**Problem 10-33**: Determine the moment of inertia \( I_x \) of the shaded area about the \( y \) axis.**Diagram Explanation**:The provided diagram displays a composite shape which consists of a rectangle with a semicircle cut out and a trapezoidal section attached to it. Here are the details of the dimensions given in the diagram:- The horizontal base has a total length of 350 mm, divided into sections: 100 mm, 100 mm, and 150 mm.- The vertical height from point \( O \) (origin) is divided equally with two lengths of 150 mm.- There is a circular cutout located 100 mm from the bottom edge and centered vertically within the rectangle part. The radius of this circle is 75 mm.- The left side of the shape starts from the origin on the bottom left at \( O \) and extends vertically upward to a height of 300 mm before tapering off diagonally to the end of the top horizontal segment.The goal is to calculate the moment of inertia about the \( y \) axis, taking into account the areas of both the composite rectangular and trapezoidal section, and incorporating the effect of the semicircular cutout.

Dividing the diagram into three parts Part 1 :- A rectangle OABC, Part 2:- A triangle BCD and Part 3:- A circleMoment of inertia of rectangle about y-axis, I1 A1 = Area of rectangle = 200 x 300 = 6 x104 mm2 y1 = centroid of rectangle from y-axis = 100 mm ICG1 = Moment of inertia of the rectangle about the CG of rectangle = 300 x 200312 = 200 x106 mm4 So, moment of inertia of rectangle about y-axis, I1 = ICG1 + A1y12 (Applying parallel axis theorem) I1 = 200 x106 + 6 x104 x1002 = 800 x106 mm4 Moment of inertia of triangle about y-axis, I2 A2 = Area of triangle = 12x300x150 = 22500 mm2 y2 = centroid of triangle from y-axis = (200 + 1503) = 250mm ICG2 = Moment of inertia of the triangle about the CG of triangle = 300 x 150336 = 28.125 x106 mm4 So, moment of inertia of triangle about y-axis, I2 = ICG2 + A2y22 (Applying parallel axis theorem) I2 = 28.125 x106 + 22500 x2502 = 1434.375 x106 mm4 Moment of inertia of circle about y-axis, I3 A3 = Area of circle = πx752 = 1.7671 x104 mm2 y2 = centroid of circle from y-axis = 100 mm ICG3 = Moment of inertia of the circle about the CG of circle = πx7544 = 24.8505x106 mm4 So, moment of inertia of circle about y-axis, I3 = ICG3 + A3y32 (Applying parallel axis theorem) I3 = 24.8505 x106 + 1.7671 x104 x1002 = 201.5605 x106 mm4 Now, moment of inertia of the shaded area about y-axis is Iy = I1 + I2 - I3 Iy = 800 x106 + 1434.375 x106 - 201.5605 x106 Iy = 2435.9355 x106 mm4 Thus, the moment of area of the above shaded are is 2435.9355 x106 mm4

Answered: 10-33. Determine the moment of inertia…

Engineering

Mechanical Engineering

10-33. Determine the moment of inertia I of the shaded axis. area about the y 100 mm --100 mm 150 mm 150 mm 75 mm 150 mm

Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

ChapterMA: Math Assessment

Section: Chapter Questions

Problem 1.1MA

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