Solutions for Homework #9

Transcription

1 Solutons for Hoewor #9 PROBEM. (P. 3 on page 379 n the note) Consder a sprng ounted rgd bar of total ass and length, to whch an addtonal ass s luped at the rghtost end. he syste has no dapng. Fnd the natural odes of vbraton he left support s gven an ntal vertcal dsplaceent a and s then released. Fnd the response. Wrte down the pulse response functons for ths syste. Careful: he rgd bar has rotatonal nerta (the ass of the bar s unforly dstrbuted). Rgd Bar!!!! SOUION he syste has two degrees of freedo for the vertcal translaton u C and rotaton θ. We frst consder a general defored shape for the syste n ters of u C and θ, as shown below. u c θ hen, we draw the free body dagra and set up the equatons of oton as follows. J θ + u B ü B u A u c u B where J= /. Choosng as degrees of freedo the rotaton of the rgd bar θ and the translaton of the center of ass, u C, we have: ua = uc θ ub = uc + θ {} he equatons of oton are forulated by defnng equlbru of vertcal forces and oents around the center of ass, as follows: F z = 0 : u A+ u B + u u C + B = 0 u C + θ + uc = 0 {} M C = 0 : u A + u B + Jθ + u B = 0

2 u C + θ + θ = 0 {3} 3 Express the above equatons {} and {3} n the followng atrx for. 0 u C u C Pz + = {4} θ 0 θ Mθ 3 Manpulate equaton {4} such that we shft the factor fro the second coluns to the rotatonal coponent θ, and then dvde the second row by to obtan the followng for. / u C 0 uc P z / /3 θ + = {5} 0 / θ Mθ / or sybolcally Mu + Ku = P {6} Note that u C P z u =, and P = {7,8} θ Mθ / o fnd the natural odes of vbraton, we establsh the followng egenvalue proble fro equaton {5}. 0 3 φ z 0 ω λ =, wth λ = {9} φθ 0 hen, the solutons for equaton {9} are gven as λ = 0.735, and λ = hen, t follows that ω = λ and ω = λ f = 4 4λ, and = 4 4λ λ f {0,,,3} λ Next, consder the response for the case that the left support at B s gven an ntal vertcal dsplaceent a and s then released. In connecton wth equaton {7}, ths ntal condton can be expressed n the for. uc0 a / u 0 = =, wth θ 0 = a/ {4} θ 0 a Also, n ters of ode shapes, u 0 s wrtten n the for. u = q f + q f {5} Fro the orthogonalty condton, we obtan f Mu0 q 0 = f Mf, and fmu0 q 0 = f Mf Fnally, we can express u = qf where = {6,7} q () t = q0 cosω t, and f = 4 4λ {9,0} λ However, reeber that u = { u θ } C. herefore, ore precsely, we can express {8}

3 u {} C = q0 cosω t 4 4λ θ = λ he pulse response functons for ths syste are expressed n the for. u = qγ f where = q () t = sn ω t ω γ Snce u = { u θ } u, for Pz = andmθ = 0 = 4 4λ, forpz = 0andMθ = λ C, we can express snω t C = γ 4 4λ θ = ω λ {} {3} {4} {5} 3

4 PROBEM. (P. 33 on page 380 n the note) Consder a luped ass structure whch can be odeled as a shear bea (= a chan of close-coupled lateral sprngs and asses,.e. the sple syste that we often use to odel hgh-rse buldngs). he dstance between floors (.e. nter-story heght) s constant. he asses are nubered fro the top down, and whle arbtrary, you should consder the as beng nown. he stffness, however, are not nown. However, fro a vbraton test, you have deterned experentally the fundaental frequency ω as well as the fundaental ode f, and you have found that ths ode s a straght lne,.e. f ={N N- N- 3 } a) Deterne the values of the sprngs n ters of the fundaental frequency and asses. u b) he overturnng oent at the base s the oent exerted by the nerta forces, whch s u N M = uz = n whch z s the heght of the th ass above the ground. Show that because the frst ode s a straght lne, only that ode contrbutes to the overturnng oent (notce, by the way, that the foundaton ust be able to resst ths oent). N u N N c) Suppose that there s a wall that runs parallel to the structure, and that there are sprngs of rgdty r = α connectng the asses and the wall, n whch α s a nown constant (.e, the sprngs are proportonal to the asses). If you new all the frequences and ode shapes of the orgnal syste wthout the added sprngs, could you fnd the frequences and ode shapes for the structure wth the added sprngs? r r d) If N= (.e. two floors only), fnd the second frequency and ode shape, agan n ters of the nown paraeters (for the orgnal syste wthout added lateral sprngs!). Assue that both asses are equal. e) Deterne the proportonal dapng atrx for the -dof syste n d) that would produce the sae value of dapng ξ n both odes. Fro ths atrx, fgure out the values and physcal confguraton of the dashpots. f) Agan for the -dof syste consdered n d), assue that t s subected to a unt pulsve load actng on the botto floor. Fnd the response n each floor, assung no dapng. N N r N 4

