Solution Manual For Dynamics of Structures, 4th Edition

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CHAPTER 1Problem 1.1Ifkeis the effective stiffness,fkuSe=fSfSk u2k u1k1k2uEquilibrium of forces:fkkuS=+()12Effective stiffness:kfukkeS==+12Equation of motion:muk up te&&( )+=

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Problem 1.2Ifkeis the effective stiffness,fkuSe=(a)fSk1k2uIf the elongations of the two springs areu1andu2,uuu=+12(b)Because the force in each spring isfS,fkuS=1 1fkuS=22(c)Solving foru1andu2and substituting in Eq. (b) givesfkfkfkSeSS=+12⇒11112kkke=+⇒kk kkke=+1212Equation of motion:muk up te&&( )+=

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Problem 1.3k1k2uk1k3k2+1k3mmmkeFig. 1.3(a)Fig. 1.3(b)Fig. 1.3(c)⇓⇓This 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 inFig. 1.3(b). Second, using the result of Problem 1.2, theseries arrangement of springs in Fig. 1.3(b) is replaced bya single spring, as shown in Fig. 1.3(c):111123kkkke=++Therefore the effective stiffness iskkkkkkke=+++()123123The equation of motion ismuk up te&&( )+=.

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Problem 1.41.Draw a free body diagram of the mass.OLmθTmgsinθmgcosθ2.Write equation of motion in tangential direction.Method 1: By Newton’s law.0sinsinsin=+=−=−θθθθθmgmLmLmgmamg&&&&(a)This nonlinear differential equation governs the motion forany rotationθ.Method 2: Equilibrium of moments aboutOyieldsθθsin2mgLmL−=&&or0sin=+θθmgmL&&3.Linearize for smallθ.For smallθ,θθ≈sin, and Eq. (a) becomes00=⎟⎠⎞⎜⎝⎛+=+θθθθLgmgmL&&&&(b)4.Determine natural frequency.Lgn=ω

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Problem 1.51.Find the moment of inertia about O.From Appendix 8,ImLmLmL0222112213=+⎛⎝⎜⎞⎠⎟=2.Draw a free body diagram of the body in an arbitrarydisplaced position.mgyxθL/23.Write the equation of motion using Newton’s second lawof motion.MI00=∑&&θ−=mg LmL2132sin&&θθmLmgL2320&&sinθθ+=(a)4.Specialize for smallθ.For smallθ,sinθ≅θand Eq. (a) becomesmLmgL2320&&θθ+=&&θ+=320gL(b)5.Determine natural frequency.ωngL=32

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Problem 1.61.Find the moment of inertia about about O.Ird AL020=∫ρ===∫ραραrrdrLmLL2042412()xLαr2.Draw a free body diagram of the body in an arbitrarydisplaced position.mgyxθ2L/33.Write the equation of motion using Newton’s second lawof motion.MI00=∑&&θ−=mgLmL23122sin&&θθmLmgL22230&&sinθθ+=(a)4.Specialize for smallθ.For smallθ,sinθ≅θ, and Eq. (a) becomesmLmgL22230&&θθ+=or&&θ+=430gL(b)5.Determine natural frequency.ωngL=43In each case the system is equivalent to the spring-mass system shown for which the equation of motion is0g=+⎟⎟⎠⎞⎜⎜⎝⎛kuuw&&wukThe 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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Problem 1.7Draw a free body diagram of the mass:müp(t)ufSWrite equation of dynamic equilibrium:mufp tS&&( )+=(a)Write the force-displacement relation:fAELuS=FHGIKJ(b)Substitute Eq. (b) into Eq. (a) to obtain the equation ofmotion:muAELup t&&( )+FHGIKJ=

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Problem 1.8Show forces on the disk:OθfSRWrite the equation of motion using Newton's secondlaw of motion:−=fISO&&θwhereIm RO=22(a)Write the torque-twist relation:fGJLS=FHGIKJθwhereJd=π432(b)Substitute Eq. (b) into Eq. (a):IGJLO&&θθ+FHGIKJ=0or,mRd GL242320FHGIKJ+FHGIKJ=&&θπθ

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Problems 1.9 through 1.11In each case the system is equivalent to the spring-mass system shown for which the equation of motion iswukugFHG IKJ+=&&0wukThe spring stiffness is determined from the deflectionuunder a vertical forcefSapplied at the location of thelumped weight:Simply-supported beam:uf LEIkEILS=⇒=334848Cantilever beam:ufLEIkEILS=⇒=3333Clamped beam:uf LEIkEILS=⇒=33192192

