A simple stressed beam¶

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

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.

Variable	Distribution
R	LogNormal( $\mu_R= 3 \times 10^6$ , $\sigma_R=3 \times 10^5$ ) [Pa]
F	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:

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

The exact $P_f$ is

P_f = 0.02920.

API documentation¶

class AxialStressedBeam

Data class for the axial stressed beam example.

Attributes:

dimThe dimension of the problem: dim=2.
DConstant: Diameter D = 0.02 (m)
modelSymbolicFunction: The limit state function.
muRConstant: muR=3.0e6, yield strength mean
sigmaRConstant: sigmaR = 3.0e5, yield strength variance
distribution_RLogNormalMuSigma distribution of the yield strength: ot.LogNormalMuSigma(muR, sigmaR, 0.0).getDistribution()
muFConstant: muF=750.0, traction load mean
sigmaFConstant: sigmaR = 50.0, traction load variance
distribution_FNormal distribution of the traction load: ot.Normal(muF, sigmaF)
distributionJointDistribution: The joint distribution of the input parameters.

Examples

>>> from openturns.usecases import stressed_beam
>>> # Load the axial stressed beam
>>> sm = stressed_beam.AxialStressedBeam()