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 (lx) The radius of the circle is 0.75 ft. NOTE: Use only the equations at the end of this test. (Hint: 4 segments) C. у 1 ft 1 ft х 3 ft 3 ft

Answer 1

Solution.

Solution : Manent of curea methool in uned to find the controid of the given figure There are four figures in the above given figure i y 2114 8 = 0.75 P1 11th se K 31->k-334 1) Triangle 2) Rectangle 3) Semicircle 4) circle let ū be the x-component of centroid Total area A = A + A₂+Az - Ay A 2+6+ 3x2 + x(2 1(1.5) 4 2 = 3 +6 +1:57-4.46 A = 8.81.64? Locate the dintente of centerxiool of all figersey form x and axin y 0.42 ft 0.23847 GIH 4.5H eft F 247 2.0-1/87 3 st 311 기 cs Scanned with CamScanner

grea of Moment of figure = Sum of Sum of Moment of area of all figures Algeball Ах = A x +A₂X2 +A₃X3 - Auxu 8.81* - 3x2 +6x<r5+ 5x0=10)36 –Excus) 6.424 X6488 2 8.81 x = 6 + 2779.42-11.35 8.81 x x 31.07 Ž= 3.58.ft 8-81 T For i AY = AN, TA2 Y₂ + A3 Y₂ - Aq Yy 3x 0-6+ +6X 1 +7x28x2 - 61.5641. 8.81 T - 2.61 + 6 +1757 -1.76 8.81Y = 4.82 Ý 0.887.ft The centroid of given (3.58,0.887) → Firat Find the moment of inertia about centoidal axia and then find Ix uning parallel axis theorem 3x2? TG1 = 0.667" = 2th 163 = 0.3926 444 Tx, = IG,+AN Ix3 = I 63 +A₂ 13² = 0.667+ 3x(0:67)? 20-3926+ = 2.0137444 = 1.963414 figure in 3 TG2 = IG 3x2 Tos= (– Sina)(234 36 12 2. 2 Ix2 = TG2 + A2/2 2 +6x12 Ix2= 8 + 4 CS Scanned with CamScanner

T (1.594 2 I64 64 -0.248 Txu = IG4 + AY = 0.248+ I (15)?x20,2 464 = 0.248 +1767) Txy = 2.015 ft4 Ix = 1x1 + Ix2 +Ix3 - Ix4 Ix =2.0137 + 8 + 1.963 -2-015 Ix = 9.9617 ft4 CS Scanned with CamScanner