I know a thin walled beam is the one in which the length is one order more than other crossectional dimensions. Please explain why it is widely used over other beams?
The "slender beam", i.e. the one that has a length (L) to be much greater than the dimensions of its cross section (b,t). Thus L>>>b and L >>> t. In principle we should have L >>> sqrt(A).
In general, the total elastic energy stored in the beam (which will eventually prescribe the deformation) is due to two major sources: Bending and Shear.
The theory of Bernoulli-Euler Beam, Only the bending is considered and the shear is ignored. In Timoshenko beam theory, both are considered.
However, for slender beams (L/sqrt(A))>>>1, the contribution of the bending (from the bending moment) is much stronger than that of the shear. Thus, the shear contribution can be safely ignored.
The advantages of applying the "slender beam" is that the deformations and curvatures can be easily calculated (they are even tabulated in some texts) in comparison to Timoshenko beam.
In Numerical modelling, You can verify a 2D plane stress problem wit hthe analytical results of the slender beam.
Hope that helps
If you mean "thin-walled open section beams" they are beams with cross section sape as U, H, or more complex open section shapes.
Due to their low weight, thin-walled open section beams are widely applied in many modern metal structures
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