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Question: Interpolation of Two Variables with Maple

Question:Interpolation of Two Variables with Maple

moses 119
July 26 2014
0

Good Afternoon Everyone,

Would anybody be kind to help me with the interpolation of two variables in a table to obtain a value for f18.1 (Stiffness) and f18.2 (Damping) given SR_ratio and Lamda_ratio.

 

Regards

Moses


 

NULL

restart

with(CurveFitting)

with(ExcelTools):

``

Slenderness Ratio (Variable 1):

SR__ratio := 84.01

``

Ratio of modulus of elasticity of Pile vs shear modulus of soil (Variable 2):

`λ__ratio` := 229.58

``

Slenderness ratio of pile from stiffness of pile table:

SR__stiff := [10.8696, 21.7391, 32.6087, 43.4783, 46.7391, 54.3478, 65.2174, 76.0870, 86.9565, 100.0]

``

Ratio of Young's Modulus of Pile to Shear modulus of soil:

EG__ratio := [250.0, 500.0, 1000.0, 2500.0, 5000.0, 10000.0]

``

Vertical stiffness coefficient for floating pile as per Novak (1983) f18.1.250:

`f__18.1.250` := [0.332e-1, 0.509e-1, 0.571e-1, 0.582e-1, 0.582e-1, 0.582e-1, 0.582e-1, 0.582e-1, 0.582e-1, 0.582e-1]

``

Vertical stiffness coefficient for floating pile as per Novak (1983) f18.1.500:

`f__18.1.500` := [0.187e-1, 0.301e-1, 0.364e-1, 0.405e-1, 0.416e-1, 0.416e-1, 0.416e-1, 0.416e-1, 0.416e-1, 0.416e-1]

``

Vertical stiffness coefficient for floating pile as per Novak (1983) f18.1.1000:

`f__18.1.1000` := [0.104e-1, 0.166e-1, 0.218e-1, 0.26e-1, 0.27e-1, 0.281e-1, 0.291e-1, 0.301e-1, 0.301e-1, 0.301e-1]

``

Vertical stiffness coefficient for floating pile as per Novak (1983) f18.1.2500:

`f__18.1.2500` := [0.52e-2, 0.83e-2, 0.104e-1, 0.125e-1, 0.135e-1, 0.145e-1, 0.166e-1, 0.177e-1, 0.187e-1, 0.197e-1]

``

Vertical stiffness coefficient for floating pile as per Novak (1983) f18.1.10000:

`f__18.1.10000` := [0.21e-2, 0.31e-2, 0.42e-2, 0.42e-2, 0.52e-2, 0.52e-2, 0.62e-2, 0.62e-2, 0.73e-2, 0.83e-2]

``

Slenderness ratio of pile from damping of pile table:

SR__damp := [10.8696, 16.3043, 21.7391, 27.1739, 32.6087, 38.0435, 43.4783, 48.913, 54.3478, 59.7826, 65.2174, 70.6522, 76.0870, 81.5217, 86.9565, 92.3913, 100.0]

``

Vertical damping coefficient for floating pile as per Novak (1983) f18.2.250:

`f__18.2.250` := [.1032, .1137, .1126, .1095, .1053, .1021, 0.989e-1, 0.979e-1, 0.979e-1, 0.979e-1, 0.979e-1, 0.979e-1, 0.979e-1, 0.979e-1, 0.979e-1, 0.989e-1, 0.989e-1]

``

Vertical damping coefficient for floating pile as per Novak (1983) f18.2.500:

`f__18.2.500` := [0.558e-1, 0.695e-1, 0.811e-1, 0.832e-1, 0.811e-1, 0.789e-1, 0.758e-1, 0.737e-1, 0.726e-1, 0.716e-1, 0.705e-1, 0.695e-1, 0.695e-1, 0.695e-1, 0.695e-1, 0.705e-1, 0.705e-1]

``

 Vertical damping coefficient for floating pile as per Novak (1983) f18.2.1000:

`f__18.2.1000` := [0.295e-1, 0.421e-1, 0.495e-1, 0.537e-1, 0.568e-1, 0.589e-1, 0.579e-1, 0.568e-1, 0.558e-1, 0.537e-1, 0.526e-1, 0.516e-1, 0.516e-1, 0.505e-1, 0.505e-1, 0.505e-1, 0.495e-1]

``

 Vertical damping coefficient for floating pile as per Novak (1983) f18.2.2500:

`f__18.2.2500` := [0.126e-1, 0.179e-1, 0.232e-1, 0.263e-1, 0.305e-1, 0.326e-1, 0.347e-1, 0.368e-1, 0.379e-1, 0.379e-1, 0.379e-1, 0.379e-1, 0.379e-1, 0.368e-1, 0.358e-1, 0.358e-1, 0.337e-1]

``

   Vertical damping coefficient for floating pile as per Novak (1983) f18.2.10000:

`f__18.2.10000` := [0.32e-2, 0.53e-2, 0.74e-2, 0.84e-2, 0.105e-1, 0.116e-1, 0.137e-1, 0.147e-1, 0.147e-1, 0.168e-1, 0.168e-1, 0.179e-1, 0.189e-1, 0.189e-1, 0.200e-1, 0.211e-1, 0.211e-1]

``
 

NULL

``

``

`f__18.1` := ArrayInterpolation(SR__ratio, `λ__ratio`, SR__stiff, EG__ratio, `f__18.1.250`)

`f__18.2__ _` := ArrayInterpolation(SR__ratio, `λ__ratio`, SR__damp, EG__ratio, `f__18.2.250`)

NULL

NULL


Download Dynamic_Table.mwDynamic_Table.mw

Good Afternoon Everyone,

Would anybody be kind to help me with the interpolation of two variables in a table to obtain a value for f18.1 (Stiffness) and f18.2 (Damping) given SR_ratio and Lamda_ratio.

 

Regards

Moses
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