The radial force is the outward force that comes from a spinning mass. It is equal and opposite to the reaction force at the root of the blade that keeps the blade connected to the hub. It can also be thought of as the mass times the radial acceleration.
You might remember from your Dynamics course that radial acceleration is equal to,
{latex} $$a_r=\ddot{r}-r {\dot{\theta}}^2$$ {latex} |
Here,
{latex} $$\ddot{r} $$ {latex} |
The radial force is simply equal to,
{latex} $$F_r = m a_r$$ {latex} |
Substituting the radial acceleration and expressing angular velocity as
{latex} $\omega$ {latex} |
{latex} $$F_r = -m r \omega^2$$ {latex} |
In this expression, m stands for the total mass of the blade and r stands for the distance in the radial direction where this mass resides. In this case, r will be location of the blade's center of mass in the radial direction. The blade mass and center of mass will be found later in the tutorial using ANSYS. We can however provide the results now for the sake of this calculation. The blade weighs 22,473 kg and its center of mass (X, Y, Z) is located at (-14.232 m, -0.213 m, 0.160 m). As you know, the blade is oriented so that the x-axis points along the radial direction of the blade. Plugging in these values in the radial force formula we obtain:
{latex} $$F_r = -m r \omega^2$$ $$F_r = -(22,473 kg)(-13.232 m) (-2.22 rad/s)^2$$ $$F_r = 1.4655*10^6 N$$ $$F_r = 1465.5 kN$$ {latex} |
Please follow along to start Part 2 of this project! We will start by defining the material for the blade.
<iframe width="640" height="360" src="//www.youtube.com/embed/P1B_hEW0pZI" frameborder="0" allowfullscreen></iframe> |
Summary of steps in the above video:
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