DSQRT(CV(I))
r---
I I I I I
I I
•I II
I I I I I I I
•I
I I I I
TEMP = TEMP
L ___
+
CMU(K.J)*U(I .K)
Postmultiply CMU by U i, II\ in the first of Eqs. (3 .6 11 l
II
Calculate angular frequencies from the first of Eqs. (3.6- 12).
2
r1
I
CMU(l ,J) = TEMP
Put result back into CM lJ
•I
Check si ze of eigenval ue .
1130
I l11w1 hnr t fur l'rour nr11 l>YNAl' I
I
C V(!)
+
= DSQRT(CV(I))
App,
c:
I lowohurt lnr Proumm lJYNAPI
Calculait: 1111g11J11r lh· qucncics from rlw 111NI
c. Normalize Eigenvectors with Respect to Largest Values
ofEqs. (3.6 19),
I I I
,----I
3
I
I
CV(!)
•L __
= 0. 0
I
Set small cigcnvullll' equal to zero.
I
I
b. Back-Transform Eigenvectors
I I
r----
I I
1
•II
r----
I I I I I I
1
I
1
I I I
Initialize SUM to ~crn
= SUM
+
U(K,J)*PHl(K ,I)
Back-transform cigcnv1'l'l111 ,, as in the second of Eq.~. ( I !I I r,
6
Search for largest value in vector I.
I I I I
CI = DABS(PHl(J,I)) C2 = DABS(BTG)
I
I
I I
C I > C2
I
I
PHl(J,I) = SUM
B IG = C l
If C J exceeds C'2, n:placc BIG wit h (' I .
CONTINUE
r----
I
I I I
P Hl(J, I) = PHl(J,l)/BIG
9
I I
Set up constants C I and C2.
8
L __
+ L __
•I L ____ _
r----
1
•I
L ___ _
I I
Initialize BIG to zero.
I
4
I I
I I S UM I I I I I L __ _
531
L ___ _
10
Put SUM into PIii.
7 RETURN EN D
Nom1alize vector I with respect to B IG.
.. I l11w1 hurt
101
I'I ou111111 IIYNl\l' 1
"''
/\pp, (
I lowoh111 t for f'to111 nrtt tJYNI\P I 533
7. SUlll'ROGRAM 1rnswr FOi{ su1wno(;IUM VIIWI' a. Reorder Angulur l<'rcc111cnclcs 1111<1 Elgcravcl'lcu',~
I
I
I lDCMP = I
If lite
I
I I
lllllSS 11 11111 iX Wu,
decomposed. µo lo I
I
I
NDP2 = NDF/ 2
Divide rhe n11111lwr 111 degrees of f recd om hy
.. 1
I I I
r---
,
I
I
J = NDF - I
+
I ; TEMP = CV(!) CV(!) = CV(J); CV(J) = TEMP
I
I I
I
Interchange sy111111c111t·11lly
I
placed angular frcquc11l'il''
I
I I
I I
I
I I
I I
•I
I l
I I
L __ _
If K corresponds to a free displacement, go to 4.
+ I
EE(J) = 0.0
I
I I I I
For a restrained displacement, put a zero into the expanded eigenvector.
4
t
EE(J) = PHI(K ,1)
I
5
L_
For a free displacement transfer a nonzero term.
I
TEMP = PHI(K ,l) PHI(K, IJ = PHJ(K,J)
I I
Interchange sy111111ctric11lly placed eigenvectors.
I I
PHI (K,JJ = TEMP
I
2
I
I
MODE (!) ANGULAR FREQUENCY (CV(!)) NODE DJI DJ2
6
I
3
Write headings for modal outpu1.
+ I L __ _
J,EE(2*J - I).EE(2*J)
Write cigc11vec1or for
mode I. 7 CONTIN UE
b. Write Angular Frequencies and Expanded Eigenvectors c. Normalize Eigenvectors with Respect to the Mass Matrix
r--1
I
I I
I I I
r------
,rI I
I
1
Set index K equal 10 ll)(J ).
I I t
,----
1
I I
t
I
I
I
Initialize SUM to zero.
1134
I low1 hull for l'1ou1n111 l>YNAl'I
A1111
App. C
1lowchort for Program DYNAPT
535
8. SUBPROGRAM DYLOPT FOR PROGRAM DYNAPT a. Dynamic Parameters I
I
I I I I I
•I
I
•I r--I I I I
I
8
L
SUM = SUM
L ____
I
I I
I I
I
I I
I I
Prcmultiply ihl.! 1111M 11111 trix by the trnm,po,l·cl eigenvector: ISOLVE,NTS,DT,DAMPR
EE(J) = SUM
Put SUM into Iii: (1111 temporary Monigc).
Read and write dynamic parameters.
ISOLVE,NTS.DT,DAMPR SUM= 0.0
r--I
L_
SUM = SUM
r--L_
b. lnitial Conditions
10 J = I.NDF
+ EE(J)*PHl(J,I)
SUM = DSQRT (SUM )
I
Reinitiali1:e SUM to 1<•111
10
•I I
+ PHl(K ,l)•CMS(K ,J)
9
I I I I
Heading: LOADING NO. (L N) OF (NLS) Subheads: DYNAMI C PARAMETERS !SOLVE NTS OT DAM PR
8 K = l,NDF
II
Postmultiply by thl.! l'lp envector. Calculate nonnalil11t1n11 constant.
J = 1,NDF
r- -
I J = I ,NND
I
I I DO(J) I L_
= 0.0; VO(J) = 0.0
Clear initial displacement and velocity vectors.
II
L _ ______
PHl(J, I) = PHl(J. 1)/ SUM
Divide eigenvector hy 1h1 normalinuion con~t,1111
Heading: I NITIAL CONDITIONS Subhead: NNID NNIV
12
NNID,NNIV RETURN END
Read and write initial condition parameters.
536
Flowchart for Program DYNAPT
App. C
App.C
537
Flowchart for Program DYNAPT
(I) Initial Displacements
Heading: INITIAL VELOCITIES Subhead: NODE VOi V02
If there are no initial displacements, go to 4.
r-I
Heading: INITIAL DISPLACEMENTS Subhead: NODE DO! DOZ
I
r- 1
I I
Read and write initial displacements.
J,D0(2*J - l),D0(2*J)
5 I = 1,NNlV
J, V0(2* J -
I I I I I L __
Read and write initial velocities.
1), V0(2*J)
J,V0(2*J - 1),V0(2*J)
5 CONTINUE
I
I I
J,DO(Z*J - l),DO(Z*J)
I
r--
I I I I I
2
L __
CONTINUE
r--
Set index K equal to ID(J).
K = ID(J)
K .;; NDF
I
1
I I I I I I I L __
6 J = l ,NND
VO(K) = VO(J)
If K corresponds to a free displacement, put VO(J) into VO(K).
I Set index K equal to
11)( I)
I
6
l __
K .;; NDF
DO(K) = DO(J)
lf K corresponds to ;i free displacement, p 111 DO(J) into DO(K).
CONTINUE
c. Applied Actions
3
7
r--
CONTINU E
1
L__
(2) l11i1ial Velocities
If' 1hcrc un.: no ini1i11I vcloci1ic>. go 10 I
8
8 AS(J)
=
0.0
Heading: APPLIED ACTIONS Subhead: NLN NEL
Clear action vector.
ftJff
Flowchart for Program DYNAPT
NLN,NEL
Read and write load parameters.
App.C
r----
L___ _
1
I I I I
PBS(I) = (2.0* BLI + BL3)*EL(l)l 6.0 PBS(2) = (2.0*BL2 + BL4) *EL(l)16.0 PBS(3) = (BLI + 2.0*BL3)* EL(l}l 6.0 PBS(4) = (BL2 + 2.0*BL4)*EL(l)l6.0 JI = 2*J N(I) - l;J2 = JI + I Kl = 2* KN(l ) - I; K2 = Kl + I AS(JI) = AS(JI) + PBS(!) AS(J2) = AS(J2) + PBS(2) AS(KI) = AS(KI) + PBS(3) AS(K2) = AS(K2) + PBS(4)
I I I I
9 l = l,NLN
J,AS(2* J - 1),AS(2* J)
Read and write scale factors for line loads.
l,BLI ,BL2 .BL3 ,BL4
I I I I I
•I
Heading: NODAL LOADS Subhead: NODE AJ I AJ2
r--
l ,BLI ,BL2,BL3,BL4
I I I
If there are no loaded nodes, go to 10.
J = l,NEL
II
1
I
( I) Nodal Loads
539
Flowchart for Program DYNAPT
App. C
Calculate equivalent nodal loads, as in Eqs. (4.10-4).
II CONTI NUE
Read and write scale factors for nodal loads.
d. Ground Accelerations J.AS(2*J -
I I
L __
l), AS(2*J)
12 Headi ng: GROUND ACCELERATIONS Subhead: !GA
9
Read and write acceleration parameter. (2) Elc,ncnt Louds
If' there arc no clements with line loads, go to 12.
[
If there are no ground accelerations, go to 16. I ll•ndin)l: U NE LOADS S11hlw11d : El.EM . IJLI ... BI A Heading: ACCELERATION FACTORS Subhead: GAX GAY
1140
Flowchart for Program DYNAPT
App.C
App. C
•I I
Read and write acccll•1ti tion factors.
GAX,GAY
GAX,G AY
( J) Fill Vecto r GA
I
Flowchart for Program DYNAPT
r---
+ L1 I
14 SUM = SUM
I I I I
with Ground Accelerations
13 I = 1,NN
I
rJI
= !0(2*1 -
I I I I I
L__
Augment AS(I) with the negative of SUM.
15 CONTINUE
= ID(2*1)
Calculate displacernl'III indexes JI and J2 .
17
1
I
Set the displacement index K equal to ID(]).
I JI ,a; NDF
J2 ,a; NDF
I
I I
I); J2
GA(J I ) = GAX
GA(J2)
= GAY
If JI is free, put GAX
into GA(J I).
If 12 is free , pul GAy mto GA(J2).
I I I I I
L_
K ,a; NDF
AS(K)
t
I I I I
AS(J)
lf K is free, move the corresponding action up the list.
17 CONTINUE
e. Read Forcing Function
CONTINUE
N = NLN
+ NEL + IGA;NFO
r ----
'
=
13
(l) Multiply /\ccclcrntions by Mass Matrix
I I I
Multiply mass and acceleration terms.
CMS(J,K)*GA(K)
16
1
I
+
AS(l) "'. AS(!) - SUM
L_____ _
r--
541
=
0
Add the parameters requiring a forcing function. If N = 0, skip reading a forcing function .
Sci lhc displucc11wn1 index J cquul 10 11 )(I) If' J is l'CSlrai11cd, N~lp the 111uliiplit·111it111.
l11i1i11lin· SlJM 11 , it•iu
Heading: FORCING FUNCTION Subhead: NFO
Read and write number of function ordinates.
llnwil11111 tor l'1u111iu11 l>YNAl'I
i
,-•
18 J
Rcud and write sulN·11pl , lime, a11d f'w1c1iou 01<11111111
K,T(K),FO(K)
I
I L-
643
DYNAfJI
21 I
=
l ,NTSS
'
1,NFO
1
I
l11011rt1111
I
I I
lk11d 11111, l •l lN( IION tll
r---..-
I lowoh111t lot
App.C
I I I I I I I I t I
STEP(J) = FO(K)
L __
I
1s
Increase J by J.
(I -
l)* DST EP
Detennine step and ramp for lime increment wilhin a piecewise-linear segment.
21
I I L____ _
r - - -K-.'_r(_ K..L) ,F· 0-(K-)--~
+
RAMP(J)
= DSTEP
22
RETURN END
I'. ( 'nknlntc Step und Ramp for Each Time Increment
r--1
I
20 J
1.NTS
9. SUBPROGRAM TRANOR FOR SUUl'ROGAM NOUMOD
a. Read and Write Number of Modes STl iP(J )
I I
L __
0.0; RAMP(J)
Clear slcp and r,1111p vectors.
