Moment of Inertia Calculator
Calculate moment of inertia (second moment of area) for rectangles, circles, hollow circles, triangles, and I-beams.
Moment of Inertia Calculator
Calculator
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Formula: Ix = bh^3/12 (rectangle) | I = pi*r^4/4 (circle)
Additional inputs: Web Thickness (mm).
Worked example — Ix = 66.667 x 10^6 mm^4 | Iy = 16.667 x 10^6 mm^4 | Sx = 666,667 mm^3
Formula
Ix = bh^3/12 (rectangle) | I = pi*r^4/4 (circle)
The moment of inertia depends on the cross-sectional shape. For composite sections, the parallel axis theorem (I = I_centroid + Ad^2) is used. Section modulus S = I/c where c is distance to extreme fiber. Radius of gyration r = sqrt(I/A).
Worked Examples
Example 1: Rectangular Beam Cross-Section
Problem:Calculate the moment of inertia for a rectangular beam with width 100 mm and height 200 mm.
Solution:Ix = bh^3/12 = 100 x 200^3 / 12 = 100 x 8,000,000 / 12 = 66,666,667 mm^4 Iy = hb^3/12 = 200 x 100^3 / 12 = 200 x 1,000,000 / 12 = 16,666,667 mm^4 Area = 100 x 200 = 20,000 mm^2 Sx = Ix / (h/2) = 66,666,667 / 100 = 666,667 mm^3 rx = sqrt(Ix/A) = sqrt(66,666,667/20,000) = 57.74 mm
Result:Ix = 66.667 x 10^6 mm^4 | Iy = 16.667 x 10^6 mm^4 | Sx = 666,667 mm^3
Example 2: I-Beam Cross-Section
Problem:Calculate the moment of inertia for an I-beam with 150 mm flange width, 12 mm flange thickness, 200 mm web height, and 8 mm web thickness.
Solution:Total height = 200 + 2(12) = 224 mm Flange area = 150 x 12 = 1,800 mm^2 each Web area = 8 x 200 = 1,600 mm^2 Total area = 2(1,800) + 1,600 = 5,200 mm^2 Flange Ix = 150(12)^3/12 = 21,600 mm^4 Web Ix = 8(200)^3/12 = 5,333,333 mm^4 Flange d = (200+12)/2 = 106 mm Ix = 2(21,600 + 1,800 x 11,236) + 5,333,333 = 2(20,246,400) + 5,333,333 = 45,826,133 mm^4
Result:Ix = 45.826 x 10^6 mm^4 | Area = 5,200 mm^2 | Efficient bending resistance
Frequently Asked Questions
What is moment of inertia and why is it important?
Moment of inertia, also called the second moment of area, is a geometric property that quantifies how a cross-sectional shape resists bending about a particular axis. It is measured in units of length to the fourth power (mm^4 or in^4). The moment of inertia is critically important in structural engineering because it directly determines how much a beam will deflect under load and the maximum stress it will experience. A larger moment of inertia means greater resistance to bending, which is why structural shapes like I-beams are designed to maximize this property by placing material far from the neutral axis. Understanding moment of inertia is essential for designing safe and efficient structural members in buildings, bridges, machinery, and all load-bearing applications.
What is the difference between Ix and Iy?
Ix and Iy represent the moment of inertia about the x-axis (horizontal) and y-axis (vertical) respectively, both measured through the centroid of the cross-section. For most structural applications, Ix (the strong axis) determines resistance to vertical bending loads, while Iy (the weak axis) determines resistance to lateral bending. The distinction is crucial because most cross-sections are not symmetric about both axes, resulting in different stiffness properties in different directions. For example, a rectangle oriented with its long dimension vertical has a much larger Ix than Iy, meaning it resists vertical bending much better than horizontal bending. This is why beams are typically oriented with their strong axis aligned with the primary loading direction to maximize structural efficiency.
What is the section modulus and how is it used?
The section modulus (S) is derived from the moment of inertia by dividing it by the distance from the neutral axis to the extreme fiber of the cross-section (S = I/c). It is measured in units of length cubed (mm^3 or in^3). The elastic section modulus is used directly in the flexure formula (stress = M/S) to calculate the maximum bending stress in a beam under a given moment. A larger section modulus means lower bending stress for the same applied moment. The plastic section modulus (Z) represents the shape factor for fully plastic conditions and is used in plastic design methods. Structural steel design codes use the section modulus extensively for sizing beams, checking capacity, and ensuring adequate safety margins against bending failure.
