# Statically Indeterminate Beam to the Sixth Degree

As the title says, I have a statically indeterminate beam to the sixth degree and I'm attempting to use the superposition method (aka force method) to solve for the reactions. My additional equations will be the angle at points A, B, C, and D as well as the deflection at points B and C.

Is this the correct method to solve this, or is this the wrong approach? My thought is that this might not be right because the equations which are used to solve for the angle and deflection assume a linear differential equation, which I _think_ is not the case here. SteamKing
Staff Emeritus
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The knife edge supports at B and C can allow a vertical reaction to form, but the beam is free to rotate at these points. Consequently, the boundary conditions for this beam are that the deflection = 0 at A, B, C, and D and the slope = 0 at A and D. For small deflections, the differential equation governing the beam's behaviour can be linearized. This configuration is known as a continuous beam, and there are several techniques which can be used to solve for the unknown reaction forces and moments.

The knife edge supports at B and C can allow a vertical reaction to form, but the beam is free to rotate at these points.
Ah, then I used the wrong support representation; the support method used at points B and C in the physical item will not allow for rotation, so a moment will be produced there.

SteamKing
Staff Emeritus