# A Simple Solution to a Complicated Equivalent

A previous post posited on the equivalence of discrete flexural springs (moment-rotation) with integration of continuous moment-curvature response.

To find the answer, we can use the principle of virtual forces (PVF) and numerical integration of the internal virtual work:

${\displaystyle \int_0^L \kappa(x)m(x)\: dx \approx \sum_{i=1}^N \kappa(x_i) m(x_i) w_i}$

where $m(x)$ is the “virtual” moment and $\kappa(x)=M(x)/EI$ is the “real” curvature, based on the “real” bending moment, $M(x)$, and the flexural stiffness, $EI$. The flexural stiffness is constant for this example, but that is not a necessary assumption for the PVF:

${\displaystyle \sum_{i=1}^N \kappa_i m_i w_i = \sum_{i=1}^N \frac{M_i}{EI} m_i w_i }$

where $m_i \equiv m(x_i)$, with similar definitions for $\kappa_i$ and $M_i$. Now, we can move the integration weight to the denominator as part of the flexural stiffness and see the rotation of each spring:

${\displaystyle \sum_{i=1}^N \frac{M_i}{EI} m_i w_i = \sum_{i=1}^N \frac{M_i}{(EI/w_i)} m_i = \sum_{i=1}^N \theta_i m_i}$

So, the correct answer is (D) EI/wi, the flexural stiffness divided by the integration weight associated with the spring location.

The point of all this is that if you are attaching zero length fiber sections to your column elements, you need to account for hinge length.

I have an OpenSees script that builds a column model out of rotational springs and “rigid” beam-column elements like what was proposed in the previous post, but the script is too complicated to explain easily. Plus, I know the rules of the internet–if the script is posted, someone will think it’s a valid way to model beam-column elements!

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