Column Buckling Calculator

import math

def euler_buckling(E, I, L, K, A=None):
    """
    Euler critical buckling load for a slender column.

    E: Elastic modulus (MPa)
    I: Minimum moment of inertia (mm⁴)
    L: Unsupported length (mm)
    K: Effective length factor
    A: Cross-section area (mm²) – optional

    Returns dict with Pcr, Le, sigma_cr, lambda
    """

    if E <= 0 or I <= 0 or L <= 0 or K <= 0:
        raise ValueError("E, I, L, K must be greater than zero.")

    Le = K * L
    Pcr = (math.pi**2 * E * I) / (Le**2)

    result = {"Pcr_N": Pcr, "Le_mm": Le}

    if A and A > 0:
        r = math.sqrt(I / A)
        result["r_mm"] = r
        result["lambda"] = Le / r
        result["sigma_cr_MPa"] = Pcr / A

    return result


# Example
E = 200000
I = 2500000
L = 3000
K = 1
A = 2500

results = euler_buckling(E, I, L, K, A)
for k, v in results.items():
    print(f"{k}: {v}")

function results = euler_buckling(E, I, L, K, A)
    % Euler critical buckling load for a slender column.
    %
    % E = Elastic modulus (MPa)
    % I = Minimum moment of inertia (mm^4)
    % L = Unsupported length (mm)
    % K = Effective length factor
    % A = Cross-section area (mm^2) – optional

    if E <= 0 || I <= 0 || L <= 0 || K <= 0
        error('E, I, L, K must be greater than zero.');
    end

    Le = K * L;
    Pcr = (pi^2 * E * I) / (Le^2);

    results.Pcr = Pcr;
    results.Le = Le;

    if nargin >= 5 && A > 0
        r = sqrt(I / A);
        results.r = r;
        results.lambda = Le / r;
        results.sigma_cr = Pcr / A;
    end
end

% Example
E = 200000;
I = 2500000;
L = 3000;
K = 1;
A = 2500;

r = euler_buckling(E, I, L, K, A);
disp(r);