To calculate the bending stress in structural members (beams), a property called SECTION MODULUS is used to express the bending moment/stress relationship.Įach point within a cross-section of a beam will have a section modulus, this being the ratio of the second moment of inertia to the distance between a point within the section and the relevant axis. The formula for a rectangular section is: However, a simple formula has been derived for a rectangular section, which is the most important section in this subject. Both board have the same cross-sectional area, but the area distributed differently about the horizontal axis.Ĭalculus is usually used to find the moment of inertia (I) of an irregular section.
The board that is supported on its 50 mm edge is considerably stiffer than that supported along its 200 mm edge. If two boards with actual dimensions of 200 by 50 mm were laid side by side - one on the 50mm side and the other on the 200 mm side. Let us look at two boards to intuitively determine which will deflect more and why. The second moment of area (I) is an important figure that is used to determine the stress in a section, to calculate the resistance to buckling, and to determine the amount of deflection in a beam. = the distance between the centroid of the object and the = the second moment of area (moment of inertia) around the xx-axis The second moment of area or moment of inertia (I) is expressed mathematically as: The second moment of area (I) about a given axis is the sum product of the area and the square of the distance from the centroid to the axis. Second moment of area (I) or moment of inertia The centre of gravity is found by dividing the specific area moment (x- and y-direction) by the total area. The easiest way is to organise all data in a table as shown below: Dimensions
Note: The centre of gravity is not necessarily within the body of the shape, it can fall outside as with most angular shapes.Ī more precise procedure to find the centre of gravity is the first moment of area method. The principle is shown in the Figure below. The line of action will always pass through the centre of gravity of the particular shape. The centre of different shapes cut out from a card board can be found by hanging it from a string. A useful analogy that helps understanding this idea may be found by considering the centre of gravity or centre of mass. If it is supported at this point it is in a state of equilibrium and should not fall off. The following table, includes the formulas, one can use to calculate the main mechanical properties of the circular section.The centre of an area, or centroid, of a shape is the point at which it is in equilibrium. For a circular section, substitution to the above expression gives the following radius of gyration, around any axis, through center:Ĭircle is the shape with minimum radius of gyration, compared to any other section with the same area A. Small radius indicates a more compact cross-section. 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 the same axis and A its area.
Radius of gyration R_g of any cross-section, relative to an axis, is given by the general formula: The area A and the perimeter P, of a circular cross-section, having radius R, can be found with the next two formulas:
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