Question

QUESTION 1 If the cross section of the beam in the previous question is "T" shape with the below dimensions, find the centroid and moment of inertia of the complex shape by constructing the table, do the required calculations and answer the following questions. TFw TFIT BFt BFw TFw= 80mm TFt= 38.4mm Wh= 50mm Wt= 10mm BFt= 32mm BFw= 100mm The y-coordinate of the centroid of the complex "I" shape =

Answer 1

Solution: < 80mm - mm om 32 mm е (обww -> A - 20438.4 - 2070 mm 1x, = 10+40=50mm 38.4 mm 2 x2=45+5= 50mm A = 50x10 = 500 mm² x3= 100 = 50mm Az= 100x 32 = 200 m² y = 50+32+ 384 = 101.2 mm Centroid (& 7): [Y2= 32+ 50 g = 57mm d... . Aix, + A₂ X2 +A323 18 mm - Ait Azt Az 7- (3072*50)+(500x50) +(3200x60 : 5=50mm Ang (3072 +500+3200) V = CAIY,)+(Azy?)+ A3y2 ( 3072x 101:2)+(50x10x57)+(32x160x16) (Ait Az+ A3) ( 3072 +500+3200) T = 5706766 mm Aus > 12 12 W X Moment of inertia: Ix= Ix,+Ix2 + Ixz 3 Ix = 80x38-43+ (80x38-4) (25+19.2927 + 10x50 37 + [ 100.32% (100432) (250+ 16)*] [2x = 121135x105 mm." Any Iy = (2x + 2y = + [43] 3.97( 954 280soxto?, szy 100 37 [Iy = 43.092 x 105 mm 4 Any itly2 + 1y3 N

If any doubt please ask and do rate possitive, Thank you!