For a circular tube section, substitution to the above expression gives the following radius of gyration, around any axis:Ĭircle is the shape with minimum radius of gyration, compared to any other section with the same area A. Like in bending stress, shear stress will vary across the cross sectional area. Looking again at figure one, it can be seen that both bending and shear stresses will develop. Shear stress however results when a load is applied parallel to an area. Small radius indicates a more compact cross-section. Normal stress is a result of load applied perpendicular to a member. It describes how far from centroid the area is distributed. The dimensions of radius of gyration are. Where I the moment of inertia of the cross-section around a given axis and A its area. Radius of gyration R_g of a cross-section is given by the formula: Where, D, is the outer diameter and D_i, is the inner one, equal to: D_i=D-2t. Įxpressed in terms of diamters, the plastic modulus of the circular tube, is given by the formula: The last formula reveals that the plastic section modulus of the circular tube, is equivalent to the difference between the respective plastic moduli of two solid circles: the external one, with radius R and the internal one, with radius R_i. The moment of inertia (second moment of area) of a circular hollow section, around any axis passing through its centroid, is given by the following expression: The total circumferences (inner and outer combined) is then found with the formula:
Its circumferences, outer and inner, can be found from the respective circumferences of the outer and inner circles of the tubular section. Where D_i=D-2t the inner, hollow area diameter. In terms of tube diameters, the above formula is equivalent to: Where R_i=R-t the inner, hollow area radius. The area A of a circular hollow cross-section, having radius R, and wall thickness t, can be found with the next formula:
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