A simple stressed beam

We consider a simple beam stressed by a traction load F at both sides.

use case geometry

Beam geometry

The geometry is supposed to be deterministic; the diameter D is equal to:

D=0.02 \textrm{ (m)}

By definition, the yield stress is the load divided by the surface. Since the surface is \pi D^2/4, the stress is:

S=\frac{F}{ \pi D^2/4}

Failure occurs when the beam plastifies, i.e. when the axial stress gets larger than the yield stress:

R - \frac{F}{ \pi D^2/4} \leq 0

where R is the strength.

Therefore, the limit state function G is:

G(R,F) = R - \frac{F}{\pi D^2/4},

for any R,F \in \mathbb{R}.

The value of the parameter D is such that:

D^2/4 = 10^{-4},

which leads to the equation:

G(R,F) = R - \frac{F}{10^{-4} \pi}.

We consider the following distribution functions.




LogNormal( \mu_R= 3 \times 10^6, \sigma_R=3  \times 10^5 ) [Pa]


Normal( \mu_F=750 , \sigma_F=50) [N]

where \mu_R=E(R) and \sigma_R^2=V(R) are the mean and the variance of R.

The failure probability is:

P_f = \text{Prob}(G(R,F) \leq 0).

The exact P_f is

P_f = 0.02920.

Load the use case

We can load this classical model from the use cases module as follows :

>>> from openturns.usecases import stressed_beam as stressed_beam
>>> # Load the use case axial stressed beam
>>> sb = stressed_beam.AxialStressedBeam()

API documentation

See AxialStressedBeam.