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Radius of Gyration of an Area

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Equations

(Eq1)     kx =  √ Ix A	radius of gyration of an area with respect to the x-axis
(Eq2)     ky =  √ Iy A	radius of gyration of an area with respect to the y-axis
(Eq3)     kO =  √ JO A	radius of gyration of an area about the origin
(Eq4)     kO2 = kx2 + ky2	equation relating radii of gyration

Nomenclature

Ix	moment of inertia with respect to the x-axis
Iy	moment of inertia with respect to the y-axis
JO	polar moment of inertia
A	area

Details

Consider an area A which has a moment of inertia I_x with respect to the x-axis as shown.



Imagine that the area is concentrated into a thin strip parallel to the x-axis as shown:



If the area A, thus concentrated, is to have the same moment of inertia with respect to the x-axis, the strip should be placed at a distance k_x from the x-axis, where k_x is defined by the relation:

I_x = k_x²A

Solving for k_x:

(Eq1)

kx =  √ Ix A

The distance k_x is referred to as the radius of gyration of the area with respect to the x-axis. In a similar way, the radii of gyration k_y and k_O can be defined. The following image represents an area with the same moment of inertia with respect to the y-axis:



For k_y :

I_y = k_y²A

Solving for k_y :

(Eq2)

ky =  √ Iy A

Following is an image of a circular area centered about the x- and y-axes that has the same moment of inertia with respect to the x- and y-axes:



Similarly, for J_O :

J_O = k_O²A

Solving for k_O :

(Eq3)

kO =  √ JO A

If Eq2 from the Polar Moment of Inertia lesson is rewritten in terms of the radii of gyration, it is found that:

(Eq4)

kO2 = kx2 + ky2