The moment of inertia for a beam is used in further calculations that determine how much a beam will bend before it breaks and what forces will cause it to break. Structural engineers aim for greater moments of inertia for main beams in new buildings and design them knowing the weight they will bear and how they will be supported. Calculating a beam’s moment of inertia is dependent on its cross-sectional area and shape, and whether it is solid or hollow. Resulting units will be (length)^4.
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Define the beam application. In this example, a structural steel beam consists of a solid steel bar with a cross section that is 2 inches wide by 4 inches high. For the purpose of calculating the beam’s moment of inertia, only its cross-sectional dimensions are necessary and not its length.
Determine the applicable moment of inertia formula. For a classic horizontal rectangular beam supported on both ends and bearing a single load in the middle, the moment of inertia of the beam is defined by the formula I = (bh^3)/12 where I equals moment of inertia, b is the cross-sectional width of a horizontal beam, and h is the cross-sectional height of the beam.
Substitute your example variables in the formula to solve for the beam’s moment of inertia. I = 2 inches x (4 inches)^3 / 12 = 2 x 64 / 12 = 10.67 inches^4.
Define your round beam (shaft) application. For example, a load is placed in the middle of a round steel axle that is supported by a wheel on each end. The axle is 3 inches in diameter.
Determine the formula for a solid round beam. Here, I = pi x d^4 / 64 where d is the diameter of the beam.
Insert the diameter's length into the formula. I = 3.14 x (3 inches)^4 / 64 = 3.14 x 81 / 64 = 3.97 inches^4.
Define the cylinder application. This example will compare solid axles to tubular axles, so the example of Section 2 with a 3-inch diameter solid axle will be used, except that the axle is hollow and has a wall thickness of 0.5 inches. The inner diameter is therefore 2 inches.
Determine the applicable cylindrical beam formula, which is I = pi x (outer diameter^4 - inner diameter^4) / 64.
Insert your dimensions to solve. I = 3.14 x ( (3 inches)^4 - (2 inches)^4) / 64 = 3.14 x (81-16) / 64 = 3.19 inches^4.
Compare the solid 3-inch axle to the hollow 3-inch axle. 3.19/3.97 = .80. This axle has 80 per cent of the rigidity and strength of the solid axle with the same outer size, while only containing 55.6 per cent of the metal.
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- The Central London Area Group/Russ Elliott: Deflection of Beams, Moment of Inertia
- University of Oregon: Centroids and Moment of Inertia
- Engineering Toolbox: Area Moment of Inertia/Cylindrical Cross Section
- University of Denver/J.Calvert: The Remarkable Theory of Beams/moment of inertia
- University North Carolina Charlotte/JD Bowen: Introduction to Beams