Differential inertia terms in a local main axis coordinate frame

At first all inertia terms are derived in a local coordinate frame $X_i$, with axes $X_2$, $X_3$ corresponding to the principal axes of the cross-section, i.e. $I_{23}=0$.

Figure 80: Cylindrical beam with arbitrary cross-section (principal cross-section axes $X_2$, $X_3$)

[width=]u1_mass_derivation (-190,112)red $\textrm{d}J^S_{1} \ddot{\varphi}_1$ (-60,140)red $\textrm{d}J^S_{22} \ddot{\varphi}_2$ (-130,140)red $\textrm{d}J^S_{33} \ddot{\varphi}_3$ (-139, 90)red $\textrm{d}m \ddot{u}_1$ (-80,110)red $\textrm{d}m \ddot{u}_2$ (-114,120)red $\textrm{d}m \ddot{u}_3$ (-285,0)$\hat X_1$ (-340,47)$\hat X_2$ (-345,12)$\hat X_3$ (-110,75)$S$ (-240,25) $\hat{\boldsymbol{X}}_S$ (-17,48)red$X_1$ (-162,50)red$X_2$ (-70,50)red$X_3$

With the moments of inertia

$$\textrm{d}I_{33} = X_2^2 \textrm{d}A$$

$$\textrm{d}I_{22} = X_3^2 \textrm{d}A$$

$$\textrm{d}I_p = (X_2^2 + X_3^2) \textrm{d}A$$ (28)

the inertia forces of a differential cross-section $\textrm{d}M = \rho A \textrm{d}X_1$ can be derived by integration of the above quantities over $A$:

$$\textrm{d}F_1 = -\int_A \ddot{U}_1 \textrm{d}m = -\rho A \ddot{U}_1 \textrm{d}X$$

$$\textrm{d}F_2 = -\int_A \ddot{U}_2 \textrm{d}m = -\rho A \ddot{U}_2 \textrm{d}X$$

$$\textrm{d}F_3 = -\int_A \ddot{U}_3 \textrm{d}m = -\rho A \ddot{U}_3 \textrm{d}X$$

$$\textrm{d}M_1 = -\int_A \ddot{\psi}_1 \rho (X_2^2 + X_3^2) \textrm{d}A \textrm{d}X_1 = -\rho I_{p} \ddot{\psi}_1 \textrm{d}X_1$$

$$\textrm{d}M_2 = -\int_A \ddot{\psi}_2 \rho X_3^2 \textrm{d}A \textrm{d}X_1 = -\rho I_{22} \ddot{\psi}_2 \textrm{d}X_1$$

$$\textrm{d}M_3 = -\int_A \ddot{\psi}_3 \rho X_2^2 \textrm{d}A \textrm{d}X_1 = -\rho I_{33} \ddot{\psi}_3 \textrm{d}X_1$$ (29)