![]() We have dA= the width which is dx*y, and we are going to integrate from x=0 to x = b, we are going to substitute y value by writing y=(-h*x/b)+h, our dA is y* dx, So we are going to multiply this value, y,* (-h*x/b)+h it by dx. We are intersecting at the left corner of that triangle, again our expression of the line BC is y= (-h/b)*x+h, for the vertical strip. We are going to use, this time, a vertical strip with width=dx and height = Y, and this strip is a distance=x from the y-axis, for the x-axis and y-axis intersection. ![]() We have our triangle ABC with base =b and height =h and d. In this lecture, we are going to talk about how to determine the moment of inertia-Iy in the y-direction for the right-angle triangle case-1. Moment of inertia-Iy for the right-angle triangle-Case-1. Using a horizontal strip as one option to estimate the moment of inertia-Iy value.Polar moment of inertia at the left corner for right-angle triangle-for case-1.Polar moment of inertia at the CG for right-angle triangle-for case-1.Moment of inertia-Iy at the CG of the right-angle -triangle.Step-by-step guide for the calculation of moment of inertia-Iy through the slides.Moment of inertia-Iy for the right-angle triangle-Case-1.
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