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Dynamics of Structures, 5th Edition Solution Manual - Page 1

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Dynamics of Structures, 5th Edition Solution Manual

Dynamics of Structures, 5th Edition Solution Manual is your ultimate textbook solutions guide, providing answers to the most difficult questions.

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Dynamics of Structures, 5th Edition Solution Manual - Page 1 preview image1CHAPTER 1Problem 1.1Starting from the basic definition of stiffness, determinethe effective stiffness of the combined spring and write theequation of motion for the spring–mass systems shown inFig. P1.1.Figure P1.1Solution:Ifkeis the effective stiffness,fkSeu=fSfSk u2k u1k1k2uEquilibrium of forces:fkkuS=+(1)2Effective stiffness:+2kfukkeS==1Equation of motion:muk up te( )+=
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Dynamics of Structures, 5th Edition Solution Manual - Page 2 preview image
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Dynamics of Structures, 5th Edition Solution Manual - Page 3 preview imageProblem 1.22Starting from the basic definition of stiffness, determinethe effective stiffness of the combined spring and write theequation of motion for the spring–mass systems shown inFig. P1.2.Figure P1.2Solution:Ifkeis the effective stiffness,fk u=(Sea)fSk1k2uIf the elongations of the two springs are1uand,2uuuu=+(b12)Because the force in each spring isfS,fk uS=1 1fk uS=22(c)Solving foru1andu2and substituting in Eq. (b) givesf111ffkkkSeSS=+12⇒12kkke=+⇒kk kkke=+1212Equation of motion:mu+keup(t=).
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Dynamics of Structures, 5th Edition Solution Manual - Page 4 preview image3Problem 1.3Starting from the basic definition of stiffness, determinethe effective stiffness of the combined spring and write theequation of motion for the spring–mass systems shown inFig. P1.3.Figure P1.3Solution:k1k2k3m⇓Figure P1.3ak1k3k2+1m⇓Figure P1.3bumkeFigure P1.3cThis problem can be solved either by starting from thedefinition of stiffness or by using the results of ProblemsP1.1 and P1.2. We adopt the latter approach to illustratethe procedure of reducing a system with several springs toa single equivalent spring.First, using Problem 1.1, the parallel arrangement ofk1andk2is replaced by a single spring, as shown in Fig.1.3b. Second, using the result of Problem 1.2, the seriesarrangement of springs in Fig. 1.3b is replaced by a singlespring, as shown in Fig. 1.3c:111123kkkke=++Therefore the effective stiffness iskkkkkkke=+++()123123The equation of motion ismu+=keup(t).
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Dynamics of Structures, 5th Edition Solution Manual - Page 5 preview image4Problem 1.4Derive the equation governing the free motion of a simplependulum that consists of a rigid massless rod pivoted atpointOwith a massmattached at the tip (Fig. P1.4).Linearizetheequation,forsmalloscillations,anddetermine the natural frequency of oscillation.Figure P1.4Solution:1.Draw a free body diagram of the mass.mOθLTmg sinθmg cosθ2.Write equation of motion in tangential direction.Method 1: By Newton’s law.-=mmg sinθa-=mmg s inθLθ(a)mLθθ+=mg sin0This nonlinear differential equation governs the motion forany rotationθ.Method 2: Equilibrium of moments aboutOyields2gsinmLm Lθθ= -org sin0mLmθ +θ =3.Linearize for smallθ.For smallθ,sinθ ≈θ, and Eq. (a) becomesgmLmθ +0θ =g0Lθθ+=()|│()(b)4.Determine natural frequency.gωn=L
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Dynamics of Structures, 5th Edition Solution Manual - Page 6 preview image5Problem 1.5Consider the free motion in thexyplane of a compoundpendulum that consists of a rigid rod suspended from apoint (Fig. P1.5). The length of the rod isL, and its massmis uniformly distributed. The width of the uniform rod isband the thickness ist. The angular displacement of thecenterline of the pendulum measured from they-axis isdenoted byθ(t).(a)Derive the equation governingθ(t).(b)Linearize the equation for smallθ.(c)Determine the natural frequency of small oscillations.Figure P1.5Solution:1.Find the moment of inertia about O.From Appendix 8,2122011223LImLmmL=+=()|│()2.Draw a free body diagram of the body in an arbitrarydisplaced position.mgyxθL/2ymgxL/23.Write the equation of motion using Newton’s second lawof motion.00MIθ=∑21gsin23Lmθθ-=mL)2gsin032mLm Lθθ+=(a4.Specialize for smallθ.For smallθ, sinθ≅θand Eq. (a) becomesθ2g032mLm Lθ +=b)3 g02Lθθ+=(5.Determine natural frequency.Lng23=ω
