>>13868437>The top half of the beam on the right experiences the same type of bending as the beam on the left because it is symmetric across the horizontal axis and that can be thought of as a fixed boundary condition.No, it is pinned. The correct analogy in this case is related to the buckled shapes.
>Therefore the Euler buckling on the right has half of the length as the one on the left.The reason for the doubled or halved effective length, and thus doubled or halved buckling load, is that the displacement diagram of the left case (case I) has the shape of half of the displacement diagram of case II; case I is half of a half-sine wave, and case II is a half sine-wave.
The buckling load depends on the buckling mode, which depends on the inflection point of bending in the buckled state.