5 SOUION We frst forulate the ass and stffness atrx of the structure: M =... N + K =... N N N + N u u he fundaental egenvalue and odal shape are an egenpar, satsfyng the egenvalue proble, naely: K f = ω M f {} We substtute the ass and stffness atrx n equaton {} and perfor the algebrac operatons n the left and rght hand sde of {}. N N u N N N N ( N ) ω M f = ω = ω N N {} N + N K f = =... N N N + N N N {3} Substtutng {} and {3} n equaton {}, we get the followng syste of N equatons wth N unnown stffness coeffcents,... N : N = ω N ( N ) = ω N + ( N ) = ω N N N N = ω N + ( N ) N ( ) ( )... ( ) = ω N + N + + N {4} N he overturnng oent at the base s the oent exerted by the nerta forces s M = uz =. We want to prove that uz = 0. N = We now that the response of the th degree of freedo for the th ode s gven by: u( t) φ q ( t) herefore the acceleraton of the th degree of freedo for the th ode s gven by: u ( t) =φ q ( t). So the overturnng oent at the base, for the th ode s evaluated as follows: =. 5

6 N M = q φ z {5} = he nter-story heght, h,.e. the dstance between floors, s constant. herefore, the poston of the asses can be wrtten n vector for as follows: { Nh ( N ) h... h} h { N N... } h z = = = f {6} We now wrte equaton {5} n atrx for as follows: M = qf Mz= q h f M f = 0 {7} due to the orthogonalty of the odal shapes wth respect to the ass atrx. herefore, fro odal superposton, the total oent at the base s evaluated as follows: N N = = M M = M = q h f M f = qh f f = qhµ {8} Suppose that there s a wall that runs parallel to the structure, and that there are sprngs of rgdty r = α connectng the asses and the wall, n whch α s a nown constant (.e, the sprngs are proportonal to the asses). All the frequences and ode shapes of the orgnal syste wthout the added sprngs are nown. he presence of the lateral sprngs, results n the followng stffness atrx for the structure: * + K = + a = K + a M {9}... N... + N N N N he new egenvalue proble s forulated as follows (assue that the odal shapes are dentcal to the prevous structure and successvely verfy): * * K f = ( ω ) M f ( + α * ) f = ( ω ) ( ) K M M f ( ) Kf M f M f * = ω α = ω {0} herefore, the new natural frequences of the syste are: ( ) ω = ω + α {} * and the odal shapes of the odfed structure rean the sae. 6

7 For the degree of freedo syste, the ass and stffness atrces are: K 0 = M = + 0 Fro equaton {4}, we calculate the stffness coeffcents, as a functon of the asses and the fundaental frequency, as follows (N = ): = ω [ ] = ω + = 3 ω herefore, the stffness atrx s now: K = ω 5 {} Fro the orthogonalty propertes of the odal shapes wrt. the ass atrx, we now that: 0 f M f = 0 { } = 0 φ = 0 {3} φ We also verfy the orthogonalty propertes of the second ode wrt. the stffness atrx: f K f = 0 { } ω = 0 5 {4} he second frequency of the syste s calculated usng the Raylegh quotent, whch wll gve the exact frequency, when the second odal shape s used: R = ω { } ω f K f 5 30 ω = = = = 6 ω f 0 5 M f { } 0 {5} he Raylegh dapng atrx for the dof syste, s forulated as follows: C = α0 M + α K We now fro the class notes that the coeffcents α 0 and α are evaluated as follows: ξ ω ω α0 = ξ ω ω α {6} Settng n equaton {6} ξ = ξ = ξ, ω = 6 ω, we evaluate the coeffcents as follows: ( 6 6) ω α 0 ω ξω ξ = = α 6 6ω ξ 6ω 5 ω {7} 7

8 herefore, the dapng atrx s now: ( 6 6) ξ 0 6 C = α0 M + α K = ω ω 5 + = 0 ω 5 ξ ω ( 6 6) ( 6 ) = + = ξ ω 5 {8} he dapng atrx can be wrtten as follows: C.58.6 c + c c 3 = ξ ω = c c+ c Due to the way the Raylegh dapng atrx was forulated (ass and stffness proportonal), the physcal confguraton of the dashpots s shown n the fgure below: c c 3 c he force vector for a unt pulsve load actng on the botto floor, s the followng: p = { 0 δ ( t) } where: ( t) δ s the Drac Delta functon {9} he odal propertes of the structure, naely odal asses and forces are calculated below (dapng s gnored): 0 0 µ = { } 5 µ { } 5 0 = = = π = { } = δ ( t) π = { } = δ δ ( t) δ ( t) ( t) {0} herefore, the structure response s evaluated usng odal superposton, as follows: u u f h f h t t ( ω ) ( ω ) = = f + f = sn + sn 6 u 5 ω 5 6ω where h and h are the pulse response functons correspondng to the st and nd ode. 8