Centre of mass and radius of gyration
The inertial behaviour of a diaphragm having a mass Σ(m_{i}) is idealised by a ring of equivalent inertial massdistribution. The ring has a total mass Σ(m_{i}), its centre is theCentre of Mass C_{M} and its radius theRadius of Gyration l_{s}.
Figure 5.4.2: The centre of mass C_{M} and the equivalent mass inertial ring with radius of gyration l_{s} Figure 5.4.2: The centre of mass C_{M} and the equivalent mass inertial ring with radius of gyration l_{s}
The coordinates of the centre of mass C_{M} of a diaphragm consisting of many mass points, linearly distributed, and surface masses linearly distributed derive from the expressions: The radius of gyration l_{s} of the diaphragm with respect to the centre of mass C_{M} derives from the expression: where Χ_{i} and Υ_{i} are the coordinates of each mass point m_{i} of the diaphragm, whereas I_{pi} is the polar moment of inertia of each mass m_{i} with respect to the centre of mass C_{M}. According to the strength of materials, the polar moment of inertia of a mass m, whose centre is located at distance L from the centre of mass C_{M} of the diaphragm, depends on the distribution of the mass:  For a point mass: I_{p}=m·L^{2}
 For a mass linearly distributed along a segment of length l: I_{p}=m· (l^{2}/12 + L^{2})
 For a surface distributed mass in a rectangular shape of dimensions b,l: I_{p}=m· [(b^{2}+l^{2})/12 + L^{2}]
 For a surface distributed mass in a triangular shape or a circular segment: the equivalent principal moment of inertia in respect to the centre of mass of this segment is calculated and the Steiner term m·L^{2} is added.
In large floor plans, the radius of inertia can be calculated only from the reaction forces of the beams on the floor columns, while in case that the particular linear loads (e.g. of the walls) are distributed uniformly throughout the floor, the radius of inertia is calculated using the formula where L _{x}, L _{y} are the dimensions of the floor plan. Radius of gyration : [g=10 m/sec^{2}, therefore force F=1 kN corresponds to mass m=0.1 t.] Slab: m_{1}=6.0m ·5.0m·0.71t/m^{2}=21.3 t b_{1}=6.0 m, l_{1}=5.0 m, L_{1}=0.0 m → I_{p1}=21.3t ·(6.0^{2}+5.0^{2})m^{2}/12=108.3 t·m^{2} Beam connecting C_{1}C_{2}: m_{2}=6m·1.0t/m=6.0 t, l_{2}=6.0 m, L_{2}=2.5 m → I_{p2}=6.0· (6^{2}/12+2.5^{2})=55.5 t·m^{2} Beam connecting C_{3}C_{4}: likewise m_{3}=6.0 t, I_{p2}=55.5 t·m^{2} Beam connecting C_{1}C_{3}: m_{4}=5.0m·1.0t/m=5.0 t, l_{4}=5.0 m, L_{4}=3.0 m → I_{p4}=5.0· (5.0^{2}/12+3.0^{2})=55.4 t·m^{2} Beam connecting C_{2}C_{4} likewise m_{5}=5.0 t, I_{p5}=55.4 t · m^{2} Columns: m_{6}=0.1 ·(4.00+4.00+6.0+4.5)=1.85 t, L_{6}=√ (3.0^{2}+2.5^{2})=3.905 m → I_{p6}=1.85 ·3.905^{2}=28.2 t·m^{2}.
