Question

Determine the polar radius of gyration of the area of the equilateral triangle of side b = 16 in. about its centroid C. Answer: kz =

Answer 1

Solution: 1. Consider the schematic diagram of the triangle. b Calculate the height h by using the Pythagoras relation: - Vo +25

Calculate the moment of inertia of the triangle about the x axis. _31364 36x8 364 96 Calculate the moment of inertia of the triangle about the y axis. I = 2 B I, = 2 +

Calculate the polar moment of inertia by using the following relation: I = 14 +1, - 136 1364 I= 96 96 48 Calculate the area of cross section of the triangular plate. Calculate the polar radius of gyration by using the following relation:

1264 48 k= 215 16 * 2113 k= 4.618 in Therefore, the polar radius of gyration of the triangular plate is k = 4.618 in. Consider the following schematic diagram of a semi-circular ring:

Write the relation for elemental area of the strip. dA=r.de. dr Write the x, and y in terms of polar coordinates. x=rcos e y=rsin Write the formula to calculate the moment of inertia of the shaded area about the r - axis. 1, = [y’da | | (rsino);r.do.de 0 0.55a

sin 20 a* -(0.5 4 = 0.11351a4 Write the relation to calculate the total area of the shaded section. A = (x - T) = (a? -(0.55a)) = 0.3487a Write the formula to calculate the moment of inertia of the shaded area about the y-axis. 1, = 1 Calculate the radius of gyration about x and axis:

0.11351a Ikx = k, = V0.3487a? k = k, = 0.571a Therefore, the radius of gyration of the shaded area about x and y- axis is k, = k, = 0.571a. Write the relation to calculate the polar moment of inertia about point O. I. =1, +1, = 0.11351a* +0.11351a* = 0.227ra Write the relation to calculate the polar radius of the gyration about point o. k. =0.227ra" 10.34870 k = 0.807a

Therefore, the polar radius of gyration of the shaded area about point O is k, = 0.807a. 3. - - - a Calculate the height h by using the Pythagoras relation:

Calculate the moment of inertia of triangle (1) and (3) about the x- axis: bh = 443905.67 mm Calculate the moment of inertia of triangle (2) about the x-axis:

= 665858.516 mm Calculate the moment of inertia of the semi-circular part about the x-axis: T(30) 8 = 318086.25 mm Calculate the moment of inertia of the given part about the x-axis: 1 = 1,3 +12-14 = 443905.67 +665858.516-318086.25

= 791677.936 I=0.7916x 10 mm Divide the composite figure in to three sections as shown: Here, G is the centroid of the rectangle, G is the centroid of the semicircle, and G is the centroid of the circle. Calculate the moment of inertia of the rectangle. ī bh I = 12

1 = 27x6 12 = 486 in The distance from the base a-a to the centroid of the rectangle is, = 3 in Calculate the area of the rectangle: A = bxh 4 = 27 x6 = 162 in Calculate the moment of inertia of the rectangle about the base a-a. 1x1 = 1x1 + Ad 1,1 = 486 +162(3) = 1944 in

Consider the schematic figure represented the centroidal axis of the semicircle is as follows: 102 d26 Calculate the moment of inertia of the semicircle. 8 #*74 8 =942.87 in Calculate the distance from the base of the semicircle to the centroidal axis. 377 d₂ = 30 4(7)

= 2.97 in Calculate the area of the semicircle is, TR² A = 2 4, +7 (7) 2 = 76.969 in Calculate the moment of inertia of the semicircle about centroidal axis. Tv2 = 7,2 – Azdı? [x2 = 942.87 – 76.969(2.97)? = 263.934 in Calculate the distance from the base a-a to the centroid of the semicircle is, d2 =6+ 4R 37 d, = 6+ 4(7) 3л = 8.97 in Determine the moment of inertia of the semicircle about the base

a-a. 142 = 1x2 + A d2 142 = 263.934 +76.969(8.97) = 6456.93 in Calculate the moment of inertia of the circle. Here, r is the radius of the circle. = 63.617 in The distance from the base a-a to the centroid of the circle is, da = 6 in Calculate the area of the circle is, Az = ar?

Az = 2(3) = 28.27 in? Determine the moment of inertia of the circle about the base a-a. 1x3 = 7x3 + Azdz 1x3 = 63.617 +(28.17)(6 =1077.737 in Calculate the moment of inertia of the composite figure about the base a-a. 14-a = 1x1 +1,2 - 113 la-a = 1944 +6456.93 - 1077.737 I = 7323.193 in- Therefore, the moment inertia of the section about the base a-a is Ia-= 7323.193 in.