Vertical deflection of a tube

Description

We consider the deflection of a tube under a vertical stress.

simply supported beam

A simply supported beam

The parameters of the model are:

  • F : the strength,

  • L : the length of the tube,

  • a : position of the force,

  • D : external diameter of the tube,

  • d : internal diameter of the tube,

  • E : Young modulus.

The following figure presents the internal and external diameter of the tube:

tube diameters

Tube diameters

The area moment of inertia of the cross section about the neutral axis of a round tube (i.e. perpendicular to the section) with external and internal diameters D and d are:

I = \frac{\pi (D^4-d^4)}{32}.

The vertical deflection at point x=a is:

g_1(X) = - F \frac{a^2 (L-a)^2}{3 E L I},

where X=(F,L,a,D,d,E). The angle of the tube at the left end is:

g_2(X) = - F \frac{b (L^2-b^2)}{6 E L I},

and the angle of the tube at the right end is:

g_3(X) = F \frac{a (L^2-a^2)}{6 E L I}.

The following table presents the distributions of the random variables. These variables are assumed to be independent.

Variable

Distribution

F

Normal(1,0.1)

L

Normal(1.5,0.01)

a

Uniform(0.7,1.2)

D

Triangular(0.75,0.8,0.85)

d

Triangular(0.09,0.1,0.11)

E

Normal(200000,2000)

References

Load the use case

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

>>> from openturns.usecases import deflection_tube as deflection_tube
>>> # Load the tube deflection model
>>> dt = deflection_tube.DeflectionTube()

API documentation

See DeflectionTube.

Examples based on this use case