Last Updated on September 8, 2026 by Maged kamel
6-Solved problems for the estimation of the Moment of inertia.
The First solved problem for a given rectangle.
We will solve our first problem, problem # 1. We have the x- and y-axes as shown on the next slide, and a rectangular section with a width of 0.40 m and a height of 0.60 m. This rectangle is offset from the y-axis by 0.20 m.
It is required to estimate the following for the given rectangle—the Moment of inertia Ix, Kx, and the radius of gyration about the x-axis. For part b, we want to estimate the Moment of inertia about the y-axis, Iy, and the radius of gyration about the y-axis, Ky.
We have our X and Y external axes, as we know. Ix = I = xg + g^2, the product of Area * (y^2).
For the Moment of inertia about the CG of a rectangular section, we have the Expression (b*h^3/12). Then we have this x-axis. We have to add the product of Area multiplied by y^2. First, our Ix g = b*h^3/12 = (0.40*0.60^3)/12, which gives us 0.0072 m^4.
Our Area is estimated at (0.40 × 0.60), which equals 0.24 m2. For item A* y^2, we have A = b*h and y̅ =, all raised to the power of 2. So we have b*h = (0.40*0.60)*(h/2)^2; h/2 = 0.30/4, giving us 0.0216 m^4.
Our required Moment of inertia about the x-axis, Ix, will be the summation of these two items. The first item is estimated as Ixg = 0.0072, plus the second item, which is ( A*y^2), whicl to 0.0216. Adding both values gives us 0.0288 m^4.
To determine k^x, we will divide Ix by the Area of the rectangle.
the value of kx^2 will be equal to Ix/A=0.0288/0.24=3/25, then the Kx value =sqrt(3/25)=sqrt(3)/5.
For part b of the first solved problem, to find Iy, add Iyg + A*xbar^22. Iyg=h*b^3/12=0.60*?(0.40)^3/12=0.0032 m4.
Ax^2g=(0.40)(0.60)(0.20+0.50.40)^2=0.0384m4. The final Iy=(0.0032+0.0384)=0.0416m4.
k^2y=Iy/A=0.0416/(0.40*0.60)=13/75. ky=sqrt(13/75)=0.4163 m2.
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The second solved problem for a given section.
We can consider the L-shape to consist of two rectangles. The first rectangle is 1.0 m wide and 4 m high, and the other rectangle is 2 m wide and 1 m high; if we join them, they will form an unequal angle with the same width, as we can see. An unequal angle of three by four and a width of1.0 m can be evaluated as composed of two rectangles.
The first one, A1, has a width of 1.0 m and a height of 4.0 m, so its Area would be (1*4)=4.0 m^2, and its X1 will be=1/2 m. y1 from the x-axis will be. 4/2, which is 2 meters. For the second Area, we have A2 = (3 – 1)*. This is the breadth; the height will be 1.0 m, so the Area is A2 = (1*2) = 2.0 m^2.
Our X2, measured to the external y-axis, will be 1/2 of the difference between (3-1.0), which gives 2 meters; oy2 = y2 = 0.50 m.
The Ix for the unequal angle is the sum of the Ix for the first and second rectangles.
The I_x^2 value of the solved problem.
The final Ix = 22.00 m^4, which is part a) of the solved problem.
For part c) of the solved problem, we estimate the product of inertia about the external axes for each rectangle and then add them together.
The final I = 6.00 m^4m4 is part b) of the solved problem.
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For the next post, the Moment of inertia for the right-angle triangle, Ix- Case-1.
This is a useful external link on the Moment of inertia: The Moment Of Inertia Of Rectangle.”