This is also correct for a cylinder (think of it as a stack of discs) about its axis. c) The couple moment produce by the single force acting on the body. that formula will then give the moment of inertia of a cube, about any axis through its center. b) The moment of inertia of the body about any axis. It is determined from the cross-sectional area of the beam and the central axis for the direction of interest. What does the moment of the force measure in the calculation of the Mohr circle’s data for the moments of inertia a) The tendency of rotation of the body along with any axis. Which is again what we expect from the standard table for a circular segment. Knowing the area moment of inertia is a critical part of being able to calculate stress on a beam. And we will integrate $x$ from 0 to $r_0\cos(\alpha/2)$. Mathematically, we can write polar moment of inertia i.e. Polar moment of inertia will be displayed by J. For a a given position along the $x$ axis, the limits of $y$ range from $0$ to $x\tan(\alpha/2)$. Polar moment of inertia of a plane area is basically defined as the area moment of inertia about an axis perpendicular to the plane of figure and passing through the center of gravity of the area. The difficulty is just in getting the correct limits of the double integral. Where $\rho$ is the mass density per unit area, which looks simple enough. moment of inertia with respect to x, Ix I x Ab 2 7.20 106 12.72 103 81.8 2 92.3 106mm4 Sample Problem 9.5 The moment of inertia of the shaded area is obtained by subtracting the moment of inertia of the half-circle from the moment of inertia of the rectangle. I'm going to use Mathematica to do the brute force algebra and integration). Since this is clearly a homework problem, I'm going to skip the algebra steps and just show you the core parts of the problem (i.e. Now, only looking at the top half we can break the piece up into two sections: $I_1$ is on the left and is a triangle and $I_2$ is on the right and is a right triangle with a circular hypotenuse. To start with, we will recognize that the symmetry about the $x$ axis lets us only work on the top half and then multiply by a factor of 2 in the end. Calculating the moment of inertia about the $x$ axis is a fair deal more complicated than calculating it about the $z$ axis as in my other answer. Since you actually asked for the moment about the $x$ axis.