Q A2 Locate the centriod E of the channel section and then d

Q A-2. Locate the centriod (E )of the channel section and then determine the moments of inertia and \'y, and polar moment of inertia le 40 mm 40 mm 40 mm 40 mm 120 mm 40 mm 02008 Pearson Persie Hal Inc

Solution

X\' = [(160x40x20)+(120x40x100)+(120x40x100)] / [(160x40)+(120x40)+(120x40)] = 68mm

Y\'=[(160x40x80)+(120x40x20)+(120x40x140)] / [(160x40)+(120x40)+(120x40)] = 80mm

So, the centroid is (68,80).

I_X\' = (80³x40/12)+2x[(40³x160/12)+{(40x160)x(40+80)²/4}] = 49493332 mm⁴

IY\' = (40³x80/12)+[(40x80)x(50-20)² ] + (2x160³x40/12) + [(2x160x40)x(50-80)²] = 36949332 mm⁴

So, the polar moment of inertia, I_c= I_X\' + I_Y\' = 86442664 mm⁴