Statically Indeterminate Beam to the Sixth Degree


by 6Stang7
Tags: beam, degree, indeterminate, sixth, statically
6Stang7
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#1
Jul6-11, 10:49 AM
P: 212
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.

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SteamKing
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#2
Jul6-11, 10:02 PM
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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.
6Stang7
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#3
Jul7-11, 10:37 AM
P: 212
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
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Jul7-11, 12:03 PM
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Statically Indeterminate Beam to the Sixth Degree


If a moment reaction can develop at B and C but the loading is only applied to segment BC, then it would appear that segments AB and CD are not affected by the load F. Whatever the case, additional boundary conditions for B and C are required in order to determine all the reacting forces and moments in the beam. As long as deflections are assumed sufficiently small, the linearized Euler-Bernoulli equation will still apply.


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