Equivalent beam rigidity for a 1D lattice structure

To model the behavior of continua, often discrete lattice models with nodes joined by 2-point spring elements (which resist tensile forces) and 3-point beam elements (which resist bending moments) are employed.

Consider a 1D structure of length $L$ is divided into $N$ elements of length $\Delta x$. The continuum structure has an elastic modulus $E$. In the discretized model, neighboring nodes are joined by springs. The spring stiffness in the continuum is $$k_s = \frac{EA}{L} \,.$$

By equating the potential energy $E_s$ of the continuum model and discrete model, we may find an equivalent discrete spring stiffness $k_s^\Delta$:

$$E_s = \int_0^L k_s (X - L) \mathrm d X = \int_0^L N k_s \left(\frac{X}{N} - \frac{L}{N} \right) \mathrm d X = N^2 \int_0^{\Delta x} k_s (x - \Delta x) \mathrm d x = N E_s^\Delta \,,$$

where $\Delta x = L/N$ and the substitution $x = X / N$ has been made. Thus,

$$E_s^{\Delta} = \int_0^{\Delta x} N k_s (x - \Delta x) \mathrm d x \equiv \int_0^{\Delta x} k_s^{\Delta} (x - \Delta x) \mathrm d x \,,$$

and we see that $k_s^{\Delta} = N k_s = k_s (L / \Delta x) = E A / \Delta x$.

Another way of seeing this is that numerically, the strain energy between nodes $i$ and $i+1$ is $$E_{s,i}^\Delta = \frac{1}{2}k_s^\Delta \left( || \mathbf x_{i} - \mathbf x_{i+1} || - \Delta x \right)^2$$ and the total strain energy is $$E_s^\Delta = \sum_i E_{s,i}^\Delta = N E_{s,i}^\Delta = E_s\,.$$

Now, I want to determine a similar equivalent stiffness coefficient for a discrete model that also has 3-point beam elements. The potential energy of a beam is:

$$E_b = \int_0^L k_b \left( \kappa(X) \right)^2 \mathrm dX = \int_0^{\Delta x} Nk_b \left( \kappa(Nx) \right)^2 \mathrm dx = (N-1) E_b^{\Delta} \,,$$ where $k_b=EI$ and $I$ is the area moment of inertia of the cross-section. This part is really confusing me. Also, the form of the discrete bending energy employed in this case is: $$E_{b,i}^\Delta = \frac{1}{2} k_b^\Delta \left[ \mathbf{\hat z} \cdot (\mathbf x_{i+1} - \mathbf x_i) \times (\mathbf x_i - \mathbf x_{i-1}) \right]^2$$

For this problem, the structure should ideally be incompressible. Thus, I obtain

$$E_{b,i}^\Delta = \frac{1}{2} k_b^\Delta \left[ \Delta x^2 \sin\theta_i \right]^2$$

where $\theta_i$ is the angle made by the vectors $\mathbf x_{i+1}-\mathbf x_i$ and $\mathbf x_i-\mathbf x_{i-1}$.

I'm not sure how to proceed from here. I thought that a good first step would be to linearize the expression. If the mesh is very fine, than the angle formed by two consecutive elements is approximately $\pi$, but I couldn't get very far with this either. I would appreciate it if someone could help me figure this out.