problem9_sol - 2.314/1.56/2.084/13.14 Fall 2006 Problem Set...

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2.314/1.56/2.084/13.14 Fall 2006 Problem Set IX Solution Solution: The pipe can be modeled as in Figure 1. Lumped mass B m B P B A m A P A R Pipe Beam with the same support Geometry and properties: L=3m, R=0.105m and t=0.007m E=200GPa, ρ =8500kg/m 3 and ρ water =750kg/m 3 m A = m B = π ( R 2 R i 2 ) L ρ + ( R i 2 + 1.1 R 2 ) L water = 133.724 Kg 2 The governing motion eqution for dynamic response of this pipe can be expressed as: [ M ]{ u } + [ D ]{ u } + [ K ]{ u } = { F } where: m A 0 [M]=mass matrix, M = 0 m B [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 theory: Suppose the displacements of point A (L/3) and B (2L/3) are u A and u B , respectively. Assuming that there is only a force P A acting on point A (at L/3), we can calculate the corresponding displacements of points A and B, satisfying the boundary conditions: 1
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u (0) = 0 u ( L ) = u (3) = 0 θ (0) = du ( z ) = 0 y dz z = 0 For L>z>L/3 : dM M = V (3 z ) = − y (3 z ) y x dz Solving this equation with the boundary condition M y (L)=0 , we get M y = C 1 (3 z ) For z<L/3 : dM M y = 2 P A + V (3 z ) = 2 P A y (3 z ) x dz Solving this equation: M y = C 2 (3 z ) + 2 P A Because of continuity at z=L/3 , we have C 1 = C 2 + P A Meanwhile, M y = σ z xdA = E ε z xdA = EK y x 2 dA = EK y I , where K y = d dz = d dz 2 u 2 A A A π 0.105 where I = x 2 dA = 2 0 cos 2 d 0.098 r 3 dr = 2.3023 × 10 5 A So that C 1 K y = M y = EI (3 z ),1 < z < 3 EI C 2 (3 z ) + 2 P A ,0 < z < 1 EI EI C 1 (3 z ) 2 9 C 2 4 P A + + y = K y dz = EI 2 2 EI EI at z=0, θ y =0 C 2 (3 z ) 2 + 2 P A z + 9 C 2 EI 2 EI 2 EI C 1 (3 z ) 3 9 C 2 4 P A 13 P A 9 C 2 u ( z ) = y dz = EI C 2 6 (3 + z ) 3 2 + EI P A + z 2 EI + 9 z C 2 z 3 EI 9 C 2 2 EI EI 6 EI 2 EI 2 EI At last, we have another boundary condtion, u(L)=0 So that 2
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3 9 C 2 + 4 P A 13 P A 9 C 2 = 0 2 EI EI 3 EI 2 EI 23 C 2 = − P A 27 4 C 1 = C 2 + P A = P A 27 Then, we obtain u C 2 8 + P A + 9 C 2
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This note was uploaded on 02/24/2012 for the course MECHANICAL 2.084J taught by Professor Mujids.kazimi during the Fall '06 term at MIT.

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problem9_sol - 2.314/1.56/2.084/13.14 Fall 2006 Problem Set...

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