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10-8 10-9 10-12 10-13 10-14 10-16 10-29 10-32 10-45 - AM/TW...

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Unformatted text preview: AM/TW B124: Tut 12, 05/05/2010 % y *10—8. Determine the moment of inertia of the area about the y axis. The moment of inertia of the area about the y axis will be determined using the rectangular differential element in Fig. 0. This area is 44 = y at = 2x4 a; 2m ' 1m 1y =I .1;sz = x2(2.x4dr) A 0 x i m 2 II! = 21:6 air = [—191 = 0.2361124 0 y 010—9. Determine the polar moment of inertia of the area about the z axis passing through point 0. The moment of inertia of the area about thex and y axes will be determined using the rectangular differential element in Figs. a and b. The area of these two elements are 4 y [£4 2m y=2x d4=(l-x)dy= l—[E] cb'andcfl=ydx=2x4dx. 2m 1! 4 2m 1/4 2 2 2 1 l4 Iszydd=I y l-* dy=J y —[—] 3’9 dy A 0 2 0 2 m 3 n4 _ Y__ i i 1314 = 4 a} [2] [13]., l 2”" Thus, the polar moment of inertia of the area about the z axis is J0 = Ix + 1., = 0.2051+ 0.2857 = 0.491 m4 x *10—12. Determine the moment of inertia of the area about the x axis. Differential Element : The area of the differential element parallel to y axis is dA = ydx. The moment of inertia of this element about x axis is dlx = d1] + CIA;2 2 1 y =—d 3 d — 12(x)y +y 36(2) 1 3 (2 — 2363)3 dx 2m 1 = 3 (—8x9 + 24x6 — 24363 + 8) dx Moment of Inertia : Performing the integration, we have 1m 1 = J d1x= (—sx9+24x6—24x3+8)dx 0 1 — —£x1° + £367 —6x4 +836 3 5 5 1m = 1.54 m4 0 010—13. Determine the moment of inertia of the area y about the y axis. 1’ 7:. Differential Element : The area of the differential element parallel to y axis is dA = ydx = (2 — 2x3) dx. Moment of Inertia : Applying Eq. 10—1 and performing y 2 2 _ 2x3 the integration, we have 2 m 2 m 1 m I=Jl xsz=J 3622—2363 dx ,_, , A 0 < > ,&.1 b 2 1 m X F‘- = _x3 _ 1x6 3 3 0 = 0.333 m4 Ans 10—14. Determine the moment of inertia of the area about the x aXis. Solve the problem in two ways, using rectangular differential elements: (a) haVing a thickness of dx, and (b) haVing a thickness of dy. a) Differential Element : The area of the differential element parallel to y aXis is dA = ydx. The moment of inertia of this element about x aXis is de = d1} + CIA;2 2 1 y = — dx 3+ dx — 12 ( )y y [2] 1 y = 4 — 4x2 = — (4 — 4362)3 dx 3 4m 1 — 3( 64x6 : 192x4 192x2 : 64) dx Moment of Inertia : Performing the integration, we have 1 1m 1 1x: I 611,: 3! §(—64x6+192x4—192x2+64)dx 1m 1m 71m 1m _ 1 64x7:192x5 192x3 :64x 3 7 5 3 ,1m = 19.5 m4 Ans 1 b) Differential Element : Here, x = — J 4 — y .The area of the differential 1} 2 element parallel to x aXis is dA = 2xdy = ‘l 4 — y dy. Moment of Inertia : Applying Eq. 10—1 and performing the integration, we have 4m J. 4m ,1 T611" 1x= IA yZdA= JO y2\/4—ydy K 2y2 2 8y 2 16 2 4m _ 4 3 _ 4 3 4_ 3 1m 1m |: 3 ( Y) 15( Y) 105( Y) 0 (b) = 19.5 m4 Ans *10—16. Determine the moment of inertia of the triangular '10-29. Determine the moment 0f inertia of the beam’s area about the x axis, cross-sectional area about the y axis. 30 mm 30 mm yz—b—(b—X) ' gbg mduwmmaded) M-mwhctessb-z-yJenee. “'w'ib‘t’m 1y = % (20)(60)3 + 2[%(40)(10)3 + 10(40)(15)2] = 54.7 (104) mm4 Ans *10—32. Determine the moment of inertia of the composite area about the x aXis. % 7% 75mm IOOMM Composite Parts: The composite area can be subdivided into three segments as shown in th.a. Since segment (3) is a hole, it contributes a negative moment of inertia. The perpendicular distance measured from the centroid of each segment to Ihex axis is also indicated. Momentof Inertia: The moment of inertia of each segment about the x axis can be determined using the parallel - axis theorem. Thus. Ix =13; “my? = L(3oox2003)+l(3ooxzoo{fi]2 + [i(300)(2003)+ 30(1200x100)2]+[—£(754)+(—rzt752)}(100)2] 36 2 3 12 4 =m106)m4 Ans. 010—45. Determine the moment of inertia of the composite area about the x axis. Composite Parts: The composite area can be subdivided into three segments as shown in F134;. The perpendicular distance measwed from the centmid of each segment to the x axis is also indicated Moment of Inertia: The moment of inertia of each segment about the x axis can be determined using the parallel - axis theorem. 11-1115, 1,, =1} may? = [113(150x3m31+300(150x0)2]+ [%(150)(1503 )+ 15q150x732]+[;16.(1so)(1503)+ _;(150x150)(50)2] =54£{|05)mm4 Ans- ...
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10-8 10-9 10-12 10-13 10-14 10-16 10-29 10-32 10-45 - AM/TW...

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