What is the parallel axis theorem and when do I use it?
The parallel axis theorem, also known as Steiner's theorem, states that the moment of inertia about any axis parallel to an axis through the centroid equals the centroidal moment of inertia plus the product of the area and the square of the distance between the two axes (I = I_centroid + A*d^2). This theorem is essential when calculating the moment of inertia of composite cross-sections made up of simpler shapes. For example, an I-beam can be decomposed into three rectangles (two flanges and a web), and the parallel axis theorem is used to transfer each rectangle's own moment of inertia to the common centroidal axis of the entire section. Without this theorem, it would be extremely difficult to calculate properties of complex cross-sections that are standard in structural engineering practice.
How does the radius of gyration relate to moment of inertia?
The radius of gyration (r) is defined as the square root of the moment of inertia divided by the cross-sectional area (r = sqrt(I/A)). It represents the distance from the axis at which the entire area could be concentrated to produce the same moment of inertia. The radius of gyration is particularly important in column design because it directly relates to the slenderness ratio (L/r), which determines whether a column will fail by material yielding or by elastic buckling. A larger radius of gyration means a more efficient cross-section for resisting buckling. For columns, designers check the slenderness ratio about both the strong and weak axes, and the critical case is typically the weak axis where the radius of gyration is smallest. This is why wide-flange sections are preferred for columns over narrow deep sections.
What is the difference between the second moment of area and the mass moment of inertia?
They share a name but describe different physics. The second moment of area, computed here, is purely geometric, depends only on the shape of the cross-section, and carries units of length to the fourth power such as mm^4; it governs bending stiffness and deflection. The mass moment of inertia sums mass times distance squared, carries units of kilogram metre squared, and governs angular acceleration under a torque. A beam designer wants the first; someone sizing a flywheel or a motor wants the second, and substituting one for the other is dimensionally impossible.
Why do I-beams put nearly all their material in the flanges?
The parallel axis theorem adds A times d squared for every piece of area, so material far from the neutral axis contributes in proportion to the square of its distance while material at the axis contributes almost nothing. In the worked I-beam here, each flange sits 106 mm from the centroid and the two flanges supply roughly 40 million of the 45.8 million mm^4 total, with the deep web adding only about 5.3 million. The same effect explains why doubling the depth of a rectangle multiplies Ix by eight, since Ix varies with h cubed. The web is retained mainly to carry shear and to hold the flanges apart.
What is the polar moment of inertia, and when do you need it instead of Ix?
The polar moment J measures resistance to twisting about an axis perpendicular to the cross-section, and the perpendicular axis theorem gives it for a plane section as J = Ix + Iy. For a solid circle that is pi r^4/4 plus pi r^4/4, or pi r^4/2. You need J rather than Ix whenever the load is torsion rather than bending, for instance sizing a drive shaft using the torsion formula shear stress = T r / J. An important caution: J is the correct torsion constant only for circular and hollow circular sections; open shapes such as I-beams warp and require a separate torsional constant.
Why is a hollow tube almost as stiff in bending as a solid bar of the same outer diameter?
For a hollow circle I = pi/4 times (R^4 - r^4), and the material removed lies near the neutral axis where its contribution is smallest. Boring a 40 mm radius hole through a 60 mm radius bar removes (40/60) squared, about 44 percent of the area, but only (40/60) to the fourth, about 20 percent, of the moment of inertia. The section keeps roughly 80 percent of its bending stiffness for 56 percent of the weight, which is why tubes are used for bicycle frames, scaffolding, and drive shafts.
How does moment of inertia affect how far a beam deflects?
Deflection is inversely proportional to the product of the elastic modulus E and the moment of inertia I, the quantity known as flexural rigidity. For a simply supported beam with a central point load the midspan deflection is P L cubed divided by 48 E I, so doubling I halves the deflection while doubling the span multiplies it by eight. Because stiffness scales with I and strength scales with the section modulus S = I/c, a member that passes a stress check can still fail a serviceability limit on deflection, which is often what actually sizes a floor beam.
References
Reviewed for accuracy by Manoj Kumar, Mathematics Educator · Editorial policy
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