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Dynamics of Structures, 5th Edition Solution Manual - Page 7 preview image6Problem 1.6Repeat Problem 1.5 for the system shown in Fig. P1.6,which differs in only one sense: its width varies from zeroatOtobat the free end.Figure P1.6Solution:1.Find the moment of inertia about about O.IrL020=∫ρd Aαr)===∫ρραrrdLmLL2042412(xLαrLr2.Draw a free body diagram of the body in an arbitrarydisplaced position.mgyxθ2L/3xymg2L/33.Write the equation of motion using Newton’s second lawof motion.0022gsin32MILmθθθ=-∑1mL=0sin3g222=+θθLmmL(a)4.Specialize for smallθ.For smallθ, sinθ≅θ, and Eq. (a) becomes22g023mLm Lθθ+=or4 g03Lθθ+=(b)5.Determine natural frequency.Lng34=ωIn each case the system is equivalent to the spring-mass system shown for which the equation of motion is0guku+=()||()wwukwThe spring stiffness is determined from the deflectionuunder a vertical forcefSapplied at the location of thelumped weight:Simply-supported beam:uf LEIkEILS=⇒=334848Cantilever beam:uf LEIkEILS=⇒=3333Clamped beam:uf LEIkEILS=⇒=33192192
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Dynamics of Structures, 5th Edition Solution Manual - Page 8 preview image7Problem 1.7Develop the equation governing the longitudinal motion ofthe system of Fig. P1.7. The rod is made of an elasticmaterial with elastic modulusE; its cross-sectional area isAand its length isL. Ignore the mass of the rod andmeasureufrom the static equilibrium position.Figure P1.7Solution:Draw a free body diagram of the mass:müp(t)ufSWrite equation of dynamic equilibrium:( )Smufp t+=(a)Write the force-displacement relation:b)SAEfuL=()|│()(Substitute Eq. (b) into Eq. (a) to obtain the equation ofmotion:=( )AEmuup tL+()|│()
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Dynamics of Structures, 5th Edition Solution Manual - Page 9 preview image8Problem 1.8A rigid disk of massmis mounted at the end of a flexibleshaft (Fig. P1.8). Neglecting the weight of the shaft andneglecting damping, derive the equation of free torsionalvibration of the disk. The shear modulus (of rigidity) of theshaft isG.Figure P1.8Solution:Show forces on the disk:OθfSRWrite the equation of motion using Newton’s second lawof motion:SOfIθ-=where22Om RI=(a)Write the torque-twist relation:SGJfLθ=()|│()whereJd=π432(b)Substitute Eq. (b) into Eq. (a):θ=0OGJILθ+()|│()or,θ=240232mRd GLπθ+()()|│|│()()
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Dynamics of Structures, 5th Edition Solution Manual - Page 10 preview image9Problems 1.9 through 1.11Write the equation governing the free vibration of thesystems shown in Figs. P1.9 to P1.11. Assuming the beamto be massless, each system has a single DOF defined asthe vertical deflection under the weightw. The flexuralrigidity of the beam isEIand the length isL.Solution:In each case the system is equivalent to the spring-mass system shown for which the equation of motion is0guku+=()|│()wwukwThe spring stiffness is determined from the deflectionuunder a vertical forcefSapplied at the location of thelumped weight:Simply-supported beam:uf LkEILS=⇒=3348EI48Cantilever beam:I3333Sf LukEIL=⇒=EClamped beam:33192192Sf LukEIL=⇒=EI
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Dynamics of Structures, 5th Edition Solution Manual - Page 11 preview image10Problem 1.12Determine the natural frequency of a weightwsuspendedfrom a spring at the midpoint of a simply supported beam(Fig. P1.12). The length of the beam isL, and its flexuralrigidity isEI. The spring stiffness isk. Assume the beam tobe massless.Figure P1.12Solution:LEIwkSimplysupportedwFigure 1.12aStatic EqulibriumDeformed positionuuδstUndeformed positionFigure 1.12bp(t)wmu..fsfsp(t)wFigure 1.12c1.Write the equation of motion.Equilibrium of forces in Fig. 1.12c givesa)( )Smufp t+=+w(whereb)Sefk u=(The equation of motion is:c)( )emuk up t+=+w(2.Determine the effective stiffness.d)ukfes=(wheree)springbeamuδδ=+(springbeambeamSfkkδδ==(f)Substitute for theδ’sfrom Eq. (f) and forufrom Eq. (d):beamSSSefffkkk=+beambeamekkkkk=+()334848ekEI LkEIkL=+3.Determine the natural frequency.mken=ω
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Dynamics of Structures, 5th Edition Solution Manual - Page 12 preview image11Problem 1.13Derive the equation of motion for the frame shown in Fig.P1.13. The flexural rigidity of the beam and columns is asnoted. The mass lumped at the beam ism; otherwise,assume the frame to be massless and neglect damping. Bycomparing the result with Eq. (1.3.2), comment on theeffect of base fixity.Figure P1.13Solution:Compute lateral stiffness:h3EI /hc31column333622ccEIEIkkhh=×=×=Equation of motion:mukup t( )+=Base fixity increaseskby a factor of 4.
