Rearranging we have: I x k x 2A I y k y 2A The radius of gyration is the distance k away from the axis that all the area can be concentrated to result in the same moment of inertia.
There is no need to use the transfer formula of moment of inertia since the centroid of all basic shapes coincide with the centroid of the compound shape. distributed as far away as possible from the centroid. Integrating curvatures over beam length, the deflection, at some point along x-axis, should also be reversely proportional to I. Solve for the moment of inertia of the complex figure by subtracting the moment of inertia of area 2 (A2) from area 1 (A1). Ix and Iy are moments of inertia about indicated axes Moments of Inertia: h c b D I R b h h Z I c b h is perpendicular to axis 3 2 12 6 I D R Z I c D R 4 4 3 3 64 4 32 4 Parallel Axis Theorem: I Moment of inertia about new axis I I +A d 2 centroid d new axis Area, A I Moment of. Therefore, it can be seen from the former equation, that when a certain bending moment M is applied to a beam cross-section, the developed curvature is reversely proportional to the moment of inertia I.
Making similar considerations, the moment of inertia of the angle, relative to axis y0 is: