Last Updated on September 1, 2026 by Maged kamel
Moment of inertia Iy-at Cg for a triangle.
Procedure to find the Moment of inertia Iy at Cg for a triangle.
1-To calculate the Moment of inertia Iy at Cg, or Iyg, we must first estimate the x-bar of the triangle, which is the distance from the triangle’s center of gravity to the Y-axis.
How do we get the X-bar distance for the triangle?
2-To get the x̅ bar distance from the centroid of the triangle to the Y-axis, by using the first Moment of Area expression, estimate the A*x̅, which is the Area of the triangle that will be equal to A1*x̅1 +A2*x̅2 or the multiplication of the first Area by x1 its cg from the Y-axis, plus the product of the second Area by its cg distance from the same y axis.
The value of the first Moment of Area for the triangle can be estimated as 1/2(b*h) multiplied by x̅, which is the distance of the Cg of the first triangle to the y-axis, where the Y bar passes by the external left edge. The first Moment of the Area can be expressed as the product of an equal initial Area and its CG.
From the y-axis, the CG distance is equal to 2/3a. The first area is (1/2)* a*h*(2/3)*a, and we add the product of the second area by (2a+b/3). The first Moment of Area for the second triangle equals (1/2)(b-a)*(h)*(2a+b/3).
Add the sum of the first moments of area of the right-angle triangles. To calculate the final value of the x-bar, divide the estimated sum by the triangle’s area as follows. Please see the following slide images for further information.
The final x-bar distance for the triangle from the y-axis is 1/3 (a + b). This distance is measured from the Y-axis passing through the left corner.
The calculations to get the moment of inertia Iy at the Cg.
3-We can use the parallel axes theorem to get the value of the moment of inertia Iy at Cg. We know that Iy=Iyg +(A*xbar^2). That means, to find the Iy at a point, add the value of Iy at the CG to the product of area by the square of the horizontal distance from that point to the CG.
4-To calculate the moment of inertia Iy at Cg, subtract the product of (Axbar^2) from the value of Iy, which equals (bh/12)(b^2+a^2+ab).
The moment of inertia Iy at Cg for a triangle is calculated as Iyg=1/18*(b^2-a^2-ab), where a represents the distance from edge point a to point d and b represents the distance between points a and b.
The square of the radius of gyration about the y-axis at the CG can be calculated by dividing the moment of inertia Iy at Cg by the area, which we call K^2y.
The square of the radius of gyration in the y direction for the triangle at the CG is equal to 1/18*(b^2+a^2-ab), where a is the horizontal distance from the left corner to the perpendicular, while b is the base length of the triangle. This is the inertia list for triangular shapes, quoted from the NCEES tables of inertia. From the table, we can see that the calculated moment of inertia Iy at the Cg matches the given value. Please refer to the following slide image for more details.
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For an external resource, the definition of the Moment of inertia with solved problems is the 2nd Moment of inertia.
This is a link for the next post, post 15: Product of inertia Ixy– for the triangle.