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Problem 1.12LEIwkFig. 1.12aStatic EqulibriumDeformed positionuuδstUndeformed positionFig. 1.12bp(t)wmu..fsfsFig. 1.12c1.Write the equation of motion.Equilibrium of forces in Fig. 1.12c gives)(tpwfums+=+&&(a)wherefs=keu(b)The equation of motion is:)(tpwukume+=+&&(c)2.Determine the effective stiffness.ukfes=(d)wherebeamspringuδδ+=(e)beambeamspringskkfδδ==(f)Substitute for theδ’s from Eq. (f) and forufrom Eq. (d):()3348/48LEIkLEIkkkkkkkkfkfkfebeambeamebeamsses+=+=+=3.Determine the natural frequency.mken=ωsimplysupported

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Problem 1.13Compute lateral stiffness:1h3EI /hc3kkEIhEIhcolumncc=×=×=223633Equation of motion:mukup t&&( )+=Base fixity increaseskby a factor of 4.

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Problem 1.141.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.uuu12310===,k21k31kEIhEIhkkEIhccc1133213122 363====uuu21310===,k22k23kEIhEIhEIhkEIhEIhkEIhcccccc22321223425223=+====()()uuu31210===,k13k23k33kEIhEIhEIhkEIhEIhkEIhcccccc33231323425223=+====()()Hencek=⎡⎣⎢⎢⎢⎤⎦⎥⎥⎥EIhhhhhhhhhc3222263335354.Determine lateral stiffness.The lateral stiffnesskof the frame can be obtained bystatic condensation since there is no force acting on DOF 2and 3:EIhhhhhhhhhuuufcS32222123633353500⎡⎣⎢⎢⎢⎤⎦⎥⎥⎥⎧⎨⎪⎩⎪⎫⎬⎪⎭⎪=⎧⎨⎪⎩⎪⎫⎬⎪⎭⎪First partitionkaskkkkk=⎡⎣⎢⎢⎢⎤⎦⎥⎥⎥= ⎡⎣⎢⎤⎦⎥EIhhhhhhhhhctttt3222200006333535

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where[ ][]⎥⎥⎦⎤⎢⎢⎣⎡===222230030355336hhhhhEIhhhEIhEIcctcttkkkThen compute the lateral stiffnesskfromTttttk01000kkkk−−=Sincek001245115−=−−⎡⎣⎢⎤⎦⎥hEIcwe get[][]kEIhEIhhhhEIEIhhhkEIhkEIhcccccc=−⋅−−⎡⎣⎢⎤⎦⎥ ⋅⎡⎣⎢⎤⎦⎥=−=63324511533633333335.Equation of motion.m uE Ihup tc&&( )+=33

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Problem 1.151khIcIcI = I /bc2h2Define degrees of freedom (DOF):123Form structural stiffness matrix:uuu12310===,k11k21k31kEIhEIhcc11332 1224==kkEIhc213126==uuu21310===,k12k22k32kEIhEIhEIhEIhEIhcbccc2244245=+=+=()kEIhEIhbc32222==()kEIhc1226=uuu31210===,k13k23k33kEIhEIhEIhEIhEIhcbccc3344245=+=+=()kEIhEIhbc23222==()kEIhc1326=Hence⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡=22212212356566624hhhhhhhhhEIckThe lateral stiffnesskof the frame can be obtained bystatic condensation since there is no force acting on DOF 2and 3:⎪⎭⎪⎬⎫⎪⎩⎪⎨⎧=⎪⎭⎪⎬⎫⎪⎩⎪⎨⎧⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡0056566624321222122123ScfuuuhhhhhhhhhEIFirst partitionkas⎥⎦⎤⎢⎣⎡=⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡=000022212212356566624kkkkkTttttchhhhhhhhhEIwherekttcEIh=324ktcEIhhh0366=⎥⎥⎦⎤⎢⎢⎣⎡=2221221230055hhhhhEIckThen compute the lateral stiffnesskfromkttttT=−−kkkk00010Since⎥⎥⎦⎤⎢⎢⎣⎡−−=−559942121100cEIhk

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we get[]33321213311120)1114424(66559946624hEIhEIhhhEIEIhhhhEIhEIkcccccc=−=⎥⎦⎤⎢⎣⎡⋅⎥⎥⎦⎤⎢⎢⎣⎡−−⋅−=This result can be checked against Eq. 1.3.5:⎟⎟⎠⎞⎜⎜⎝⎛++=412112243ρρhEIkcSubstitutingρ==IIbc41 8gives3381813111201152441211224hEIhEIhEIkccc=⎟⎠⎞⎜⎝⎛=⎟⎟⎠⎞⎜⎜⎝⎛++=Equation of motion:)(111203tpuhEIumc=⎟⎟⎠⎞⎜⎜⎝⎛+&&

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