Question

Determine the location of the centraid (x, y) in inches and the second maments of area 7 and 7 in in4 with respect to the centroidal axes for the cross-sectional area of the angle. (For the location of the centroid, enter your answers to at least two decimal places.) 9.6 in 3.2 in 3.2 in 9.6 in in in in4 in

Answer 1

*aink Y Area of rectangle CAD=12,843:2 = 40.96372 Area of rectangle ☺ (Az)= 6.4% 3.2 = 20.48102 960 % / 6450 W, = 3,2 = 1.6 10 *= 3.2+619 = 6.476 8 = A + A45 - skal 31201 40.6x16+2048x64 40196+2048 Ai+A2 я=32 0 оде Parallel is theorem Ti = vy, = 3.2 - 106 = 1.6in t2 = 9/2= 614-312 =3.2 in moment of inertia about X-X axis Ixx= [RO3 + A1t] + (808 + A2h53] - foto 1 40961063] + (92Merce+ 2048X Candy Dxx = 419,43 int] % = X2 = 1218 3.2 = 614 in = 116 in ✓ = Aixit A212 Ai+A2 = 40.966.4+ 20148x16 40.96+20,48 T = 418 in from Lus [left side]

Parallel axes theorem his india 6.4-418 = 16 in Te = ~12 = 48 – 1,6 = 3.2 in moment of Inertia about Y-Y axis Tipy a pode + A16i] + [DeS + As ton] = $12X120° + 40.26860.677 + [64x380 + 2048x(227 12 Iyy = 891129 int Results Location of the centroid (se) = 4.8in from Left side (9) = 3.2 in from base mom Ait Of Inertia about xx axis ( Ixx) = 419:43 int moment of Inertia about y-y axis (Tyy) = 891129107
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