20 NMODES CONTINUE
N l'OMI
,---
= 0.0
NFO
22 K
I: J
0
NORMAL-MODE SOLUTION NMODES = (NMODES)
lniti,liizc NFOM I 1111t1 .I
l ,NFOM I
b. Calculate Transformation Operator
1
I
I I
~
·s·_s_·_ < _.,_.
I
I
r----
2 l = l ,NMODES
If' NTSS
'~----
2 J = l ,NDF
0, skip 1111• ~11•11
anti nunp cakul111ion,.
I
•I
C.1icula1c number ol 1i111r
slcp~ for a picccwisl' I incur M .lgnicnl.
llST/\JI
( IU(K
I I)
-i-
l'O(K)J/ NTSS
Culcullllc 1hc nuc ol chnngc ol nnli11u1l·, wilh 1c~pcc1 lo linlL' ,
1
I
•I
lnilialize SUM to zero.
1144
Flowchart for Program DYNAPT
I
App.C
•
r-- -
I
•I L' I I
SUM = SUM
+ PHl(K ,l)*CMS(K,J)
Calculate ;
1
=
IM.
2
-,L__
L - - -
App.C
TR_o_r_u_.J..,.. l _= _s_u_M_-..J
l
I I I I I
Flowchart for Program DYNAPT
r--1
5
L
SUM = SUM
L _____ _
I I I
I
AN([) = SUM
I
I I I I
•I
RETURN END
COMMON.
Initialize DSUM and VS UM to zero.
4 I = I.NMODES
1
I
+ TROP(l ,J)* DO(J) + TROP(l,J)* VO(J)
Premultiply DO and VO with TROP, as in Eqs. (4. 3 ~).
I I
OM EGA = CV(I)
( I ) Response Co nstant s
I
•I
l)NO(I)
DS UM; YNO(l) = VSUM
I
Put DSUM and VSUM i1110 ONO an VNO.
4
I
I
(1
I
I I
I I
1,NMODES
For a rigid-body mode , go to I.
I I
I
cl. 'lh 111Nfor111 A1>pllccl Aclfons
Put CV(!) into OMEGA.
I I
3
L __ _
l __ __ _
OM EGA2 = OMEGA*OM EGA OMEG AD = OMEGA* DSQRT(l.0 - DAMPR* DAMPR) CN = DAMPR* OMEGA: EX = DEXP( - CN* DT) C = DCOS(OMEGAD*DT); S = DSIN(OMEGAD*DT) C l = EX*(C + CN *Si OMEGAD) C2 = EX* S/OMEGAD; C3 = ( 1.0 - C l) /OMEGA2 C4 = (DT - C2 - 2.0* CN* C3) /(0MEGA2* DT) CS = - OMEGA2 *C2; C6 = Cl - 2.0*CN*C2; C7 = C2: CS = C3/DT
For a vibrational mode, determine the constant's CI through C8 given by Eqs. (4 . 10-3).
•I
1
I
•I
-r AF equivalent to AS in
IO. SUBPROGRAM TIHIST FOR SUBPROGRAM NORMOD a. Calculate Response for Each Mode
r------
DSUM = DSUM VSUM = VSUM
r- ---
Put SUM into vector AN.
Put SUM into the transformat ion operator TROP.
r---
•I I'
Premultiply vector At with
4 I = l ,NMODES
DSUM = 0.0; VSUM = 0.0
I
+ PHI(J,l)*AF(J)
6
c. 'lhmsfonn Initial Conditions
,r----
545
l11 i1i11litl· SlJM 10
11•1, 1,
I I
I I
Cl
C5
1.0; C2 0.0; ('6
DT; C3 1.0: C7
DT• DT/2.0: C4 = DT * DT/ 6.0 C2; C'8 C3 !l)T
For a rigid-body mode, use contants from Eqs. (4 .9 9) and (4.9- 10).
546
Flowchart for Program DYNAPT
App, C
Flowchart fo r Program DYNAPT
App.C
54'
(2) S1cp-hy-S1cp Rc,pon,c
•I
2
I I
DNJ = DNO(I); YNJ = VNO(l)
I
lni1ializc DNJ and VN.1
12. SUBPROGRAM NUMINT FOR PROGRAM DYNA a. Read and Write Integration Parameters
r----
I I I
ALPHA.BETA,GAMMA
1
•I
I I
I
•L___ _
I I I
DN(l,J) = Cl *DNJ + C2*VNJ + C3•AN(I) *STEP(J) + C4*AN{l)*RAMP(J) VNJ = C5*DNJ + C6*VNJ + C7*AN(I) *STEP(J) + C8*AN(J)*RAMP(J) DNJ = DN(l,J)
I I
Calculate modal n:sponscs using Eqs. (4. IO I l and (4. 10-2).
DIRECT NUMERICAL INTEGRATION ALPHA = (ALPHA) BETA = (BETA) GAMMA = (GAMMA)
3
CONTINUE
I
b. Calculate Structure Damping Matrix
4
L _____ _
(I) Delcrminc Factor SA
r-----
,
RETURN END
I II. SUBPROGRAM TRABAC FOR SUBPROGRAM NORMOD
r-----
1
2 J = 1.NTS
I I I
I
r-----
I
1 Initialize SUM to , cro .
I
r--1
SUM = SUM + PH l(K,l )*DN(l,J)
Back-1ransfonn cfo,phtn• mcnt, with Eq . (4 I ~)
2
L_ _ _ _ _ DF(K,J) - SUM
IWl'l JRN l•N I>
SUM = SUM + CMS(J,K)*PHl(K.I )
•
I = I.NMODES
1
•L I I
I I
Initialize SUM to zero.
I L
r---
Compute Cl = C)iJ .
•t - - - - -
1
I I I I I
Cl = DSQRT(2.0*DAMPR*CV(I))
I
Place SUM intu thl' ll\'t' di~phtCCllll' lll lllllll IX I)( •
I I
I
L____ _
SA(J.1) = SUM*CI
2
Multiply M by N. Complete factor SA = MNew needed for Eq . (4.6- 17).
548
f l11w11l110 I 101 1'11111111111 l>YNAI' I App. C
Flowchart for Program DYNAPT
549 (2) Mul11pJy SA 111111 II• l\ 1II INfHl1<1•
r-------
r----
1
1
l I I
I
r----
'
I I I I I
I I
r-----
I I L 1
r---
•I L'
I l
3
I I I I
SUM
= SUM
+ SA(K, J)*SA(J,J)
CS(K ,J)
6 SUM = SUM - SS(J ,J )*DO(J) - CS(l,J)*VO(J)
+ Apply Eq. (4.6- 17).
AA(!) = SUM
I
l ______ _
+
AF(l)*STEP(J )
7
l
Set up right-hand vector in Eq. (5 . 1- 4).
= SUM
4
L___ _
[
SOLYER(SA,AA.AO,NDF)
Solve Eq. (5.1 - 4) for inirial accelerarions.
c. Calculate Initial Accelerations
r-
r-
I I
'
I- I I I I
L_
9 I = 1,NDF
1 AO(]) = 0.0
I
If rhe mass matrix is singular, use zero inilial accelerations.
l_
SA(l,J ) "' CMS(I.J)
Copy mass 111a1rix lo SA. 5
d. Calculate and Factor Equivalent Stiffness Matrix FAC.Vf'OR(SA, Nl)/~&8)
F11c1or 11111~s 111atrix. If' ir is si,1guhu; go 10 8. CI C3
= 1.0
+
ALPHA; C2
= 1.0/( BETA*DT*DT)
= GAMMA/( BETA*DT)
Detennine consrants for Eqs. (5.5- 15) and (5.5-33).
680
I l11woh1111 ror l'r our IHn llYNI\I' I
I\JJp, C
I. NI> ~
I
I
t II J
I
1,NDF
I
AA([)
= SUM
I
SA(],J) = Cl*SS(l,J) + C3*CS(!,J)
+ I L __ _
+
C2*CMS(l,J)
&1s.(5.5 15)111111 (\, 111 SOLVER(SA,AA ,DD,NDF)
1
I I
I
DF(l, I) = DO(!) + DD( !) AO(!) = AO(!) + C5*DD(I) - Q(I) VO(]) = VO() ) + C6*DD(I) - R(I)
I
CI = 1.0/ (BETA*DT) ; C 2 = 1.0 / (2.0*BETA) C3 = GAMMA / BETA; C4 = (GAMMA*C2-l.O)*DT C5 = Cl/OT; C6 = GAMMA*CI
(5.5- 13). f. Calculate Responses for Subsequent Time Steps
Q(J) = C I* VO(J) R( I) = C3*VO(I )
+ +
C2*AO( l) C4*A0(1)
r-------
Obtain Qand 1 Vl'l'lui. from Eqs. (5 .5 10) and (5 . 5- 1 I ).
12
r-------
,---
'
I I I I I I I
I I
I I
•I r----IJ K
I. ND!'
1
L{
SlJM
.~lJM I C'MS(l ,K)' Q(K) I C'S(I,K)• R(K)
t
Add tc 1111,
lllllSS
lllld d11111pl11p
111 llq . (~ , 111)
I I I
I
1
+ I I
=
Q(l) C l *VO( I) R(I ) = C3*VO(I)
L __ _
I
r --- --
15
L __ _
Dete rmi ne cons1u111s 1111 Eqs. (5.5- 10) 1h101111h
Apply Eqs. (5.4- 16), (5.4- 17), (5.4- 18), (5.5- 12), 1111d (5 .5 13) 10 find lhc 10111! rcsponsl!s.
12 I = l,NDF
I
L__ _
Solve Eq. (5.5- 14) for the incremental displacements.
r---
Factor cquiv11lc111 still ncss matrix . If 11 I, s ing ula r, go to IOI ,
c. Calculate Response for First Time Step
I I I
Complete the calculation of terms in Eq. (5 .5- 16).
CONTINUE
Calcul111c l!qulv11ll1 111 stiffness n1u11 ix hy
II
FACTOR(SA ,NDF,& 101)
r---
+ AF(!)*RAMP(I)
14
L______ _
I I
551
l
,---~ L ____
I lowol11111 for Program DYNAPT
+ +
C2* AO(!) C4*A0(1)
Use Eqs. (5.5-)0) an9 (5.5-J I) to obtain vectors Q and R.
552
Flowchart for Program DYNAPT
App.C
App. C
553
Flowchart for Prog ram DYNAPT
13. SUBPROGRAM RES2PT* FOR PROGRAM DYNAPT
a. Read and Write Output Selections
tI tI I I I I I I I I
t
I I
I
r----~ 17
I
L
I
r~~~~~~~~-'--~~~~~--, SUM = SUM + CMS([ ,K)*Q(K) + CS(l ,K) *R(K) + ALPHA*SS(l ,K)*DD(K)
I I I I
AA(l) = SUM + AF(l)*(STEP(J) + RAM P(J) - STEP(J - I) - RAMP(J - I))
L _____ _
Heading: OUTPUT SELECTIONS Subhead: !WR IPL NNO NEO
Apply Eqs. (5.5- 16) and (5.5-34) to de termine the incremental load vector.
IWR. IPL, NNO,NEO
I I I
(JNO(J), J
t
I I I
SOLVER(SA,AA,DD,NDF)
Solve Eq . (5.5- 14) for the incremental displacements.
(IEO( l), I
DF(l,J) = DF(l,J - I) + DD(l) AO(I ) = AO(!) + C5*DD(1) - Q(l) VO([) = VO(l) + C6*DD(I) - R(I)
IL __ _
=
Read and write nodes selected.
l ,NNO)
NODES: (JNO(J), J = l ,NNO __ ) __ 6
,r--I I I
6
CONTINUE
I I
6
18
I I I
Read and write output parameters.
IWR ,IPL,NNO,NEO
=
Read and write elements selected.
l,NEO)
Use Eqs. (5.4- 16), (5.4- 17), (5.4- 18), (5. 5- 12), and (5.5- 13) to o btain the total responses.
ELEMENTS: (IEO(l), I = 1.NEO)
6
19
CONTINUE
I
7
I
l ______ _
20 I , write parameters
to plotter.
b. Transfer Displacements from DF to DS
RETURN
IOI
If equivalent sti ffness FACTORIZATION OF EQUIVALENT STIPrESS MATRIX f'AILED
matrix is singular, write this message; and stop cal cu lat ions.
