Question

Determine the location of the horizontal centroidal axis and moment of inertia.

2.08 ft 16.67 ft 2.08 ft 14.58 ft

Answer 1

3.08 ft 16.67 ft G donde I 2. 14.58f are There turo reitingule components in the given figure Area of component a, 08 X 16.67 a, 34.67 36 t Area of component @ = 2.08 x 14.58 = = a2 2 2 = 30.52646+ 2

location of the hori horizontal centroidal anis < = qiy, + 9242 9,892 of yo centroidal distance component 1 2.08 + 16.67 2 10.415 ft yi Yz cenferoidal distance of component a 2.08 loufr Yz <el 9. 4. & 9242 aita 34.6736 X 10.415 + 30.3264 x 1.04 34.6736 + 30. 3264 Ý 6.041 ft from horizontal anis.

% = 7-9 of two components Moment Inertia Iga + Ay 2 In this problem blem there are Therefore, I= Igia + Aig? + Iger + A₂%, I bid," + 9, 7,2 + bedre + a2 Yz 2 = 12 12 berswerft bedrero bi =2.08 ft 34.6736ft 2 =10.415-6.041 y = 4.374 ft di = 16.67 ft 1 = 8-у 9 in i b2 = 14.58f d₂=2.08ft q= 30.3264 ft Y = 6.041 -1.04 Iz = 5.000 ft Substitute the values Moment of Trentia I + 34.6736 x 4.374? 12 14.58 x 2.083 + 30.3264x5.0012 12 = 2235.718 H I = 2.08 x 16.67 + Time

THE HORIZONTAL CENTROIDAL AXIS IS LOCATED AT 6.041 ft

THE MOMENT OF INERTIA IS 2235.718 ft⁴

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