Question

For the shape below a) Locate the center of shepe (and y) b) Calculate the second moment of inertia (1x) and (1) about the axis passing from center of the shape. Nur = S(7 + Ad) v by rectangle) - =" (Cireles = Tid 6 in. Sin. Sin. r = 4 in. 12 in. IMA 19.56

Answer 1

Solution step- 2010 2017 ozhin 1-4 in 112in This structure this as can shown assumed equivalent like belows above step-2 Let A1 = Area of reactangle Yo = cig of reactangle from x-azpis x = c.g of reactangle from y-aris. 5 14-20 74, = 20 = 101n step-3 A = 20x14 = 280 in² di = 14 1 2 = Tin y = 20 = 10 in Az - Area of circle Y ₂ = c.god circle from x-aris = d.g of circle from y-aris, A2 = xr² = TX 4² = 167 in x2 = 6in 4 = 12 in. - Gin rouin

Stepy ă = EAidi EAN AX - A222 A - A2 ł = 220x7 - 16+x6 280-167 ă = 7.22 in y= EAR E Aiyi - Aiy, - Az Ya Ai- Az 280X)0 - 16 7 x 12 280 - 16 T y = 9.56in (6) Now therefore X= 7.22in j = 9.56in calculation second moment of Inertia Ix and Iy about the axis passing from the centre of the shape 47 steb-s zoin Tuin x-bris IxI= second moment of Inertig of reactangle bassing from Antse of shape parallel to IX = second moment of Inertia of reactangle about antre of reactangle parallel to 88-axy da distance between Centre of reactangle and centre of shape in a direction

anis since According to parallel theorem IX, IX + Ad² Tx = 65° 4X (2013 = inge Ix = 28000 in 4 d= - x = 7.22-7= 0.22 in. Ad² = 28080.222 = 13.552 ina therefore IxI= IX + A d² TY - 2 9 GDO + 3 SS2 zin 9346.88 in4 3 step of Is In Seco 12 in II bin I Ix2= second moment of Inertia I Docuin of circle passing from centre of shape paraled to a-aais. Id= second moment of Inertia of circle passing from untre of Circle Paralled to a-anis. d= distance between centre of reactangle and centre of shape in x-direction. Ide - IX & Ad² Ix = b y = x(49 _ 645 int A de X.- X2= 7.22-6- 1.22 in. A d² - Tx 4² x 1.22² - 74.81in" 4

Ix2= TX + Ad² Ixa = 642 + 74.81 Ix2= 275.877 in Ix= second moment of Inertia of shape about axis passing from the centre of shape in x-direction. Ix = Idi - Isla IX = 9346.88 - 275.877 IX = 9071.003 in4 similarly for Calculation of Ty. Step- for reactangle Iy = second moment of anertia of reactangle about anis passing from centre of shape parallel to y amis. Tu second moment of Inertia of reactangle about contre chreactangle Parallel to y-axis de distance between centre of reactangle and centre of shape in y-direction. Ту, - Ту + Al- Ty = he - 20x143 - 5000 in

d- y - y = 10-9.56= 0.44 in Ad²- 2808 0.44= 54.208 in 4. Iy, = Jy + Ad2 Iy= 5000 + 54.208 Ty, = 1720.875 in4 step 8 Iy, = second moment of Inetria of circle passing from centre of shape Barallel to yaxis Iy = second moment of Inertia of circle passing from centre of circle paralled to go axis de distance between centre of circle and centre of shape in y-direction. Ih - Iy + Ada Iy= Art = 7 x 4 = 64 k in" 1- 2-4 d = 0 - 1 = 12 - 9 s6 do 2.44 in. Ad? - 167x 2.44² = 299.261 in Iy - Ty + Adh Tu 14A 1293 26 | Ty = 500.323 in4

Jy second moment of Inertia of shape about anis passing from centre of shape in x-direction. IY = Iy - Iy Iy = 1720-875- 500.323. y = 1220.552 in4 Final Answer: - x = 7.22 in | 1 - 3 Số 2 Ix = 9071.in4 Jy = 1220.552 in4
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