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.

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.

Examples

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

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