Stiffness factor for member in beam

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SUMMARY

The stiffness factor (KBC) for a member in a beam is determined by its support conditions. For a far end pinned support, the stiffness factor is K = 3EI / L, while for a fixed support, it is K = 4EI / L. The discussion clarifies that if the far end of segment BC is fixed, the stiffness factor increases to 4EI / L, making it more resistant to rotation. Understanding these distinctions is crucial for accurate beam analysis in structural engineering.

PREREQUISITES
  • Understanding of beam theory and stiffness factors
  • Knowledge of structural support conditions (pinned vs. fixed)
  • Familiarity with the terms EI (modulus of elasticity and moment of inertia) and L (length of the beam)
  • Basic principles of mechanics of materials
NEXT STEPS
  • Study the derivation of stiffness factors in beam theory
  • Learn about the effects of different support conditions on beam behavior
  • Explore the application of the stiffness factor in structural analysis software
  • Investigate real-world examples of beam stiffness in civil engineering projects
USEFUL FOR

Structural engineers, civil engineering students, and anyone involved in beam design and analysis will benefit from this discussion.

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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