,---1
t STOP END
4 I = l ,NND
I
Set K equal to ID(l).
I •For convcnicm·c, i mplied itcrntivc control stntcmcnts arc u,cd tn tlm ,11hp1u11n1111 l.011ic11I output unit, urc 6 f'or printer (ltltl / IOI (lhlllt'I
064
Flowchart for Program DYNAPT
•I •I I I
Fl owchart for Program DYNAPT
Otherwise, put zero into DSO (for a restrained displacement).
DSO(l) = 0.0
r!WR = I
1
L_
DS(I.J) = 0.0
I
If IWR = I , write initial displacements.
For a free displacement, put DO into DSO.
DSO(I) = DO(K)
,---
,-
I I
1
L_
I
DS(l,J) = DF(K,J)
Also, put OF into OS.
4
L _ __ _
I I I I
I I l".
Write Displacements
,---1
I
K
I
J
The node number K is fo1111d , parameter KO is comp111cd, and the number of r'Cst111i111., is obtained.
JNO(KK); KO = 2*(K - I) NRL(KO + I) + NRL(KO + 2)
•I I I I
I
i I
6
TIME,(DSO(KO + K), K = 1,2)
7
2
I
I
J,TIME,(DSO(KO + K). K = 1.2)
Also, put zeros into OS.
IPL = I
I
•I
Initialize J and TIME to zero.
J = O; TIME = 0.0
I
I
If IPL = I, write K to plotte[
IPL =
•I I
555
free displacement, go to 2.
I
I
App. C
If K corresponds to a
I I
I
App.C
If J = 2, there arc 110 displacements al node K; so 11011c arc writhlll.
2
llc11di111:1: l)ISPLAC'EMENT TIME I II S'J ORY FOR NODE (K) Suhh,•ud : S'l'lll' 'l'IME DJ I D.12
6
I I I I I I
I IL ___
,--I I I
I
I I
I I I
I
Initialize MAX and MIN terms.
5
8 J = 1,NTS
I
•I
I I I
5 K = 1,2
DMAX(K) = DSO(KO + K) DMIN(K) = DSO(KO + K) TMAX( K) = 0.0; TMIN(K) = 0.0
I
I I I I
Similarly for IPL = I.
I I
• I I
TIME = DT*J
Compute TIME al step J.
r-I I I I I I
I
DS(KO + K. J) < DMAX(K )
DMAX(K) = DS(KO + K. J) TMAX(K) = TIME
L __
6
If displacement DS go to 6.
< DMAX ,
Otherwise, replace DMAX and TMAX with current values.
111111
1 lowuhut I lo t l't or,1111t1 l>YNAl' I
I I I
I I
I
•I I
I I I I
7 K
I I I
I
DMI N(K) TM JN(K)
Otherwise, 1q il11l'1• 111\ II N 11nd 'l'MI N w ill1 1•11111•111 value,.
DS(KO+K, J) 'J'JM E
IWR
I
J,TIME.(DS(KO I K, J), K
J- .---------.J 11'1.
I
t
I L __ _
1,2)
If IWR
[WR = I
TIME OF MINIMUM (TMIN(K), K = 1,2)
I
I
6
Write times of minima.
9
d. Calculate and Wr ite Axial Forces
J
h
I, w,IJ1• d lN pl111111 11111
TI ME,(DS(KO + K , J), K
1,2)
Similarly for 11'1 ,
8
r-- - - I I I I
t
CONTINUE
14 II = l,NEO
l = IEO( II)
!WR = I
MAXIMUM (DMAX(K). K
1,2)
(J
Wri1c 111aximu111 d1 Np h1, 1 rncnt,.
11'1.
(DMAX(K), K
1,2)
7
Sim ilarly l
I
Write 11n1c~ uf 11111, 111111
J.
Heading: AXIAL FORCE TIM E
I
IWR
[ I WI{
Find dc111c111 1w111hc1 I .
HISTORY FOR ELEMENT (I) Subhead: ST EP TIME AM I
I
I
I
I
+
1
I
' iI
7
CONTINUE
I
I I I I
1,2)
Similarly for plotter.
L____ _
I I
t
=
I
I ll\llN
I
I I
I I I I
(DMIN(K), K
I If' displ11t'l'IIK'1II I >S go 10 7.
DS(KO I K, J) -> l)MIN(K)
I I
I
IPL= l
I
L__ I
I
557
I
t
I I
Flowchart for Program DYNAPT
1,2
+
I
App. C
t
,r--
I
Ar,11 (
I
~ I NIMl lM ")M IN(K), K
L
J
l.2)J'1 W1111•1111111r1111111cllspl1111 11111111N,
I I I I I I I
+ I I i I I I I
t I I
7
IPL = I
JI= 2*JN(IJ - l ; J2 ~ JI + I Kl = 2*KN(l) + I; K2 = Kl + I SCM = E*AX(])/ EL(I) J = 0: TIME = 0.0 AMO = SCM*((DSO(K I) - DSO(Jl))*CX(I) + (DSO(K2) - DSO(J2))*CY(I))
=
IWR
I
Write I to plotter.
Compute displacement indexes , and so on.
Calculate initial axial force.
6 If !WR = I , write initial axial force.
1 IPL =_._ _ _~ -T-IM_E_,,AMD 1
: imilarl y for IPL = I.
6
668
Flowchart for Program DYNAPT
App. C
Flowchart for Program DYNAPT
559
t
• I
AMDMAX TMDMAX
I
t I I I I I I
App<
= AMO; AMDMIN = AMO 0.0; TMDMIN = 0.0
=
I
Initialize MAX and MIN terms.
r---1
I I
I + t
TIME = DT *J; AMO = SCM*((DS(K 1,1) - DS(J I ,J))*CX(l) + (DS(K2,J) - OS(J2,J))*CY([))
Calculate TIME and AMI>
!WR = I
I
I I I I
If axial force AMO < i\MllM \ go to 11.
! WR = I
AMOMAX = AMO; TMOMAX - T IME II
If axial force AMO · /\M l >t-.ll N go to 12.
AMO > AMOM IN
AMOM IN = AMO: TMOM IN = T IME
M INIMUM(AMDMIN)
I I
6
Write minimum axial force.
+
7
IPL = I
I
Similarly for plotter.
I I
Otherwise , replace AM l)M/\ \ and TMDMAX with cu11 ,•111 values.
6
Write time of maximum.
I
AMO < AMOMAX
TIME OF MAXIM UM (TMOMAX)
!WR = I
I I I
TIME OF M INIMUM (TMOMIN)
6
Write time of minimum.
L _____ _
Othe1wise. replace /\MDM IN and TMDMIN with Clllll' ltl values .
14
CONTINUE
RETURN END
12
!WR=
14. SUBPROGRAM DECOMP* FOR SUBPROGRAM STASYM I, write axi11I 111111· IPL = I
•I I
I
L __ _ _
Similarly for IPL
I
•I
I,
13
IWR
L ____ _ I I
CONTINUE
t M/\X IMUM( /\MDM/\ X)
6
I I
I
I
I I
3 l = l ,NDF
1
t I
r -----
7
Write 111uxi111u111 11x111I lrn, 1 IPL
(/\Ml)M /\X~
--r
7
I
•
SUM = A(l,J); Kl = I - .t
r---1
L_
I I
SUM = SUM - A(K,l)*A(K ,J)
Si111il11rly !or pli1111•1
•Sec Co1111i1111•r
l'ro111·11111.1'.f111· S1r111·111n,I /\1/(//1w/,1, hy W, Wcuvcr. Ii , ll V1111 Nost1 aml , l'1111t·1·to11 , N J , 1%7
Arguments: (A ,NDF, *)
11110
I lowoh1111 for flroornm IJYNAPr App. C
Flowchart for Program DYNAPT
t I I
561
t
I
I
SUM = 0.0: JMI = J - I
I
I
I I I I
A(l,l) = DSQRT(SUM)
I
+
I I I I
r--1
L
2 SUM = SUM - A (KJ)*A (K,J)
I
L__ _
A(J,I) = SUM*A(J.J)
2
A(l,J) = SUM /A(l,I)
+
RETURN
I L ___ _
END CONTINUE
16. SUBPROGRAM FACTOR* FOR SUBPROGRAM NUMINT RET URN END
A(I . J) ,s; 0.0
15. SUBPROGRAM INVERU* FOR SUBPROGRAM STASYM
r-
I I = 1,NDF
ArgumcnJs: (A,NDI •)
I A(I .I) = 1.0/ A(l .1)
NI
r----I
lI
L ___ + I
•11,111
1
I
I
L
,-----
3 I
NDF -
l,N I
I
I I
r----
I
1
+ I I I I
IPJ
.I J
I I I
11'1 ,NI)!'
I
I
I
I
I I I
+
+
SUM = A(l ,J); II = I - I
I I I
SUM
I L ___ _
= SUM
- A (K ,l)* A (K ,J)
2
A(l,J)
= SUM
I
•Sl'\' M111r/1 ll1111/1w/.1 11{ F ri11111•r/ .\'11111 11111•1, 211d ,·d • by W W1•i1Vl' I, Jo 1111,i J. M (ll•H·, V1111 Nos111111d 1{1•1 11h11Jd , N1•w Y1 11 i-, l'IKIJ
A rguments: (A ,N , *)
562
Howchart for Program DYNAPT
App
c
App. C
Flowchart for Program DYNAPT
I I
SUM = A(J,J)
+ I
! r--'II
I I
TEMP = A(K,J)/A(K ,K) SUM = SUM - TEMP*A(K,J)
I I IL__ _
3
I I I
1 -.- 3 I = I .N IL _ 3 __.____ _
A(K,J) = TEMP
X(I ) = X(]) / U(I.I)
I tL ___ _
,----
4 A(J,J) = SUM
'II
RETURN END
l = N - 11 +1 SUM = X(I) K2 = I + I
I 17. SUBPROGRAM SOLVER* FOR SUBPROGRAM NUMINT
I I
t
r----
A rguments: (U, B,X ,N)
2 I = l,N
1
I I I I I I I
t
SUM
I -
I
r--L
4
L
SUM
L ___ _
= SUM
- U(I. K)* X(K)
5 X(I) = SUM
RETURN
1
END SUM = SU M - U(K, l)*X(K)
L ___ _
' Ihid
= 8(1); KI =
I I I
,r---
2 X(I)
SUM
56:i
Ar1ewo1 11 tu l'r uhlome
565
6 2.3-7. D = \ {j()8 sin flt
Lf, 88g G cos 1271T4 2 2.3-10. D = d cos flt 35 2.3-8. D
2.3-9. D = ~; d sin flt 2.4-1. u =
Answers to Problems
13f sin (flt -
2.4-3. u,. = e
1
_ _ ( 2nfl ) 8 - tan 1 w 2 _ ff
- 8);
-n,(
M fl
\N cos wdt +
+ Nn
wd
Pl 3 2.4-4. D = 0.01445 sin(flt - 8) El
mal3
~
2.4-6. D = -3. 185 2.5-1. P (t) =
A
ut
8 = tan- C)~fln2)
8);
ma . - f3T sm (flt
2.4-2. u* =
*=
. \ sm wdt) Ml 2.4-5. 88 = 9.086 ~G cos (flt - 0) 7Tr
cos(flt - 8)
2.4-7.
/3rcs = 50, 25, . . . , 2.5
;:Pi(sin flt + i sin 3flt + . . ) P (cos flt - j cos 3flt + .. )
2.5-2. P (t) = ;
1
CHAPTER 2
2.2-1. 2.2-2. 2.2-3.
/3ii 1 /3ii T-- 2 L -Vm /mL w = L 't;;:;i;; ! = z1rL -v-;;i> 4 /3ii 2 /3ii 7Tl /mL w = L -V-;;;i; f = 7Tl -V-;;;i ; T = 2 V3Ei - l /3EI . - l /3EI. T -- 2 L V(2;i, w - L Vi;;:;i, . t - z1rL Vi;;:;i, , w 1
1r.
M£.
