2.314/1.56/2.084/13.14 Fall Problem Set IX Solution

Transcription

1 Solution: The pipe can be modeled as in Figure /1.56/.084/13.14 Fall 006 Problem Set IX Solution m Lumped mass P R m P Pipe eam with the same support Geometr and properties: L=3m, R=0.105m and t=0.007m E=00GPa, ρ=8500kg/m 3 and ρ water =750kg/m 3 m = m = π (R R i )Lρ + (π R i + 1.1πR )Lρ water = Kg The governing motion eqution for dnamic response of this pipe can be expressed as: [M ]{u&&} + [D]{u&} + [K]{u} = {F} where: m 0 [M]=mass matrix, M = 0 m [D]=damping matrix [K]=stiffness matrix [F]=vector of loads, earthquake load in our case, {F} = [M ] {S a } {u}=vector of nodal displacements First of all, we should calculate the stiffnes matrix of the pipe using beam theor: Suppose the displacements of point (L/3) and (L/3) are u and u, respectivel. ssuming that there is onl a force P acting on point (at L/3), we can calculate the corresponding displacements of points and, satisfing the boundar conditions: 1

2 u(0) = 0 u(l) = u(3) = 0 θ (0) = du(z) = 0 z=0 For L>z>L/3: M = V x (3 z) = (3 z) Solving this equation with the boundar condition M (L)=0, we get M = (3 z) For z<l/3: M + V x (3 z) (3 z) Solving this equation: M = C (3 z) + P ecause of continuit at z=l/3, we have = C + P Meanwhile, M = σ z xd = Eε z xd = EK x d = EK I, where K = d π where I = x d = 0 cos θ dθ r 3 dr = K = M = (3 z),1 < z < 3 EI EI C (3 z) + P,0 < z < 1 EI EI (3 z) 9C 4P + + θ = K = EI EI EI at z=0, θ =0 C (3 z) + P z + 9C EI EI EI (3 z) 3 9C 4P 13P 9C u(z) = θ = + EI 6 + EI EI z 3 EI EI 3 C (3 z) P + 9 C z 9C + z EI 6 EI EI EI t last, we have another boundar condtion, u(l)=0 θ d = u

3 3 9C + 4P 13P 9C = 0 EI EI 3EI EI 3 C = P 7 4 = C + P 7 Then, we obtain u C 8 P + + 9C 9C P EI 6 EI EI EI EI u = C1 1 9C 8P 13P 9C = P EI 6 EI EI 3EI EI EI Similarl, assuming that there is onl a force P acting on point (at L/3), we can calculate the corresponding displacements of points and, satisfing the boundar conditions: For L>z>L/3: M = V x (3 z) = (3 z) Solving this equation with the boundar condition M (L)=0, we get M = (3 z) For z<l/3: M = P + V x (3 z) = P (3 z) Solving this equation: M = C (3 z) ecause of continuit at z=l/3, we have = C Meanwhile, M = σ z xd = Eε z xd = EK x d = EK I, where K = dθ = d u π where I = x d = cos θ dθ r 3 dr = K = M EI < z < 3 = EI (3 z), C P (3 z) +, z < EI EI (3 z) 9C 5P θ = K = + EI EI + EI C (3 z) P 9C + z + EI EI EI at z=0, θ =0 3

4 (3 z) 3 9C 5P 9C z 19P u(z) = θ = EI EI EI EI 6EI C (3 z ) 3 P + 9C z + z 9 C EI 6 EI EI EI t last, we have another boundar condtion, u(l)=0 3 9C + 5P 9C 19P = 0 EI EI EI 6EI 13 C = P 7 14 = C = P 7 Then, we obtain u C 8 P + 9C + 9C P 0.14 EI 6 EI EI EI EI u = C1 1 9C 5P 9C 19P = P EI 6 EI EI EI 6EI EI, according to superposition when P and P act on the sstem simultaneousl, u P = u EI P In the meantime, P u P = K u We can get K = EI = EI = Secondl, we should compute the damping matrix: The damping matrix has onl two non-zero elements, located on the diagonal. These elements are equal and give two percent critical damping for vibration at the sstem fundmental frequenc. Thus, we should calculate the undamped fundmental frequenc first. [M ]{u&&} + [K ]{u} = u&& u + EI = u&& u Solving this equation b assuming u = e ω t and u = e ω t, we get 4

5 133.74ω EI EI EI ω EI = 0 (133.74ω EI )(133.74ω EI ) (EI ) = ω EIω (EI ) = 0 ω = or Then, the fundamental undamped frequencs are ω 1 = 940.8,ω = in radius/s; note that ω = πf and corresponding eigenvectors are u 1 =,u = Then, we should also calculate the critical damping matrix, Suppose that the critical damping of lumped mass is considered seperatel: 0 D cr = 0 C [M ]{u&&} + [D]{u&} + [K]{u} = 0 N and the solution is u(t) = n cosω n tu n n=1 Multipl the above equation with N T leftl, and also replace {u} with N {u}, where N = [u 1 u ] = we get ω + C ω + 5.7EI = ω + C ω +.9EI If there are nontrivial solutions for 1 and, we obtain ω + ω + 5.7EI = ω + C ω +.9EI = 0 critical damping matrix: D cr = = 0 C T D cr = N D cr N = Damping matrix in this case D = 0.0D cr =

6 t last, the force due to earthquake {F} = [M ] {u&& g } Natural frequenc ω1 ω f 1 = = and f = = π π ccording to the response spectrum of Fig 8 in note M-3, extrapolating to the calculated frequencies, we obtain the maximum displacements corresponding to S a =0.33g are S d1 S a = = m ω S d S a = = m ω So {F} = u&& g Finall, we obtain u&& u& 5.7EI 0 u T && + & + = N u&& g u u 0.9EI u u&& u& 1 5.7EI 0 u 0.64 u&& u& EI u = 1.39 u&& g Then, maximum displacement u 1,max = 0.64 S d1 u 1 = u,max = 1.39 S d u = P 0.5 1,max = [(Cωu) + (Ku) ] = EI = = 64 6

7 P 0.5,max = [(Cωu) + (Ku) ] = EI = = Now, given the forces of P and P, we calculate the forces acted on supports: F L P P F 0 ssuming the forces that act on the supports are F 0 and F L, respectivel, as in the above figure. dk 4 14 F L = = EI 7 7 z=l z=l 3 13 F 0 F L 7 7 For P 1,max : F F 1 = = P 1,max = F L Likewise, for P,max : F = F P = F = L ,max 308. gain, F = [F 1 + F ] 0.5 = 539 Newton So, the peak forces that act on the support z=0 and z=l are 539 and Newton, respectivel. 7