Calculation of Deflection of Gradient Pile Using Deflection Equation

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紹臺
紹臺 2024 年 10 月 18 日
回答済み: Rahul 2024 年 10 月 21 日
function value = Kstiff_function(ii, interval, ksi, count, EI)
value = zeros(1, count);
% 节点的局部刚度矩阵定义
if (ii == 1)
value(1, 1) = -1 * EI(ii,1) / interval^3;
value(1, 2) = 3 * EI(ii,1) / interval^3;
value(1, 3) = -3 * EI(ii,1) / interval^3;
value(1, 4) = 1 * EI(ii,1) / interval^3;
elseif (ii == 2)
value(1, 1) = 1 * EI(1,1);
value(1, 2) = -2 * EI(2,1);
value(1, 3) = 1 * EI(3,1);
elseif (ii >= 3) && (ii <= 100)
value(1, ii - 2:ii + 2) = [1*EI(ii), -4*EI(ii), ...
6*EI(ii) + ksi(ii,1)*interval^4, -4*EI(ii), ...
1*EI(ii)];
elseif ii == 101
value(1, ii - 2:ii + 2) = [1*EI(ii), -4*EI(ii), ...
(6*EI(ii) + ksi(ii,1)*interval^4 + 6*EI(ii+1) + ksi(ii + 1,1)*interval^4)/2, -4*EI(ii + 1), ...
1*EI(ii + 1)];
elseif (ii >= 102) && (ii <= count-2)
value(1, ii - 2:ii + 2) = [1*EI(ii), -4*EI(ii), ...
6*EI(ii) + ksi(ii,1)*interval^4, -4*EI(ii), ...
1*EI(ii)];
elseif (ii == count-1)
value(1, count-2:count) = [1*EI(count-2,1), ...
-2*EI(count-1,1), ...
1*EI(count,1)];
elseif (ii == count)
value(1, count-3:count) = [1*EI(count-3,1), ...
-3*EI(count-1,1), ...
3*EI(count-2,1), ...
-1*EI(count,1)];
end
end
  1 件のコメント
Rahul
Rahul 2024 年 10 月 18 日
Hi @紹臺, can you explain what you're trying to achieve and the issues that you are facing using the given function.

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回答 (1 件)

Rahul
Rahul 2024 年 10 月 21 日
Hi Praful,
I believe you are using a MATLAB function to calculate Deflection of Gradient Pile Using Deflection Equation. Here a few improvements you can consider accommodating in your given code snippet, according to your use-case:
  • Matrix Indexing with EI: There is a usage of EI(ii,1) for accessing the stiffness matrix, but in other places (for ii >= 3), EI(ii) is being used without specifying the second dimension. You can ensure that EI is either a vector or a 2D matrix. If EI is a 2D matrix, you can use EI(ii, 1) consistently, otherwise, use EI(ii) for a vector.
value(1, ii - 2:ii + 2) = [1*EI(ii), -4*EI(ii), 6*EI(ii) + ksi(ii,1)*interval^4, ...];
  • Indexing of value Matrix: Accessing parts of value(1, ...) might end up accessing indices out of bounds, since MATLAB uses 1-based indexing, and also while using ii - 2:ii + 2, especially for small values of ii. For example, when ii = 1 or ii = 2, ii 2 becomes zero or negative, which would lead to errors. Ensuring that these ranges are properly bounded can avoid index out-of-bounds errors.
value(1, ii - 2:ii + 2) = [1*EI(ii), -4*EI(ii), ...];
  • Inconsistent Access to ksi Matrix: Similar to EI, inconsistency in using ksi(ii, 1) (assuming ksi is a matrix) and sometimes just ksi(ii), could lead to errors depending on how the ksi variable is structured:
6*EI(ii) + ksi(ii,1)*interval^4
  • Dimension Mismatch in Assignments: For example, in the following part, while assigning a 3-element vector to a portion of the value array:
value(1, count-2:count) = [1*EI(count-2,1), -2*EI(count-1,1), 1*EI(count,1)];
For more information regarding usage of matrices and arrays in MATLAB, refer to the documentation links mentioned below:

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