°V-;;;E ; f 2,-2
2.2-5. w
=
2.2-6. w =
1
{3;G
r2
u V2pA ; f = 7Tl /3ii
8
L -V-;;;i ; f
2.3-1. u = [
1
=
11114
9
5
d sin
!lt
/mL
{3;G
7Tl2
J
ma I 2.3-2. u * = - [ _ (fllw )2 k 1
2ML cos 2.3-4. 80 = - - 31rr 4 G
ma1.' . cos 45111
nt
~2 - ~1r (sin flt
a0 2 5-5 u = • . k
+
/2pA
V2pA ; T = 7 V~ 4 /3ii 7Tl /mL 7Tl -V-;;;i ; T = 4 V3Ei
2.3-6. 0 "'
2.5-4. P (t) =
1r
~
L.J
1• 1
flt - ! sin 2flt + ! sin 3flt - ...)
2
3
+ !sin 2flt + ! sin 3flt + ...) 2 3
a1 cos (i flt - 8;) + b, sin(iflt - 8,) k \!(1 - i 2 fl 2 / w2 ) 2 + (2yifl/ w) 2
Pi
2.5-7. u = 13.06 k cos flt
. flt 2.5-8. u = -6.2 1 Pi k sm
T = 21r°VAE.
2
_ (~ /w) 2 ] d cos flt
PL 3 2.3-3. D '""' W sin flt 144 2.3-5. /)
ME.
= Z7r °V-;;;E ;
3_ Pi (sin
2.5-6. u = 6.70t sin flt
1T.
2.2-4. w =
2.5-3. P (t) =
n,
t'OH
Ht
2.6-1. u =
kPi(· sm flt
2.6-2. u =
t
2.6-3. u =
~ [w t, kw
2.6-4. u =
u=
P,
[1 - e- n,(cos wdt + : d sin wdt)
k (1 -
t
fl sm . wt)\[ _ (D,/ 1 w)2 - -; 1
J
J 2
2
t - 2n + e- n,(2n cos wdt - wl - n sin wdt\ ]
cos wt)
[cos w(t - t,) - cos wrl
Wd
}
(0 st s ti )
n1111
At111wor11 to Problems
I',/ 1
co.~ 1,1 (t
k
7Z lcos w(t
11 -
-i
I I
1,)
Ml
cos
11)]
u
t,
= -Pi [ - cos
wt
(0 •
wt,
sin wt - sin w(t - t ,)] +-------
wt,
k P, 2.6-6. u = -(1 - cos wt) k
[ u >I< = - a1 - 1 w2
(t
w
= - :: [cos w(t -
!1
- -2w
wt 1
2
2.6-9. u* = - ~2 [ ~ - - 2 -( t - cos u*
=-
d
w
= 0
11 10
I ii
(t 1 •
I
,,1
(t I
tr
2.7-5.
112() .., •
2,7-fl.
//1 0
2.546
Pi
k
0.3 186t
U1 0
= kPi
3.4-2. pb(t)
= {1, 3} b2L
3.4-3. K =EA[-; ~: -~] 3.4-4. M = p1sL[ : 3L 1 -8 7 -! mx1L 3.4-5. pb(t) = {I , 4, 1} bxL 3.4-6. Pb(t) = {5, 3}
6
rl
Pb(th = {6x
3.4-11 . Given I)
2
-
:x
(0
Wl1
wt +
/1)
COS W
I
I)
wt)]
(0
2.7-3. Given
"'!:. I
U1 0
2.7-JO.
11 10
I iJ
pAL pAL ; etc. M,11 = 3.28 - -; M,12 = O; M, 13 = 0.64 6 6 3EA EA EA L ; S,12 = - L; S, 13 = - L; etc. 3.5-3. S, 11 = 2 2
I)
4pAL _ . _ pAL . M,1 1 = - -; M,12 - 0, M,13 - - - , etc. 3 6 EA EA 3.5-4. S,11 = L; S,12 = O; S.,13 = -L; etc.
/1 )]
2.7-4. Given
2.7-7.
O; M,13 = 0.06 pAL; etc.
EA EA EA 3.5-2. S,11 = I.64L; S,12 = 0.48T; S,13 = -L; etc.
(t - ( >]
Pi 0.9 135 k I'1
-~] 2
60
EA
=
~
3.4-8. pb(t) = {9, 2L, 21, -3L} b2L
3.4-12. Given
= 2. 72 p~L; M,, 2
12
12
6xL, 3x 2L - 4xL2 + L3, - 6x 2
EA
M,11
P, U20 = - 4 k Pi 2.7-9. 1110 = 1.379 -( 2.7-6.
6
3.5-1. S, 11 = 0.64T; S,12 = - 0.48L ; S,13 = O; etc.
a ~{ w; ,,2[cos wt - cos w(t - t,) + wt, sin w(r w
2, 7-2,
I)
I
+ cos w(t - t,)} l.7- 1.
( r1
t 1) - cos wt]
a 1 - ao [ sin w(t - t ,) - sin
w
/ 1)
(0 •
t 1)-- -sin +sin-w(t - -- - wt] wt,
<.1.>2
+ 2b2} L
(0 s I · I i)
Wt1
2.(,-8. u* = - ao - (I - cos wt) - -a, -- 2 a- 0 - - -sin- wt) u*
I
(I i
2.6-7. u* =_ a~(!._ _ sin wt)
V1
{2b 1 + b2, b,
3.4-7. pb(t) = {L - x, x}
- ti)] u = -P, [ 1 - cos wt - -t - - t, + -sin -w(tk t2-t1 w(t2- ti) [ sin w(t - t1) - sin w(t - t2)J u = P, - - cos wt + - - - - - - - - - k w(t2 - t1) W
=
II
(IJ
sin wt) 2.6-5. u = -Pi ( l - cos wt - -t + k
CHAPTER 3
3.4-1. pb(t)
[cos w(t - t2) - cos w(t -
567
pAL pAL M,11 = - -; M, 12 = O; M,13 = 0.64--; etc. 3 6 EA EA 3.5-5. S,1 1 = I.64L ; S,12 = 0.48L; S,13 = O; etc.
I. 102 ,..
M, 11 = Zp:L ; M, 12 = O; M,,3 = 0.06 pAL; etc.
ti68 An11wurn to f'tuhl1t1111
EA 3.5-6. S, 11 = U; S,.,2 = - l; Sm == O; etc. 2 4pAL _ . _ pAL. M, , I = -3-; Ms12 - 0, M,13 - 6 ' etc.
Answers to Problems
3EA
3.5-7.
=
S,11
M,
1
'
4EJ T;
S ,1 2
pAL
T2EJ ; Ssl 3 =
=
3
8EJ T;
3.5-8. S, 11 =
s,,
=
M sll
3.5-10.
=
S , 11
S , 12
6E! u;
=
13pAL. M i2 35 , s
8EJ
T;
S , 12
W 2.4
3.7-4. (a) w 1
=
__ pAL 3 . 140 , etc.
-u;
=
Ssl3
2EJ
T;
13pAL2; etc. 420
_ 0.5412, L 1.307
@. Vp'
= 0.9393~
/E ; 1 = r-21 yp
W1,2 -
3.6-3.
W1 ,2
3.529
3.6-4. w,,2 = 0.6049, l 1.815 3.6-5.
W1.2
5.546 , 21.04 = L2
3.6-6.
W1 , 2
= 15.14, L228.98
3.6-7.
W1,2
==
<1>1 -
1 _ [ -
<1>2
<1>1 -
<1> 2
81 .98 /El ~ ypA.;
h . . modes ape 1s unity.
==
,.2
v;x;
[OJ 1
W1 ,4
(b)
W2,J
=
l.271, 1.927 L
3.7-7. (a)
W3,4
==
1.352, 2.218 L
W1,2
=
0 , 1.225 l
0.6650, 2.909 L
==
/E.
Vp'
_ [ I -
1 ] . == [ l 0.8801 ' 4 -0.6091
1
/E. Vp' /E. Vp'
4
-[
2
-
3 _ [ -
1 ]· 0.08687 ' I
] .
-0.3242 '
l] . _[
/E ; 1 = [J yp
_ [ (9 sin w1 t)/ Wt + (sin w2t)/ w2 43 . -1. D - (3 sin w,t)/ wt - (3 sin w2t)/w2
=
, 2 -
<1>
= [
3
<1>
== [
4
1 ] -0.5
. = [
'
2
I
1
r 0.1m, , 1.1· ,,)
10
J
_ [ 1.207 COS W1 t - 0.2072 COS Wz( d 3 4.3- . D - 0.5000 cos Wt t + 0.5000 cos W2f
]
0.707 1
- r c-2 sin W1t)/w1 + (2 sin W2t)/w2]Dot 4 4.3- . D (sin w, t)/w, + (4 sin w2t)/w2 5
81.98
c,,
JDoi
_ [I.961L cos w, t + 0.03923L cos Wzt] PoL2 43 2 • • • D 2. 701 cos W1 t + 0.2990 cos W2t 6EJ
[1]1
/El w2 = ~ ypA.
{ij
==
CHAPTER 4
= r-31 1
L ] . = [ L ] -0.3947 • 2 - 26.6 1
(b)
53 .28, 139.2
3
1
_ r]J 2
/El ~ ypA.; 22.74
-
; i -
1 -
3.7-2. (a) w 2 =
w, .J
/El. vpA.,
-[l]· o '
2
3.7-6. (a)
(b)
1
_ [1]. 3 '
/E. vp'
/E. yp'
2
/El. -[ -II]·, V04.· 13.32, 34.79 /El . _ [ l ] L2 V04. · -0.707 1
3.7-1. (a) w1 =
(b)
'¥1
27 cos w, t - 3 cos W2t] Pol 4.3-5. D == - [ 81 cos w 1t + cos Wzt 18EA
1
0.1m ,
1
.
3 (l.015L sin w, r)/w, - (0.0J506L sin w2r)/w2 1 D0, 4. -6. D == [ ( 0.4006 sin
, .
1
1 ] 0.10n
J @.== [ 1.344 l ]·= [ 1 J == 0.6581, L 1.974 Vp' ' -1.092
= r2.414J 1 '. = r-0.4142 1
,I>
== [
/E mode shape 1s. uruty. . T1 134 yp;
3.6-1. Answer not provided. 3.6-2.
J 16. 86
W1 , 4
etc.
3 2pAL 13pAL2 . __ pAL3 . M - ,· M sl 2 = - 420 ' Ms 13 sll -- 105 140 , etc.
<1> 4
e
== [
3
(b)
u' etc. -
L2
3.7-5. (a) w2,J == 1.485,L 1.732
M , 13 -
llpAL2. Ms1J = 210 ,
16.64, 95.21
2
(b) w 2 =
_ 6EJ. Ssl3 -
6EJ
=
(b)
6EJ 2EJ u; Ssl3 = T; etc.
Ss12 =
12£/ u;
=
I
. 0, etc.
_ M,13 -
3 2pAL 13pAL2. Ms11 = ~ ; M,12 = - ~ , 3.5-9.
w1 , J
O; etc.
pAL3 .
=105·' M,12 = ---, 140
{ii. i = [ -0.5868 e J ; cf> V04.• 53.28, 139.2 /El. = [ I J. = L2 vpA, -0.1011 • 0.9258 /E . · == ~ yp; mode shape 1s unity. ==
3.7-3. (a)
4.).7,
I) •
/
: / (} u
~·os 10,1
l
-7.365 1 2.055
]
J
670 Answers to Problems
4.3-8. D = [ 2.707 cos -1.914
COS
1.293 cos w 2 t Wit+ 0.9142 COS Wzt
Wit+
0
14£/
3
4.4-9. D t
_ [ (1.000 - 1.221c, 4.5-2. D - 1.681c, _ [ - 1.000 453 -1.000 • • · D-
[
/
\
0.1465(t - ~i sin w1~/wt
=[
+ 2(1 + 4(1 -
/3,/wt + 9{3z/ w~] Pi cos flt 3{3,/wt - 3{3z/w~ 10m
.!:_
sin w 2 ~/w~ t,m
-
- 2b2 4b
I -
2
m =
1
2
t - wz sin w2t / w2
+ 0 .2205c2)L] ~ + l.68lc2 L
4 •4•7.