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Dynamics of Structures, 5th Edition Solution Manual - Page 13 preview image12Problem 1.14Write the equation of motion for the one-story, one-bayframe shown in Fig. P1.14. The flexural rigidity of thebeam and columns is as noted. The mass lumped at thebeam ism; otherwise, assume the frame to be massless andneglect damping. By comparing this equation of motionwith the one for Example 1.1, comment on the effect ofbase fixity.Figure P1.14Solution:1.Define degrees-of-freedom (DOF).1232.Reduced stiffness coefficients.Since there are no external moments applied at thepinnedsupports,thefollowingreducedstiffnesscoefficientsare used for the columns.Joint rotation:L132EIL32EIL3EILEIJoint translation:132EIL33EIL33EILEIL3.Form structural stiffness matrix.uuu1231===,0k21k312131,0uuu===k22k23kcEIhEIhEIhcc223425=+=()kEIhEIhcc3222==()kEIhc1223=uuu03121===,k13k23k33kEIhEIhEIhcc333425=+=()ckEIhEIhcc2322==()kEIhc1323=Hencek=┌└│││┐┘│││EIhhhhhhhhhc322226333535
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Dynamics of Structures, 5th Edition Solution Manual - Page 14 preview image134.Determine lateral stiffness.The lateral stiffnesskof the frame can be obtained bystatic condensation since there is no force acting on DOF 2and 3:┌63hh3┐(u)(fEI1S)c││││││335hh22hu} =││{{0h2}│22│││││└35hhh┘(u3)(0)First partitionkaskkkkk=┌└│││┐┘│││= ┌└│┐┘│EIhhhhhhhhhctttt3222200006333535where[ ]=36hEIcttk[]=3033hhhEIctk││┘┐││└┌=222230055hhhhhEIckThen compute the lateral stiffnesskfromTttttk01000kkkk--=Sincek001245115-=--┌└│┐┘│hEIcwe get[]kEIhEIhhhhEIEIhhhcccc=---┌└│┐┘│ ┌└│┐┘│63324511533333[]kEIhc=-633kEIhc=335.Equation of motion.=m uE Ihup tc( )+33
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Dynamics of Structures, 5th Edition Solution Manual - Page 15 preview image14Problem 1.15Write the equation of motion of the one-story, one-bayframe shown in Fig. P1.15. The flexural rigidity of thebeam and columns is as noted. The mass lumped at thebeam ism; otherwise, assume the frame to be massless andneglect damping. Check your result from Problem 1.15against Eq. (1.3.5). Comment on the effect of base fixity bycomparing the two equations of motion.Figure P1.15Solution:1khIcIcI = I /bc2h2Define degrees-of-freedom (DOF):312Form structural stiffness matrix:uuu1231===,0k11k21k3111312242cEIEIkh==3ch213126cEIkkh==uuu2131===,0k12k22k3222444(2 )cbccEIEIEIEIEIkhhhh=+=+=5ch322(2 )2bcEIEIkhh==1226cEIkh=uuu311===,20k13k23k3333444(2 )cbccEIEIEIEIEIkhhhh=+=+=5ch232(2 )2bEIEIkh==ch1326cEIkh=Hence│││┘┐│││└┌=22212212356566624hhhhhhhhhEIckThe lateral stiffnesskof the frame can be obtained bystatic condensation since there is no force acting on DOF 2and 3:│)│})│(│{(=│)│})│(│{(│││┘┐│││└┌0056566624321222122123ScfuuuhhhhhhhhhEI
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Dynamics of Structures, 5th Edition Solution Manual - Page 16 preview image15First partitionkas│┘┐│└┌=│││┘┐│││└┌=000022212212356566624kkkkkTttttchhhhhhhhhEIwherekttcEIh=324ktcEIhhh0366=││┘┐││└┌=2221221230055hhhhhEIckThen compute the lateral stiffnesskfromkttttT=--kkkk00010Since││┘┐││└┌--=-559942121100cEIhkwe get[]33321213311120)1114424(66559946624hEIhEIhhhEIEIhhhhEIhEIkcccccc=-=│┘┐│└┌││┘┐││└┌---=This result can be checked against Eq. 1.3.5:││))||((++=412112243ρρhEIkcSubstitutingρIbI=c4=1 8gives3381813111201152441211224hEIhEIhEIkccc=│))|((=││))||((++=Equation of motion:)(111203tpuhEIumc=││))||((+
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Dynamics Of Structures by Anil K. C...

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