problem9_sol

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

This preview shows pages 1–4. Sign up to view the full content.

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

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
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
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

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

## 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.

### Page1 / 7

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

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online