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In moment distribution method, the bending stiffness of a beam is taken as $EI/L$. But in these kinds of stiffness tables, the flexural rigidity is usually divided by some power of $L$: enter image description here

So why is $EI$ divided by simply $L$ to get the bending stiffness, instead of some of its powers as in the tables? Could somebody provide a quick derivation? Thank you!

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"Stiffness" quantifies the level of resistance of a structural member against deformation under loads. Two types of deformations are of particular concern - translation and rotation, in structural engineering.

  • Translation deformation: The stiffness is the invert of the member flexibility ($f$), which is an index of member deflection ($\Delta$), so $K = \frac{1}{f} = \frac{1}{\Delta}$. For member subjects to uniform load, $\Delta = \frac {5wL^4}{384EI}$, $K = \frac {384EI}{5wL^4}$; for member subjects to mid-span concentrate load, $\Delta = \frac{PL^3}{48EI}$, $K = \frac{48EI}{PL^3}$, and so on for other load cases. $K$ is sometimes referred to as the member "deflection/deformation stiffness".
  • Rotational deformation: The stiffness is the invert of the flexibility of the member end rotation when subjects to the applied moment. The flexibility is the index of rotation $f = \theta = \int \frac {M}{EI} dx$, and $K = \frac{M}{\theta} = \frac{EI}{L}$. This is the stiffness used in the moment distribution method, and in the structural stiffness matrix of the finite element program. It is sometimes referred to as the "bending stiffness".

One special case, for member subjects to uniaxial loads (tension/compression) only, such as the truss member, $K = \frac{EA}{L}$.

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I/L is relative stiffness. For the same flexural member material and cross sectional characteristics, the longer the flexural member, the more it displaces/deflects. Therefore, I/L provides a relative stiffness value depending on member's length.

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