Inverse FORM - Cantilever Beam

The following example is Example 7.2 from Chapter 7 of [15].

A cantilever beam is considered to fail if the displacement at the tip exceeds the threshold \(D_0\). The performance function \(G(\textbf{U})\) of this problem is given by

\[G = D_0 - \frac{4L^3}{Ewt} \sqrt{ \left(\frac{P_x}{w^2}\right)^2 + \left(\frac{P_y}{t^2}\right)^2}\]

Where the external forces are modeled as random variables \(P_x \sim N(500, 100)\) and \(P_y \sim N(1000,100)\). The constants in the problem are length (\(L=100\)), elastic modulus (\(E=30\times 10^6\)), cross section width (\(w=2\)), cross section height (\(t=4\)), and \(D_0=3\).

First, we import the necessary modules.

import numpy as np
from scipy import stats
from UQpy.distributions import Normal
from UQpy.reliability.taylor_series import InverseFORM
from UQpy.run_model.RunModel import RunModel
from UQpy.run_model.model_execution.PythonModel import PythonModel

Next, we initialize the RunModel object. The file defining the performance function file can be found on the UQpy GitHub. It contains a function cantilever_beam to compute the performance function \(G(\textbf{U})\).

model = PythonModel(model_script='local_pfn.py', model_object_name="cantilever_beam")
runmodel_object = RunModel(model=model)

Next, we define the external forces in the \(x\) and \(y\) direction as distributions that will be passed into FORM. Along with the distributions, FORM takes in the previously defined runmodel_object, the specified probability of failure, and the tolerances. These tolerances are smaller than the defaults to ensure convergence with the level of accuracy given in the problem.

p_fail = 0.04054
distributions = [Normal(500, 100), Normal(1_000, 100)]
inverse_form = InverseFORM(distributions=distributions,
                           runmodel_object=runmodel_object,
                           p_fail=p_fail,
                           tolerance_u=1e-5,
                           tolerance_gradient=1e-5)

With everything defined we are ready to run the inverse first-order reliability method and print the results. The solution to this problem given by Du is \(\textbf{U}^*=(1.7367, 0.16376)\) with a reliability index of \(\beta_{HL}=||\textbf{U}^*||=1.7444\) and probability of failure of \(p_{fail} = \Phi(-\beta_{HL})=0.04054\). We expect this problem to converge in 4 iterations. We confirm our design point matches this length, and therefore has a probability of failure specified by our input.

inverse_form.run()
beta = np.linalg.norm(inverse_form.design_point_u)
print('Design point in standard normal space (u^*):', inverse_form.design_point_u[0])
print('Design point in original space:', inverse_form.design_point_x[0])
print('Hasofer-Lind reliability index:', beta)
print('Probability of failure at design point:', stats.norm.cdf(-beta))
print('Number of iterations:', inverse_form.iteration_record[0])