I)
rJ
I /31Pi L sin flt - I 88wrm
- 0 .05776(1
4.4-8.
I) •
(
r w lu•11• 111
0.04084 r
where
+ 0.1816(1 -
m = pAL 420
wheres= EAIL ; mi = pAL/6; m
= 3.28m1;
_ [(-1.000 + 1.053c, - 0.0532c2)l] (J 456 -1.000 - 0.4156c, + l.415c2 ' • • • D where
3.28 pAL 6
Wi
cos w 2 t)/w~
I W2
1 (tJ1
/IA/
•I '0
I
. SIil
CHAPTER 5
m
5 .2-1. and
Pi
5.2•2,
U10
= 0.9819k
5.2-4.
Ut 6
= 0.3622k
5.2-6. u20
W1 f ) / (1/11
5.2-11. Given
A
=
/\I J
sin
c1 = cos wit c2 = cos w2t
10,,)/1,d
111/I
5.2-8. 5.2- IO.
P, - 4.004k
111 0 ""
P, 0.3 146k
II 111
I , 107
I'1
1.
Pi
5.2-3.
Uzo
= -3.063k
5.2-5.
U20
= -2.595k
A
P,
5.2-7. u,o = 0.9182k 5.2-9.
U1 0
bi
= /3i/wt;
b2 = {3z/wi
c1 = cos w1 t c2 = cos W2t
b2 = /32/ Wi
o.05776b1 - o. t882b2J so, . fl 458 • • • D = [ -0.04084b 1 - 0.133lb2 --;;;- sm t b, = {3i/ WT; b2 = {3z/ Wi
3
w,,)!wr+ 0 . 1882(, ,,'., sin ,,,,,)/M t I 0, I 1.l 1(,
_!_ sin
c, = cos w, t C2 = cos W2t
5
4.4-6. D = r - 0 .001745L(l - cos w,t)/wt - 0.006828L(1 - cos w 2 t)/w?]Mi 0.0006885(1 - cos w,t)/wf pAL3 where m = - 2 10
sin w,r)/wt )
J~ cos .at
~
cos Wzt)/w'i]_!j_ cos w 2 t)/w~ 10m
where
~i
+ 1.207c, + 0.207lc2 JLO + 0.5000c1 + 0.5000c2 '
4.5-7. D* = - [ J b2 (J, sin flt
2(1 - cos w,t)/wt (1 - cos w,t)/wt h 2.72 pAL were m = 6
4.4-5. 0
J
2
=r-
(r-
(
bi = 13d wt b _ r.i. / 2 2 - JJ2 W2 (0.6s + 3m,n2)b1 - 3(1.8s + mifl2)b2J d . r> 455 D - [ - - sm ~u • • . - 3(0.6s + 3m,fl 2)b1 + (I.8s + miil2)b2 lOm
Wz
+ 0.8536(- ~
where m = pAL
4.4-4. D
* _ [ 2b,b
4.5-4. D -
J_ sin w1 t\ I wt - 0. 353/t - J_ sin w) / w~1 w,
si = s2 =
(b) D = [(0.02675{3i/ wt + 0.2878{3z/ w~)l] M 2 sin flt 0.03683/31/wr + 2.193{32 /wi mL2 pAL where m = ; (Dz)A, = (D,)A 2 210
\'
-
4.5-1. D* - [ -9s, - Sz -3si + 3s2 lOt,
4 .4 •2 • (a) D = [(0.01943/31 /wt + 0.03775{3z/w~)l] P1 sin flt 0.02675{3i/ w,2 + 0.2878{3z/ w 2 ml
O. 353/t -
3
J~
(b) D = [3(1 - cos w,t)/wt - 3(1 - cos w 2 t)/w'i]_!j_ (1 - cos wit)/wt + 9(1 - cos w 2 t)/w~ 10m 3.28 pAL where m = ; (D2)A, = (D1)A 2 6
=_
~ 2 sin w2t)/w~1___!:_ 1 ) z l08t1m 12 t - wz sin W 2 t /w2
(
pAL where m = - 6
4 .4 •1. (a) D = [ 9(1 - cos Wit)/wf + (1 - cos w 2 t)/w~]_!j_ 3(1 - cos wit)/wt - 3(1 - cos w 2t)/w~ 10m
4.4-3. D
+ 24(r-
_ - [t
3
t - (sin w2t)/w2
571
]M, L
[t + (2 sin Wzt)/wz]D01
4.3-9. D =
Answers to Problems
= J .387 kPi
s = s + 3mfl2 s = 2El/l m = pAL3 /420
572
Answers to Problems
5.2-12. (D1)20
= =
5.2-13. (D1)max
=
- 19.94~; (D2ho
-0.9992~; (D2)max s
-19.85~
S'l
Answers to Problems
5.4-4. (8) U15 = 0.3676t (b) U15 = 0.3659t
s
= 0.7950~ s
t
5.4-5. (a)
U 20
= -2.513t (b)
U20
= -2.541
5.4-6. (8)
U20
=
-3.942~ k (b)
U20
=
5.4-7. (8)
Uio
=
0.9087~ k (b) U10 = 0 .9119~ k
5.4-8. (a)
U10
=
0.3191~ k (b)
5.4-9. (a) U10
=
l.372t (b)
U10
=
l.377t
5.4-10. (a) U10
=
J.094~ k (b) U10
=
1 .098~ k
EA
wheres= L
For problem sets 5 .3 and 5 .4 ' parts (a) and (b) denote average- and linear-acceleration methods.
5.3-1. and 5.3-11. Given 5.3-2. (a) U10 = 1.031 5.3-3. (a) U20
=
t
(b)
=
U10
-3.023~ (b) k
=
U20
5.3-4. (a) U1 6 = 0.3676t (b) U16
l.016t -3 038~ . k
s
=
- 2.513t" (b)
U20
=
-2.541
5.3-6. (a)
U zo
=
-3.942t" (b)
U20
=
-3.963t"
5.3-7. (a)
U10
=
0.9087t (b) U10 = 0.9119t
5.4-13. (a) (D1)max = - 0.9974~ (b) (D1)max
s
= 0.7917~
(D2)max
5.3-10. (a)
uw
=
l.094t" (b)
Uw
uio
=
=
l.37?t
- _
l.098t"
-
s
= 0.7860~ s
5.4-1 and 5.4-11 Given
t
5.4-2. (a) U1 0 =
J
5.4•3. (11)
Pi - .1.02.\ k (h)
11 111
.03 ]
(b) U10
=
11 111
s
-
-
_
2
(A tj)1
+ -
-i>1 -
1
-
c
2
'
4
-s;
-
Aj+I = Aj+I +
1 0 lH /'i
A
c -Qj + s _Rj
AtJfr. R 1 = -D1 - MD .. - (At1 )2D..1 1 1 2
At 5.4-17. C = C +-M + ___!S· At1 2 '
1
4
'
-
AA= AA1· + MQ1 + SR1 1
where Q1 = 2DJ; RJ = -AtJDJ
5.4-18. M = M + Atic + (AiY s· AA1· = AA1 + 2 4'
l.0J 6t .
At1
2
where -Q · =
s
(D2)max
_
5.4-16. M - M +
s
= 0.7917~
= 0.7861~
-
Ati
2
Atj
1 5.3-13. (a) (D 1)ma. = -0.9976: (b) (D 1)max = -0.9994~ (D2)max
s
5.4-15. C - C + -,.-M + -S; Ai+t = A1·+ 1 + M Q·1 +SR1· 1.1tj 2 2 , .. flt .. where Qi= - D1· + D1 • R1 = -D1 - ___!D1·
(D2ho = - 19.51 ~
s
2
- 0.9993~
4 4, .. _ 2 . where Qi= (Ati)2Di + Mi Di+ Di; Ri = Ati Di + Di
5.3-12. (a) (D1ho = -20.45~ (b) (D1ho = - 20.32~ s (D2)zo = -19.47~
4
=
5.4-14. S = S + (AtJ1M + tltj C; Aj+I = Aj+I + M Qj + C Rj
0.3 191 ~I (b) U10 = 0.3176t
5.3-9. (a) U10 = l.372t (b)
(D2)max
s
-
=
s
t"
Uzo
U10
0 .3176~ k
=
5.4-12. (a) (D1ho = -21.22~ (b) (D 1)20 = -21.40~
= 0.3659t
5.3-5. (a)
5.3-8. (a)
U10
-3 .963~ k
c -Q·1 + s -R1-
1174
Annwur " to l'rublome
At,
5.4-19. S* '"' S I (A:)2 M I ~lt1 C;
1 •
A1,, I M
Q/
5
I CR/
" 6 D1 · + 2D1; ·· R/ == -;.3 Di -1 2D1 . I At D1 = (A 6 )2 D1 + -;.1 u.tj u.lj u.t1 2 5.4-20. C* = C + AtM +3 2 At1S ; Aj+I * = A1+1 + M QJ* + S RI *
where
Q/
J
2 · 1 + D-· ··1 R -* = - D. - 2At1 1 where Q* = -D D· 1 1 ' A t1 3 1
5.4-21. M *
AfJC + ~ (AtJ)2S ; =M +2
-(At1)2 D·· 6 1
-
= A~+l + CQ* + SRj *
A* j+ I
j
CHAPTER 6
6.2-1. pi:i,)
Lby-{t) pi:k{t) = Lby·(t) 78 18 l92 OJ l92; [OJ [1IL -SL
=
At ·· · R * = -D - At-D. - - --='---D'J (At-)2 .• 1 where Q.* = -D· 1 - -2D 1 1 2
;,
1
3
1
2 5.4-22. C* = C + A M + Atj S; AA/ = AA; + M Q/ + S R/ u.tj 3
Q/ = 2D··1;
where
Pbi(t) =
. - T (At)2 R/ = -AtiDi D ··j
=M +
~jc + (At
=
-ArD· R* J 1, J
where Q* J
s;
AA/= AA1 + C
Q/ + s R/
6.2-2. p;/t)
J
A
1
y
= S + /3(AtJ 2 M + /3 Ati C; 1
A
1
A
A
A
Ai+1 = A1+1 + M Qi + C R 1 1 - 2(3
°
[1~]
J
PPit) 5.5-1. S
=
=[
54 72 ]Lby-(t) 960 -25L
P~f); p;k(t) = [ 5~ -9L
3L
2 = -AtD - (AtJ 2 1) J
=[
55L
2
5.4-23. M*
234]Lby-{t) ; Pbk(t) 312 960 [
-10
to
J
Py,(t)
64\/2 ; p,,k(r)
=[
3V2L
J~f) - 54
54
J
/~,,(I) 64V2
- 9V2L
M,(t) 6.2-3. p~(t) = {O, - 6, -L, 0 , 6, - L}4L
00
where Qi= /3(A tJ Dj + /3 Ati Di+ ~ D i 2
and
y
R1 = /3 Ati D1 A
(1 - "jjy) Di o
(
y) At1D• 1
1-
213 1 5.5-2. C = C + + M + /3 yAt S; A1+1 = A1+1 + M y u.tj 1 • 1-y•• where Q·1 = - - D1 + - D 1 y At1 y
lt + s Rj
A
and 5.5-3.
- D1 - (1 - ~) Mi)1 -
G-
~)(At1)2 D 1
M=
M + y dtjC + /3(Af_j) 2 S; Aj+I = A1+ 1 + C Qj where Q1 = -D1 - (1 - y) At1Di
and 5.5-4.
R.1 =
R.1 =
-D1 - At1Di 1
C = C + -y Atj-M + /3 ydtj S; where Q1 =
;D
1;
R.1 =
G-
+s~
/3)(AtYD1
AA.j = dA; + M Qj + S
-At1Di -
G-
~)(dt1>2 D1
R.1
PM(t) = {-12, -6,
-VSL,
M,(t) 12, 6, -VSL} 4V5L
Lb2(t) 6.2-4. pi:(t) = {O, 9, 2L, 0, 21, -3L}6()
pb(t) = {- 36, 27, IOL , -84, 63, -ISL} Lb2(t) 300 Py,(t) 6.2-5. p;(r) = {O, 9, 2L, 0, 9, - 2L}-9 ' r,: Py,(t) p,,(t) = {-1 8, -9, 2V5L, -18 , -9, - 2v5L} V5 9 M,(t) 6.2-6. p,&(t) = {O, -18, -L, 0, 18, -L}~
M,(t) PM(t) = {18, -18, -V2L, -18, 18, -V2L} 8V2L Lbi(t) 6.2-7. pi:(t) = {l, 0, 0, 3, 0 , 0}12
p,,(t) = {2, - 1, 0, 6,
Lb2(t) 3, 0} 12VS
At1Nwnr" lo l'r oblumu
,,o,
' , ..I • I , p,,' (I ) I'" (I)
,,.J..z.
