If I have a beam with the left end being "A" and it is fixed at that end and the right end being "B" with a roller support, would KAB=3EI/L??
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$\begingroup$ you need to define the load, and the position you need to find the stiffness factor. $\endgroup$– NMechCommented Apr 28, 2021 at 8:06
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$\begingroup$ I'm speaking in terms of the far end formula, I want to know if hypothetically for a member which is fixed at the left, "A" and pinned/roller at the right. "B" if the stiffness for KAB with A being the reference point is 3EI/L $\endgroup$– Agassi MurrayCommented Apr 28, 2021 at 8:26
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2 Answers
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By definition, the bending stiffness $"K"$of a structural member is the moment that must be applied to an end of the member to cause a unit rotation ($\theta = 1$) of that end.
- For a beam with far end fixed,
Substituting $\theta_{A} = 1$, we get
$K_{AB} = 4EI/L$
- For a beam with far end pinned,
$= 3EI/L(\theta_{A} - 0) + (0 - 0)$
$M_{AB} = (3EI/L)\theta_{A}$
Substituting $\theta_{A} = 1$, we get
$K_{AB} = 3EI/L$
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Update to address your comment:
The stiffness factor at the far end B, is going to be
$$K=\infty$$
No matter what force you put at the end (point B), point B will never move if its simply supported.
Uniform load
Assuming that this is what you are asking (left B, right A)
then $k = \frac{185 EI}{l^3}$
This is at $x= 0.4126 \;l $
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$\begingroup$ No but I am intrigued as to how 185 if used in that formula, may I have a site for reference if possible? $\endgroup$ Commented Apr 28, 2021 at 8:28
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$\begingroup$ The OP is asking about the stiffness in matrix operation such as the direct stiffness method. $\endgroup$– r13Commented Apr 28, 2021 at 16:29