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Finite difference method for fourth order PDE |
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#1 |
New Member
Box
Join Date: Feb 2021
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How can I solve this initial boundary problem?
WhatsApp Image 2021-02-14 at 20.58.08.jpeg Are my approximations right? WhatsApp Image 2021-02-14 at 23.18.58.jpeg it's the nonlinear mathematical model of the dynamics of a rotating drill string compressed at both ends by the longitudinal force N(x3, t) u2 - displacement of the center of bending of the cross-section of the bar along the x2 axe due to bending E = 2.1e11 # Pa - Young's modulus rho = 7800 # kg/м3 - density v = 0.28 # Poisson's ratio d1 = 0.12 # м - drill string inner diameter d2 = 0.2 # м - external diameter A = 2.01e-2 # Column cross-sectional area Ix1 = PI / 64. * (d2 ** 4 - d1 ** 4) # Ring section moment of inertia Omega = 0.33 # drill string angular velocity C1 = 0.01 # constant determining the rate of deviation of the drill string axis from the initial position in the Ox2x3 plane at the initial moment time. Thank you in advance ![]() Last edited by BoxQwerty; February 15, 2021 at 23:37. |
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#2 |
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Filippo Maria Denaro
Join Date: Jul 2010
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I don't know the physics described by this PDE, the problem is not only in a consistent discretization but also on the mathematical character of the equation. A second order in time PDE could be hyperbolic but I see also fourth order spatial terms. Central discretization in time and space is consistent but I don't know if it is also stable.
Then, also the BCs for this problem should be analysed. Have you some specific reference? |
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#3 |
New Member
Box
Join Date: Feb 2021
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Material I have is all in Russian.
![]() In short, it's the nonlinear mathematical model of the dynamics of a rotating drill string compressed at both ends by the longitudinal force N(x3, t) u2 - displacement of the center of bending of the cross-section of the bar along the x2 axe due to bending E = 2.1e11 # Pa - Young's modulus rho = 7800 # kg/м3 - density v = 0.28 # Poisson's ratio d1 = 0.12 # м - drill string inner diameter d2 = 0.2 # м - external diameter A = 2.01e-2 # Column cross-sectional area Ix1 = PI / 64. * (d2 ** 4 - d1 ** 4) # Ring section moment of inertia Omega = 0.33 # drill string angular velocity C1 = 0.01 # constant determining the rate of deviation of the drill string axis from the initial position in the Ox2x3 plane at the initial moment time. |
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#4 |
New Member
Box
Join Date: Feb 2021
Posts: 4
Rep Power: 5 ![]() |
Material I have is all in Russian.
In short it's the nonlinear mathematical model of the dynamics of a rotating drill string compressed at both ends by the longitudinal force N(x3, t) u2 - displacement of the center of bending of the cross-section of the bar along the x2 axe due to bending E = 2.1e11 # Pa - Young's modulus rho = 7800 # kg/м3 - density v = 0.28 # Poisson's ratio d1 = 0.12 # м - drill string inner diameter d2 = 0.2 # м - external diameter A = 2.01e-2 # Column cross-sectional area Ix1 = PI / 64. * (d2 ** 4 - d1 ** 4) # Ring section moment of inertia Omega = 0.33 # drill string angular velocity C1 = 0.01 # constant determining the rate of deviation of the drill string axis from the initial position in the Ox2x3 plane at the initial moment time. |
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Tags |
finite difference, numerical solution, pde |
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