5/,, IH , 0, 11/ , /Hf 1,/, (I) 192
(20/,,
1>,\(1) ... {O,
44/,, 331., 390} Lb, (!) 960
I5L, 90,
4L , 20, 0, 2L , 7} p;~)
577
6.4-4. p,:(t) = {O, 2, 0, 0, 3, O} Py,(t) 5
PPi (t) = ~PPk(t) = {O, 0.2828, - 0.2828}Py,(t)
6.4-5. ph(t) = {O, -1, 0, 0, l, O} M~(t); L =
V29
PMi(t) = -pMk(t) = {-0.07773, -0.1184, - 0. 1201}M,,(t)
{ 4L , - 12L, 20VW, 2L, 6L, 7VJO} P, (i) 27VJO
1>,,(1)
tt.J.J . ,,;,(,) - {O, 3l, 18, 0, -5l, - 18} My-(t) 16L
l>M(l) ... { 3L, 3l, 18V2, 5l, - SL, - 18V2} My•(!) 16V2l t,..1.4, 1,:1(1)
{12l, 3L, 18 , 4L, - 5L, - 18} My(t) 16V2L
P.11(1) :.:: {9l , 15L, 18V2, 9L - l - l8V2}My{t) ' ' 32l
td-5. 1>J,(1) ""' {12L, -3l, - 18, 4l, 5L, 18} M;)!j_ 16 2l l>M (l) {1 5l, 9l, - J8V2, - l, 9l, J8V2}Mx(t) 32£ ,, . .1.(,. J>,;(I) {O - 3L 21 0 2L 9}lbi(t) ' ' ' ' ' 60 1>,,(1)
{ 9L, - l2l, 105, - 6l, -8L , 45} Lbi(t) 300
(t,J.7, 1>/,(1) ... {O, 1'1· (t) (,.J. ff , 1•J,(1)
'
'
' '
16
{9L, I 2l, 80, - 9l , - 12l 80} P, (t) ' 80 (0 I O O - I O} My•(t) 3 ,
t
'
{2 I O - 2 - I O}M,.-(t)
,,,4- L 1>,;(1)
{O, I, 0, 0, I, O}Lb)"(t) ; L
,I
1
'( )
'•""•• 1>,, I
,
()
,
t,,4.,1. 1>,\(t) I>/'/(/)
'
3\/5
= VJ4
1>114(/)
{O , 0, 3, 0, 0 , 2) fo,,d1)
0.5192}b,,,(t)
- ; L = v 14 6 { 0.3126, 0.2233, 0.49 12}/J,,(,) } / ,/),,(/)
I , 0, 0, I -
'
1>,,,(t)
.
4 l>11A(I) ""' { 0.3043, 0.7 161 , {()
_ { My-(t) PMk () t - 4, 0, 0, 0, - 4, 3}---is-
_ { M,,(t) 6•5•6• PMj () t - 0, -16, 12, 5, 0, 0} - 75 _ { M,,(t) PMk () t - 0, 16, -12, 0, 0, 0}75
'
l>M (I)
' . '
V29
PMi (t) = -pMk(t) = {-0.1332, -0.03805, 0.1327}My-(t) 6.4-7. PMt) = pbk{t) = {- 1.000, 0.4242, O}b(t) Pbi (t) = Pbk (t) = {-0.5992, -0.1141, -0.8988}b(t) 6.4-8. pi:)t) = 4Pi:h) = {-0 .3334, 0.6009, O}b2 (t) Pbi(t) = 4Pbk(t) = {-0.3234, 0.3637, -0.4851}bi(t) 6.5-1. Pbi(t) = {-1.057 , 2.439, -0.4269, - 1.769, - 0.49 13, 1.572}b,,(t) Pbk (t) = {-1.057, 2.439, -0.4269, 1.769, 0.491 3, - 1.572}/JAI) 6.5-2. Pbi(t) = {- 1.601, - 0.4447, 1.423, 0.6523, - 1.505, 0.26'.'ltlfh,,,(1) Pbk(t) = {- 0.3695, - 0.1026, 0.3284 , - 0.2965, 0.6842, 0. 1 l
JL J6 0 3L 16}?,(t)
'
1>1,1(/)
6.4-6. ph(t) = {O, 0, I, 0, 0, -1} M~(t); L =
•
6.5-7. Pbi(t) Pbk(t) 6.5-8. Pbi(t) Phk(t) 6.7-1.
W1
6 , 7 • 2,
W1 2
6.7-3.
W1
=
r,-;
'\(I)
(- 0.4899, 0.2449, 0.2449}/>.,(t)
'
=
{1.5, 0, 0, 0, 0.5, 0.5}bx(t)
= {1.5, 0, 0, 0, -0.5, - 0.S}b,.(t)
{- 0.01111 , 0.02222, 1.028, -0 .3000, - 0.1500, O}b 1(t) = {0.01111, - 0.02222, 0.4722, 0.2000, 0.1000, O}b 1(t)
=
15.69 L2
/El,
-V;;;.
= 5.603,
= 9.941
l2
31.19 2l
-v/El, pA
L
lfil, -
pA
;
=
[ 1.000 0.5435
1.000 -0.9364
J
1118
t,.7-4,
6.7-5. 6.7-6.
An"wnr11 to l't uhln1111
t,)1,1 . , •
W 1,2 -
W 1.2
0, 0, 22.47 /}1 I 1. 648 , 5.540L2
Jn
pA ; (I>
{iii. ; = Y"iA
{iii, = 0.9044, 3.840 L2I Y"iA;
/G] 6.7-7. w, __ 6. 149 Y"iA
u:-
6.1•8 .
[0.8844 1.000
O IJ _ 5.943 /GJ w, - u:- Y"iA =
[ 01
Index
Absolute damping, 166 Anisotropic materials, 311 Absolute system of units, 495 Antisymmetric modes , 112 Accelerations: Applied actions, 4 ground, 21: arbitrary, 42; periodic, normal-mode response, 152 38; rigid-body, 159 Applied body forces , 80 incremental, 212 Arbitrary forcing functions, 38, 172 relative , 23 Arbitrary ground accelerations, 42, Accelerometer, 22 159 Accuracy, 208 Assembly: numerical, 223 of elements, 95 Action equations of motion, 140 of substructures: in componentAction vector: mode method, 470; in tridiagoapplied, 152 nal method, 448 reduced, 282 Average-acceleration method, 219 Actual nodal loads, 98 direct linear extrapolation, 212 Adjoint matrix, 106 iteration with, 203 Algorithmic damping, 218 Axes: Amplification matrix, 223 body, principal, 420 Amplitude: rotation of, 95 forced vibration, 21 stress, principal, 315 free vibration, 16 translation of, 157, 414 suppression, 225, 234 Axial element, 82 Angular frequency: Axial rigidity, 94 dunircd free vibrnlions, 28 Axinl s1rnins, 83 u111l11111pt•d l'it•t• vih1·111 ions, l 11, l O'i co11strni111 s 11uuim.t, NO
580
Axisymmetric loads, 357 Axisymmetric shells: element AXSH3 , 394 nonaxisymmetric loads , 401 Program DYAXSH, 406 Axisymmetric solids, 357 element AXQ4, 359 element AXQ8, 359 nonaxisymmetric loads, 361 Program DYAXSO, 365
Backward substitution, 461 Beams: continuous, 243 , frequency coefficients (table), 288 Blast load, 50 Block mass submatrices, 437 Block stiffness submatrices , 437 Body axes, principal, 420 Body forces, 78 Body-oriented method, 434 Boundary nodes, 446 Bracing in tier buildings, 468 Buildings, multistory, 425
Center of mass, 418, 427, 434 Central-difference predictor, 197 Characteristic equation, 105, 498 Characteristic matrix, 106 Characteristic-value matrix, 141 Cholesky method, modified, 498 Cholesky square-root method, 106 Coefficients: damping, 165 ncxibility, 145 ror Gaussian Quadrature (table) , 328 frcqullncy, 287 , (t(lb/e), 288 lllUSS , 82 st i ffrH.'SS, 82 <'0111pom·11t rnodl' nwthod , 468 IINNl'lllhly of s11h11t1m'llll'l'N, 470
Index
dynamic coupling in, 470 for plane trusses, 471: Program COMOPT, 477 for space trusses , 477: Program COMOST, 477 substructure equations of motion, 468 Computer programs, 9 Condensation, matrix, 282 Conditional stability, 208 Consistent-mass matrix: for axial element, 84 for axisymmetric shell element with nonaxisymmetric loads , 405 for axisymmetric solid element with nonaxisymmetric loads, 364 for element, 82 for element AXQ4, 359 for element AXSH3 , 401 for element H8, 347 for element PBQ8, 378 for element Q4 , 336 for element SHQ8, 389 for flexural element, 91 for grid member, 251 for plane frame member, 246 for plane truss member, 103: in component-mode method, 474 for space frame member, 261 for space truss member, 255 for structure, 98 for torsional element, 87 Consistent systems of units (table), 496 Constant-acceleration method, 219 Constraint, modal, 500 Constraint matrix, 291, 372, 382, 394 Constraints: against axiul strains, 290 rlgid-body, 4 I I , 427 C'ontinuu: Hll'l'SSl'S
and Hll'HillN in , J 10
Index
two- and three-dimensional, 310-69 Continuous beams, 243 loads , 265 Program DYNACB, 265 Convergence, rate of, 206 Conversion factors, 496 Conversion of US units to SI units (table), 496 Coordinates: natural, 318 normal , 139 relative, 22, 159 Corrector, 203 Critical damping , 30 Critical time step, 198, 224 Curvature, 90
D' Alembert's principle, 3, 13 Damped forced vibrations, 30 Damped free vibrations, 27 Damped response of MDOF systems: to arbitrary forcing functions , 172 to periodic forcing functions, 168 Damping: absolute, 166 critical, 30 effects of, 25 in MDOF systems, 164 modal , 167 numerical, 218 proportional, 165 relative, 166 Damping coefficients, 165 Damping constant, 3, 27 modal, 165 Damping matrix, 164 Damping ratio, 31 modal , 165 Decomposition: Fourier, 36 1 spectral , 223 Dl'flution , 500
581
Degrees of freedom, number of, 105 Density, mass, 80, 497 Dependent displacements, 282 , 291 Determinant of Jacobian matrix, 330, 332 Diagonalization, 140 Differential equations of motion, 75 Dimensionless coordinates, 319 Direct iteration, 498 Direct linear extrapolation, 211 average-acceleration method, 212 linear-acceleration method, 213 Direct numerical integration methods , 195-240 Direct stiffness method, 98 Direction cosines, 96, 101 Discretization of structures, 6 Displacement equations of motion , 145 Displacements: dependent, 282 , 291 generic, 75 , 78 ground, 21: rigid-body, 157 incremental, 38, 212 , independent, 282, 291 initial, 14, 147 kinematically equivalent, 414 nodal, 75, 79: free, 99; restrained, 99 relative, 23 , 159 Displacement shape functions , 74 , 79 for axial element, 82 for element AXSH3 , 398 for element H8, 345 for element H20 , 349 for element PBQ8, 375 for element Q4, 333 for element Q8, 338 for element SHQ8, 386 for flexural element, 89 for torsional element, 85 Dissipative force, 3 Duhumcl' s integral , 39, 152, 173 Dy11n 111 il' t•q11ilibl'i11111 , I
lnda,c
ISB2
I>y11111111l' lrnt'l', I >y111111m· ,ntlucnc.·cs, 4 I >y1111111k load data:
lo, 11u11tislory plane frames (table) , •lh5 trn plane frames (table) , 269 IOI plane trusses (table), 180 IOI Program DYAXSH (table), 407 tor Pmgram DYNAPB (table) , 380 IOI l'rogrum DYNAPS (table), 341 IOI Prngram OYNASH (table) , 391 lor Progrum DYNASO (table), 352 Im :-.puce fnunes (table) , 279 t >y1111111ic loads, 1 hlnst, 50 ICl'llll'Ocul theorem, 154 I >y1111111ic reduction, 282
li:!1,wnvaluc matrix, 14 l hiHl'IIVttlue problem, 105, 314 l•w·nvulucs, 106, 498 ll'Pl'Ull'd, I 07 1'.11\l' nvcclor matrix, 140, 315 1 1),tl'llVCClors, 106, 498 11m111ulization of, 142 l•,lusticity , modulus, 497 1'111st it'ily force, 3 l•lt•111l' II( /\XQ4 , 359 i'.h·111L·11t /\XQ8, 359 l•h•1m·111 /\XSll3 , 394 hlt'llll 1l( /\XSR3, 394 l•l1•111t•nt 118, 345 (<'.lt•Jlll'II( 11 2(), 349 I •,lt•111L•nt PBQ8 , 371 I k1m·11l PQR8, J7 I l•k1m·11t ()4 , 1 n 1•11·1111•111 ()8, JJ8 I l1•1m•11t R4 , 1 U l•h•1111•11t RH , \ lH l•h•1111•111 RS8, '4.'l l•h•1m•11t RS l O, W> l•h•1m•11t SI 1()8, IH) I•I11111111111011. 1
1
1111 WIii d, ,1(,1
Clauss .Jordun, 291 Elimination method, purolld, 4~':t Equations of motion , J , 11, 7':t, 18 action, 140 displacement, 145 for finite elements, 78 incremental , 2 12 structural, 98 Equivalent nodal loads: for element, 82 nonproportional, 267 for structure, 98 for surface pressure, 366, 408 Equivalent viscous damping, 26 Euler's extrapolation formu la, 20~ Explicit formulas for extrapolation , 11) I Extrapolation , direct linear, 2 11
