# Beam bending strain field

lewis198
If I have a wide beam, parallel to the x axis, with its COM at the origin, then I want it to curve about the y axis, what would the elements of the strain tensor be?

I have come to the conlusion that the beam would, for example,contract above x-axis and expand below it. But I don't know how to describe strain relative to the y and z axes, and then how to translate this knowledge to the strain tensor.

afreiden
If I have a wide beam, parallel to the x axis, with its COM at the origin, then I want it to curve about the y axis, what would the elements of the strain tensor be?

I have come to the conlusion that the beam would, for example,contract above x-axis and expand below it. But I don't know how to describe strain relative to the y and z axes, and then how to translate this knowledge to the strain tensor.

"wide beam" to me sounds like a shell.

Reissner-Mindlin shell theory is what is used in Finite Element Analysis. They neglect the higher order terms but there is a lot of information on the subject.

If you are more interested in analytical theory that is analogous to standard beam theory, then I believe "Kirchhoff-Love" would be what you are looking for.

Hope that helps,

Studiot

Wide beam? : What sort of wide beam?

Wide flanged I beams are available for increased bending resistance and their properties are tabulated.

There are code requirements and specifications for wide RC beams.

I am going to assume a simple homogeneous rectangular wide beam of height h in the y direction and breadth b in the z direction.

Yes you need also to specify the z direction - wide beams are 3 dimensional.

A beam is considered wide when

b >> h say 5 times or more.

Under these conditions the material is not free to expand or contract in the lateral z direction under bending along the x direction about the y direction. (note I said direction not axis)

In particular εz ≈ 0 at z=0 so

$${\varepsilon _z} = \frac{1}{E}[{\sigma _z} - \nu \left( {{\sigma _x} + {\sigma _y}} \right) \approx 0$$

Since h is small σy ≈ 0 So

$${\sigma _z} = \nu {\sigma _{{x_{z = 0}}}}$$

Thus

$${\varepsilon _x} = \frac{{1 - {\nu ^2}}}{E}{\sigma _x} = \frac{{1 - {\nu ^2}}}{{E{I_z}}}{M_z}y$$

In general the strain is reduced by a factor of ${1 - {\nu ^2}}$

Edit
So you can see that the change is the insertion a modifying constant into your strain tensor, I will leave you to do this since you haven't provided any notation.
You need to be careful here since my strain is engineering strain, not tensor strain, which is a factor of 1/2 different.

You should also note that the sideways distribution may also depend upon the support conditions.

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