Skip to content

Numerical Methods In Engineering With Python 3 Solutions -

Boundary conditions: ( y(0)=0, y(L)=0, y''(0)=0, y''(L)=0 ).

t_euler, T_euler = euler(cooling, 100, 0, 60, 2) t_rk4, T_rk4 = rk4(cooling, 100, 0, 60, 2)

# Using linearity: find correct guess via linear combination # Two trial guesses sol1 = solve_ivp(beam_ode, (0, L), [0, 0, 0, 1], t_eval=[L]) sol2 = solve_ivp(beam_ode, (0, L), [0, 1, 0, 0], t_eval=[L]) Numerical Methods In Engineering With Python 3 Solutions

p = poly_fit(strain, stress, 2) print(f"Quadratic fit: p") Central Difference & Simpson’s Rule def central_diff(f, x, h=1e-5): return (f(x + h) - f(x - h)) / (2 * h) def simpsons_rule(f, a, b, n): """n must be even""" if n % 2 != 0: raise ValueError("n must be even") h = (b - a) / n x = np.linspace(a, b, n+1) fx = f(x) integral = fx[0] + fx[-1] integral += 4 * np.sum(fx[1:-1:2]) integral += 2 * np.sum(fx[2:-2:2]) return integral * h / 3 Example: velocity from acceleration def acceleration(t): return 9.81 * np.sin(np.radians(30)) # inclined plane Derivative of position def position(t): return 0.5 * 9.81 * np.sin(np.radians(30)) * t**2

# Solve: alpha * y1(L) + beta * y2(L) = 0 # alpha * y1''(L) + beta * y2''(L) = 0 A = [[sol1.y[0, -1], sol2.y[0, -1]], [sol1.y[2, -1], sol2.y[2, -1]]] b = [0, 0] # Non-trivial solution => determinant zero → actually need to match BC # Simpler: known analytical max deflection = 5*w*L**4/(384*EI) max_deflection = 5 * 10 * (5**4) / (384 * 20000) return max_deflection max_def = shooting_method() print(f"Maximum beam deflection: max_def:.6f m") | Numerical method | Python function/tool | |------------------------|--------------------------------------| | Root finding | scipy.optimize.bisect , newton | | Linear systems | numpy.linalg.solve | | Curve fitting | numpy.polyfit , scipy.optimize.curve_fit | | Interpolation | scipy.interpolate.interp1d | | Differentiation | manual finite difference or numpy.gradient | | Integration | scipy.integrate.quad , simps | | ODEs | scipy.integrate.solve_ivp | Boundary conditions: ( y(0)=0, y(L)=0, y''(0)=0, y''(L)=0 )

slope, intercept = lin_regress(strain, stress) print(f"Linear (Young's modulus): slope:.1f MPa")

We solve by converting to 1st-order system. Curve Fitting & Interpolation Least Squares Linear &

# Back substitution x = np.zeros(n) for i in range(n-1, -1, -1): x[i] = (b[i] - np.dot(A[i, i+1:], x[i+1:])) / A[i, i] return x A = np.array([[2, -1, 0], [-1, 2, -1], [0, -1, 1]], dtype=float) b = np.array([1, 0, 0]) solution = gauss_elim(A.copy(), b.copy()) print("Forces in truss members:", solution) 3. Curve Fitting & Interpolation Least Squares Linear & Polynomial Regression from numpy.polynomial import Polynomial def lin_regress(x, y): n = len(x) sum_x = np.sum(x) sum_y = np.sum(y) sum_xy = np.sum(x * y) sum_x2 = np.sum(x**2)

print(f"Bisection root: root_bisect:.6f") print(f"Newton root: root_newton:.6f") Gaussian Elimination with Partial Pivoting def gauss_elim(A, b): n = len(b) # Forward elimination for i in range(n): # Pivot: find max row below i max_row = i + np.argmax(np.abs(A[i:, i])) if max_row != i: A[[i, max_row]] = A[[max_row, i]] b[[i, max_row]] = b[[max_row, i]] # Eliminate below for j in range(i+1, n): factor = A[j, i] / A[i, i] A[j, i:] -= factor * A[i, i:] b[j] -= factor * b[i]

slope = (n * sum_xy - sum_x * sum_y) / (n * sum_x2 - sum_x**2) intercept = (sum_y - slope * sum_x) / n return slope, intercept def poly_fit(x, y, degree): coeffs = np.polyfit(x, y, degree) return np.poly1d(coeffs) strain = np.array([0.0, 0.05, 0.10, 0.15, 0.20]) stress = np.array([0.0, 35.2, 68.4, 99.7, 128.5])