Factorization: Cholesky, l06, 498 Givens, 514 Factors , conversion, (table), 496 Finite-difference formulations, 195 Flexibilities, I 0 Flexibility coefficients, 145 Flexible-body motions, l 38 Flexural deformations, 90 in plate-bending, 378 Flexural element, 88 Flexural rigidity , 94 Flexural strains, 89 Forced vibrations , 19 , 30 Forces, body , 78 Force systems, types of, 241 forcing functions: arbitrary, 38, 172 hurmon ic, 19 periodic, 35, 168 piecl'wise lincur, 46, 175, 182 Fonn f1w1or , 177 FORTRAN, 9
Fm wnrd t•li11ii11ntio11, 4<> I F111w111d lll'llllioo, 41>8 1•1111111·1 nwll1r11·11h, IC,
583
Index
Fourier series , 35, 169, 361, 401 Framed structures, 6, 241-309 Frames: plane, 243 , 244 space, 244, 259 Free nodal displacements, 99 Free vibrations, 12, 27 Frequency, 14, 28 Frequency coefficients, 287, (table). 288
Gaussian quadrature, 328 coefficients for (table), 328 Gauss-Jordan elimination, 291 Generalized acceleration method, 217 Generalized rotation matrix, 505 Generalized stresses and strains, 92 General shells: element SHQ8, 382 Program DYNASH, 390 General solids: element H8, 345 element H20, 349 Program DYNASO, 351 Generic displacements, 75, 78 Geometric center, 318 Geometric interpolation functions , 320, 324 for element AXSH3, 396 for element HS, 345 for element H20, 350 for element PBQ8, 374 for element Q4, 333 for element Q8, 338 for element SHQ8, 383 Geometric transformations, 431 Givens factorization, 514 Givens method, 509 Global axes, 95 Sram-Schmidt orthogonalization, 107, 500 Grnvitalionnl system of units, 49.5
Program DYNAGR, 270 Ground accelerations, 21 arbitrary, 42 periodic, 38 rigid-body, 159 Ground displacements, 21 rigid-body, 157 Guyan reduction, 282, 445
Harmonic forcing functions, 19 Hexahedral coordinates, 323 infinitesimal volume in, .no Hilbert-a method , 222 Householder method, 'l 11
Immobility, 1>ta1 it', I•I'I Implicit fo11n11lt1Nfrn ill' lllll1111 , IO\ Incremental ncn·kratH111s, 1 1) Incremental uclions, 21l. Incremental tlispluccments, J8 , 2 12 Incremental equations of motion, 212 Incremental impulse, 38 Incremental velocities, 38, 212 Independent displacements, 282, 291 Independent motions of support restraints, 162 Inertia, moment of, 91 Inertial body forces, 80 Inertial force, 3 Inertial moment per unit length, 87 Inertias , translational and rotational · (rotary), 91, 378, 390, 401 Infinitesimal area in quadrilateral coordinates , 329 Infinitesimal volume in hexahedral coordinates, 330 Initial conditions, 4, 14 normal-mode response, 147 Integration, numerical, 326 Integration points: for hexahedron , 332 for quadri lateral, 330 I111l•1fo1 lllldl'N, 44(1
ll84
h11 hue
Index
585
lllll'I polutioo f'1111ctio11s:
gL·onwtlie, J20, 325 (.1·<<' olso CJ co 111etric inlerpolation functions) piecewise-linear, 46, 175, 182 Inverse ilcralion, 498 Inverse of Jacobian matrix , 321 , 325 In verse of normalized modal matrix , 143 lsoparamctric elements, 310- 69 clement AXQ4, 359 elcmenl AXQ8, 359 hcxahcdron H8 , 345 hexahedron H20, 349 quadrilateral Q4, 333 quadrilateral Q8, 338 Iso tropic materials, 77 , 311 Iteration (for eigenvalues), 498 ltcralion with implicit formulas, 203 1
,lncobian matrix, 321, 324 determinant of, 321 , 325 inverse of, 321, 325 Jacobi method, 505 Joinls, 241
Kilogram (kg), 495 Kilopound (kip) , 497 Kincmalically equivalent displacements, 414 Lu~ time, 17 I,inear-acceleration method, 219
direct Iinear extrapolation, 2 13 iteration wilh , 206 Linear differential operator, 79 I ,incur extrapolation , dirccl, 2 11 I .oud oplll'ltt or, 223 Louds: 11xi sy 11111wtric, 357 rn1 rn11lin11011s hcums, 265 dy11111n il', I 11•1·ip1t w 11l tlwc111·111 , 15,1
ct1uivulcnt nodul : for elcmenl , 82 for structure, 98 nonaxisymmetric, 36 1 normal-mode, 152 on grids, 27 1 on hexahedra, 351 on plane frames, 267 on plane trusses, 179 on quadrilateral s, 340 on space frames , 278 on space trusses, 274 Local axes, 95 Lumped mass matrix, 99
Magnification factor, 20, 32, 169 Main program: for DYNA (flow chart), 226 for NOMO (flow chart) , 178 for VIB (flow chart), 11 9 Mass center of, 418 , 427, 434 of a body, 419 Mass coefficients, 82 Mass density, 80 , 497 Mass matrix: for element, 82 for structure, 98 lumped, 99 principal , 140 reduced, 283 Mass moments of inertia of a rigid body, 41 9 principal, 420 Mass products of inertia of a rigid body , 419 Master displacements, 283, I 0.2- 1<) Material properties (table), 497 Materials: anisotropic, 311 isotropic, 77, 311 orthotropi<:, J I I pmpl'I I il•s of (/o/,/1') , ,11>7 Mntd \ l 'Clll
MDOF systems, 74 Mechanical analogue, 12 Member-oriented approach, 416 Members , 241 Meter (m), 495 Modal constraint, 500 Modal damping , 167 Modal damping constant, 165 Modal damping ratio, 165 Modal matrix, 140, 505 normalized, 142: inverse of, 143 Modal truncation, 139, 143 Modes: antisymmetric, 112 flexible-body, 138 rigid-body, 138, 148 symmetric, 112 vibrational, 138 Mode shape , 105 Modified Cholesky method, 498 Modified tridiagonal method, 451 for multistory buildings, 457 Modulus of elasticity, 497 Motion, equations of, 3, 13, 75, 78 Moving load: on a beam, 266 on a plate, 381 Multiplicity (of repeated eigenvalues) , 107 Multistory buildings: modified tridiagonal method, 457 with rigid laminae , 425 Multistory plane frames: dynamic load data (table), 465 Program DYMSPF, 463 structural data (table) , 464 Natural coordinates, 318 Natural frequency, 14, 28 Natural period, 4, 14, 28 Newmark-,8 method , 2 18 Newton, 495 Nodal actions, 7<) Noda I l'i re k•s. 15 I
Nodal coordinates: for element H8 (table), 347 for element H20 (table) , 351 for element Q4 (table), 335 for element Q8 (table) , 340 Nodal displacements , 75 , 79 free , 99 restrained , 99 Nodal loads: actual, 98 equivalent, 153 for element, 82 for structure, 98 Nodal vectors: for element AXSH3, 396, 406 for element SHQ8, 385,39 1 Nodes: boundary, 446 interior, 446 of finite elements, 6, 73 Nodewise solution, 198 Nonaxisymmetric loads: on axisymmetric shells, 401 on axisymmetric solids, 361 Nonproportional equivalent nodal loads, 267 Nonna! coordinates, 139 Normalization of eigenvectors, 142 Normal-mode load, 152 Normal-mode method, 138-94 Normal-mode response: Program NOMO for, 177 to applied actions, 152 to initial conditions, 147 to support motions , 157 Nonna! stresses and strains, 75 Notation, 483 for matrices and vectors, 483 for programs, 487 Greek letters, 486 simple variables, 485 subscripts for matrices and vectors, 484 N111111•111 11I lll'l'tll ary, N 11111e•11111l d11111p111p,
.U I 1 18
~88
N11111l'lll'lll 111tt•g1allo11, I 2C, N11111t•1 il-111 stabilit y, 22.l ( )uc-dlmcnslonal clements, 82 relationships, 140 <>1 thogon11l ization, Gram-Schmidt, 107, 500 Ot thotropic materials, 311 <>ullinc of' Program DYMSPF, 463 Outline of Programs DYPFAC and DYSFAC, 299 Output selection data (table), 182 Ovcrdumping, 29 Overlay technique, 453, 461 <)1 tho).\onulily
Pnrnllcl elimination method, 455 1'11rc11t rectangle: t·lcmenl AXSR3, 394 clement PQR8, 371 lllt.lment R4, 333, 336 l'lt.lmcnt R8, 338 1'11rcnt rectangular solid: clement RS8, 345, 348 clement RS20, 349 Pt11K'td, 496 Pt•riod, 4, 14, 28 l't'riod elongation, 225, 234 l't11iodic forcing functions, 35 d11111pcd response of MDOF systems, 168 l',•1 iodic ground acceleration, 38 Ph1111c angle, 16, 32, 105 , 169 l'IIIINC plane, 14 I'it·t·cw isc-1inear forcing functions , 46 , 175, 182 l'l1111t• frumcs, 243, 244 11xi11l constraint s, 290: program DYPFAC, 299 dynnmic loud duta (table), 269 1'10g1·11111 DYNAPF, 267 1lgid bodies: Program DYRAPF, ,1}4 Nlllll'tlll'lil d11t11 (tnl>/t•), 268
l11rla11
l11clox
587
l'liuw slltss und pln11t• sl111i11 , I l I l'rogri1111 DYNAPS , MO Plane trusses, 24.l component mode mcl11od, 471 Program COMOPT, 477 dynamic load data (tabll'), 180 Program DYNAPT, 227 structural data for (tab/<:), 110 Plate-bending: element PBQ8, 37 1 flexural and shearing dcf'o rmu tions, 378 Plates, 370-410 Poisson's ratio, 497 Polar moment of inertia, 87 Pound (lb), 495 Predictor, central-difference, 197 Predictor-corrector method , 203 Pressure, surface, 366, 408 Principal body axes, 420 Principal coordinates , 139 Principal damping matrix , 165 Principal flexibility matrix, 145 Principal mass matrix, 140 Principal mass moments of inerliu , 420
Principal normal strains, 316 Principal normal stresses, 314 Principal planes of bending: for space frame member, 259 for space lruss member, 255, 477 Principal stiffness matrix, 140 Principal stress axes, 315 Program COMOPT for plane trusstis by component-mode method, 477 Program COMOST for space trusses by component-mode method , 477 Program DYAXSII for axisymmetr il' shells, 406 dynamic load dnUI (tablt'), 407 structural data (ta/J/1•), 406 Progn1111 DYAXSO for 11xisy11111Hitd cu l solids. l
Program OYMSPF for multistory plane frames , 463 Program DYMSTB for multistory tier buildings, 463 Program DYNA for dynamic response, 225 main program (flow chart), 226 Program DYNACB for continuous beams, 265 Program DYNAGR for grids, 270 Program DYNAPB for plates in bending, 379 dynamic load data (table), 380 structural data (table), 380 Program DYNAPF for plane frames, 267 dynamic load data (table), 269 structural data (table), 268 Program DYNAPS for plane stress and plane strain, 340 dynamic load data (table), 341 structural data (table), 341 Program DYNAPT for plane trusses , 227 flow chart, 519-63 Program DYNASF for space frames, 277 dynamic load data (table), 279 structural data (table), 276 Program DYNASH for general shells, 390 dynamic load data (table) , 391 structural data (table), 391 Program DYNASO for general solids, 351 dynamic load data (table), 352 structural data (table), 352 Program DYNAST for space trusses, 273 Program DYPFAC for plane frames with axial constraints, 299 Pro11ram DYRBPB for rigid bodies in plate-bending continua, 438 rigid-body data (table), 440 rigid body dynamic-load data
