Stiffness factor for member in beam

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Homework Help Overview

The discussion revolves around the stiffness factor for a member in a beam, specifically questioning the values of KBC and KAB in relation to their support conditions. Participants are examining the implications of fixed and pinned supports on beam stiffness, referencing equations related to beam deflection and stiffness.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants are exploring the reasoning behind the stiffness factors KBC and KAB, questioning whether the values should be 3EI / L or 4EI / L based on support conditions. There is a focus on understanding the definitions of "far end" and how it affects the stiffness calculations.

Discussion Status

The discussion is active, with participants sharing their interpretations and seeking clarification on the definitions and implications of support conditions. Some guidance has been offered regarding the behavior of beams under different constraints, but no consensus has been reached on the correct stiffness factors.

Contextual Notes

Participants note that their understanding is based on textbook extracts, and there is a lack of explicit definitions for terms like "far end." The discussion highlights the complexity of beam behavior under varying support conditions.

fonseh
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Homework Statement


In this question , I don't understand why the KBC is 4EI / L ...

Homework Equations

The Attempt at a Solution


I was told that for the far end pinned or roller supported , the K = 3EI / L , so shouldn't the KBC = 3EI / L.. Is the author wrong ?

In the 3rd photo , we can see that for far end pinned supported , k = 3EI / L , not 4EI / L
 

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fonseh said:
for the far end pinned or roller supported , the K = 3EI / L , so shouldn't the KBC = 3EI / L.. Is the author wrong ?
BC is only a segment of the beam ABCD. The "far end" of BC from B is D.
Think about it: if D were just a pin then the beam could rotate a bit about D, allowing it equally to rotate a little about C. But with D fixed, it is much harder for there to be any rotation about C.
 
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haruspex said:
BC is only a segment of the beam ABCD. The "far end" of BC from B is D.
Think about it: if D were just a pin then the beam could rotate a bit about D, allowing it equally to rotate a little about C. But with D fixed, it is much harder for there to be any rotation about C.
So , in this problem , the far end of AB is D , as D s pinned , so the KAB is 3EI / L ? If D is fixed , then KAB would be 4EI / L ?
 

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fonseh said:
So , in this problem , the far end of AB is D , as D s pinned , so the KAB is 3EI / L ? If D is fixed , then KAB would be 4EI / L ?
That's my understanding. But please bear in mind that my entire knowledge of the subject comes from the textbook extracts you have posted.
 
haruspex said:
That's my understanding. But please bear in mind that my entire knowledge of the subject comes from the textbook extracts you have posted.
May I know which part ? Can you highlighted it ?
 
fonseh said:
May I know which part ? Can you highlighted it ?
I have not noticed an actual definition of "far end" but just thinking about how I would expect a beam to behave it is clear that if any part of a beam is unable to rotate at all it will make the beam stiffer at all joints. Hence that is how I interpret the term.
 
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