Question

2. CENTROID AND MOMENT OF INERTIA For the shape shown below, determine the following: (Make sure to label or describe the different segments.) a. Centroid (Xbar, Ybar) b. Moment of inertia about the x-axis (1x) C. The radius of the circle is 0.75 ft. NOTE: Use only the equations at the end of this test. (Hint: 4 segments) y 1 ft 1 ft 3 ft 3 ft

Answer 1

=> x = 3^2)+(6x 4.5) + (1.57 X 6.424) -(1.766 X 6) 1.7662 3+ 6 + 1.57 10.596 => ž= 6 + 27 + 10.085- 8.80 IX -) 3.691 fut Angwen and Ý - A,9, + A, I, + A₂} - Aaja A, +Azt A3-A4 => = (3x) +(6x!) +(1.57X1) + (1.766X1) 3+ 6 +1.57-1.766 -> 2 + 6 +1:57 +1.766 8.80 =) F = 1,288 fut sum

g. 948 ftt 0 As we know, sha 3823 = 2 ft 4 12 12 3 bh? 3x23 8 ft4 - - U (Ix)z 3 4 bh (Ex)3 = a (?- sad 1.962 5x x 7259 128 149 (Iz) n = 7d7 ade 4 x X (1) 2 2 TX 1.54 XX 1.5 4 = 0.2483 +1.766 + 64 2.014 ft 4 Ix = Ix, & Ix2 + Ixz - IxA = 2 + 8 +1,962 - 2.014 11

comment :- Here it have used all site required standard formula and parallel axis theorem and then super position theorem to find moment of Inertia and centroid.