(table), 440
Program DYRBPF for rigid bodies in plane frames, 424 Program DYSFAC for space frames with axial constraints, 299 Program NOMO for normal-mode response, 177 main program (flow chart), 178 Program NOMOPT, 179 dynamic load data (table), 180 Program notation, 487 Programs for framed structures, 264 Program VIB for vibrational analysis, 118 main program (flow chart) , 119 Program VIBPT, 11 8 structural dutu (tr1bll') , I 20 Proportional dH111pinH, I (1 ~ Properties of' 11111tt·1 l11ls (tali/,•) , ,111I Proportionul du111pi11g, IM Proport ion11 I lo11ds, I8 J Pscudostutic prohhm1s, 2 I ) Quadrature, Gaussian, 328 Quadrilateral coordinates , 318 infinitesimal area in, 329 QR algorithm, 514
Radius of gyration: for cross section of grid member, 251 .for rigid body, 433 Ramp function, 43 Ramp-step function , 55 Rate of convergence, 206 Rayleigh quotient, 500 Reactions at supports , 99 Reciprocal theorem for dynamic loads, 154 Rectangular impulse, 41 Rectangular parent: element AXSR3, 394 clement PQR8, 371
688
lndox
Rectangular parent (cont.) clement R4, 333, 336 clement R8, 338 Rectangular solid parent: clement RS8 , 345, 348 clement RS20 , 349 Reduced action vector, 282 Reduced equations of motion, 448 Reduced mass matrix, 283 Reduced stiffness matrix, 282 Reduction: dynamic, 282 Guyan, 282, 445 static, 282, 420 Redundant constraints, 290 Reference point on a rigid body, 4 13,4 16 Relative acceleration , 23 Relative coordinates, 22, 159 Relative damping, 166 Relative displacement, 23 Repeated eigenvalues, 107 Resonance, 21 Response, 4 steady-state, 20, 31 transient , 25, 34 Response calculations , step-by-step , 45, 175 Response to harmonic forcing function: l'orced part, 20 free part, 20 Rt·sponse spectra, 5 1 Rt•strai necl nodal displacements, 99 Retai ned modes , 469, 472 l{t•vcrse iteration, 498 l{igid-boclies, 8, 4 11 - 43 equations of motion, 4 18,437 III finite-clement networks, 434 in frumed structures, 4 13 In plane fr u111cs, Program
l>YRBPJI, 424 111 pink lwnding rnnt i111111 , Prn g111111 DYH ll l'B, 1118 IIIIIHN
1111d
ll ll l 'iN lllllllH11ll
of im•11111
Index
589
Rigid-body center of mass, 41 8 Rigid-body constraints, 411 , 427 Rigid-body data for Program DYRBPB (table), 440 Rigid-body dynamic-load data for Program DYRBPB (table), 440 Rigid-body ground accelerations, I!'! '> Rigid-body ground displacements, 157 Rigid-body modes, 148 Rigid-body motions , 138, 153 Rigid-body reference point, 4 J 3, 4l / Rigidity: axial, 94 flexural, 94 torsional , 93 Rigid laminae in multistory build ings, 427 Ring element, 357 Rotary inertias (see Rotational incr tias) Rotating vectors, 14 Rotation of axes: for grid member, 252 for plane frame member, 24 7 for plane truss member, J02: i11 component-mode method , 47~ for space frame member, 263 for space truss member, 256 for stresses and strains, 3 14 Rotation-of-axes transformations, 1> .~ Rotation matrix , 97 generalized , 505 Rotational (or rotary) inertias, 91, 378, 390, 40 1
SDOF systems, 10 Second moment of area, 9 1 Second (s), 495 Selection data, output (/ahfr) . IH.l Series: 1:ou1 k·r, JS s11h~l111l'll11 t•11 111 ,14c, Sl•lhncks 111 1ll•t h111!dl11~11. ,l(,H 1
Shape functions, displacement, 74, numerical, 223 79 (see also Displacement shape unconditional, 208 functions) Stability criterion, 224 Shape functions and derivatives for Standard , symmetric form (of eigenelement Q8 (table), 340 value problem) , 106, 505 Shear cores in tier buildings, 468 Static determinacy, 145 Shearing deformations: Static equilibrium, 1 in plate-bending, 378 Static immobility, 145 Shearing stresses and strains, 75 Static reduction , 282, 420 Shell element SHQ8, 382 Statically equivalent actions, 41 4 Shells, 370-410 Steady-state forced vibration, 20 axisymmetric: element AXSH3, Step function , 40 394; Program DYAXSH, 406 Step-by-step response calculutions, general: element SHQ8, 382; 45 , 175 Program DYNASH, 390 Stiffness coefficients, 82 Shifting , spectral, 503 Stiffness matri x: SI units , 495 for axial clement , 84 Slave displacements, 283, 457 for axisymmctric shell clement Slug, 495 with nonaxisymmetric loads, Solid of revolution (see Axisymmet405 ric solids) for axisymmetric solid element Solids: with nonaxisymmetric loads, axisymmetric: isoparametric ele364 ments , 357; Program DYAXSO, for element, 82 365 for element AXQ4 , 359 general: isoparametric elements , for element AXSH3, 400 345; Program DYNASO, 351 for element H8, 347 Space frames, 244, 259 for element PBQ8, 377 axial constraints, Program DYSfor element Q4 , 335 FAC, 299 for element SHQ8 , 389 dynamic load data (table), 279 for flexural element, 91 Program DYNASF, 277 for grid member, 249 structural data (table), 276 for plane frame member, 244 Space trusses, 243 , 253 for plane truss member, 101: in component-mode method, 477: component-mode method, 474 Program COMOST, 477 for space frame member, 259 Program DYNAST, 273 for space truss member, 255 Spectra, response, 51 for structure, 98 Spectral decomposition, 223 for substructure in multistory Spectral matrix, 141, 315, 470, 505 building, 460 S::- ctral radius, 223 for torsional element, 87 Spectral shifting, 503 for x-beam, 428 Square-root method , Cholcsky, I 06 for y-beam, 428 Stability: fill' z l'llhlltlll , ,1W ('Olldi1101111l , ' OH pli11l'1p11l , 1·10 0
Index
590
Stiffness matrix (cont.) reduced, 282 reduced and assembled, 448 tridiagonal, 446 Stiffnesses, 11 Stiffness method, direct, 98 Strain-displacement relationships, 79, 310 Strain energy, virtual, 80 Strain energy density, virtual, 316 Strain-stress relationships , 77 Strain transformation matrix, 316 Stress-strain relationships, 77, 80, 311 Stress transformation matrix, 316 Stresses and strains, 75, 310 generalized, 92 Structural damping, 26 Structural data: for multistory plane frames (table), 464 for plane frames (table), 268 for plane trusses (table), 120 for Program DYAXSH (table), 406 for Program DYNAPB (table), 380 for Program DYNAPS (table), 341 for Program DYNASH (table), 39 1 for Program DYNASO (table), 352 for space frames (table), 276 Slructural equations of motion, 98 Slructural mass matrix, 98 Sii uctural stiffness matrix , 98 S11hpt1rametric elements, 333 Suhprogrnm CMAS, 118 S11hpmgra111 DHCOMP (flow chart), C\59 Suhprngrnm DYLO, 177, 225 S11hpm~rn111 DYi.OPT, (.flow chart),
., 1-,
S111tp10~111111 lil
Subprogram FACTOR (flow chart) , 561 Subprogram INVERU (flow chart), 560 Subprogram NORMOD, 227 Subprogram NUMINT, 227, (flow chart), 547 Subprogram RESl , 118 Subprogram RESlPT (flow chart), 532 Subprogram RES2, 179, 227 Subprogram RES2PT (flow chart), 553 Subprogram SDAT, 118 Subprogram SDATPT (flow chart), 519 Subprogram SOLVER (flow chart), 562 Subprogram STASYM, 118, (flow chart), 525 Subprogram STIF, 118 Subprogram STIFPT (flow chart), 523 Subprogram TIHIST, 177, (flow chart), 545 Subprogram TRABAC, 179, (flow chart), 546 Subprogram TRANOR, 177, (flow chart), 543 Subprogram TRAVEC, 118, (flow chart), 529 Subprogram VIB, 177, 225 Substitution, backward, 461 Substructure equations of motion, 446 in component-mode method, 4C,H Substructures, 8, 444- 82 Substructures in parallel, 455 Substrnctures in series, 445 modified tridiagonal method, •I~ I tridingonal method, 445
Supc1 purnml·lric ckml•11ts, \ \ \ S11pp1111 111111,ons, •I i111l1•p1•1ul1•111 , I <,J
Index
normal-mode response, 157 Support reactions, 99 Support restraints, independent motions of, 162 Surlace pressure, equivalent nodal loads for, 366, 408 Symmetric, standard form (of eigenvalue problem), 106, 505 Symmetric modes, 112 Systems of units: absolute and gravitational, 495 consistent (table), 496
Three-dimensional continua, 310-69 Tier buildings, 425, 461 bracing, setbacks, and shear cores, 468 Program DYMSTB, 463 Time step, critical, 198, 224 Torsional element, 84 Torsional rigidity, 93 Torsional strains, 85 Transformation matrices for framed structures (table), 415 Transformation methods for eigenvalue problems, 505 Transformations, geometric, 431 Transient response: with damping, 34 without damping, 25 Translational inertias, 91, 378, 390, 401 Translation of axes, 157, 414 Trapezoidal rule, 203 Triangular impulse, 45 Tridiagonal matrix, 509 Tridiagonal method, 445 modified, 451 Trigonometric series, 35
591
Truncation, modal, 139, 143 Trusses: component-mode method, 471 plane, 243, 471 space, 243, 253, 477 Twist, 85 Two-dimensional continua, 310-69 Types of force systems, 241 Types of framed structures, 241
Unconditional stability, 208 Undamped forced vibrations, 19 Undamped free vibrations, 12 Underconstrained frames, 290 Unit-load method, 146 Units, systems of (SI and US), 495 Unit vectors, 96
Velocities: incremental, 38, 212 initial, 14, 147 Vibrational analysis, 105 Program VIB , 118 Vibrational motion, 152 Vibrations: forced, damped, 30 forced, undamped, 19 free, damped, 27 free, undamped, 12 Virtual strain energy, 80 density, 316 Virtual work, 80 Viscous damping, 26, 164
Wilson-8 method, 220 Work, virtual, 80 Working points, 416
