Question

(10 points) For the cross section shown below, find the centroid of the section y and the moment of inertia 1. 16 in 1.0 in LO in 10 in 2 in > I in 1 in 3 in 2 in > > in 3 in 2 in

Answer 1

In this question we have to find the distance of the centroid and the moment of inertia of the given composite section.

In order to solve this question we have to divide this whole composite section into a number of sections and solve for them, then by adding them we get the answer for the whole section.

The step by step calculations are given below:-

16 in. 8 lin lin Z 10in K * ** 1 in 2 in lin 3 in zin zin. lin 3 in lin First we into have to divide this composite section. multiple sections such as and @. The is symmetric. given composite section Y-axis" about Assume au axis a-a the reference asis. The distance of the centroid would be measured from aseis.

Now we have to find find the area. of all the individual sections 3 the componte. So. Area V section ① A 1 x2 2 in2 1 similarly Az 2x 1 2 in² А 1x 11 11 in2 3 A 4 3X1 3 in? 2x 11 Ą 22 in² and since section is symmetric. 3 in 2 '6 A = a 11 in 2 A 7 3 Az Az a 2 in2 An 2 in 2 Ag g

and. the distances of of the centroids the areas. from a-a. axis be calculated centroid distance of section in. as Y = 10 Similarly, 10.5 in. y = 10+ 1 2 2 11 z = 5.5 in, 2 10+ 10.5 in. Y4 yo 11 5.5 in. dd and since section is symmetric. y = 4 10.5 in. 12 5.5 in. gy 3 yo 10.5 in, 2 10 in. Yg - y 1

Now, from the principle of moments, the formula for the centroid of the section is given & Ay EA or y (AL Y + A2Z + Azt + A Yut As It A & A + As Got Agg) (A + A + A + A + A + A + AT At Ag) 9 24(2x 20 + 3x10.5) + 22x5.54 2x(2 + 2 + 11 +3)+22 + 2x 10.5 + 11 x g. S t 121 loo 266 t 36 t 22 387 58 म 6.672 in Ansi

Now all the Moment of Thertia the sections. about be calculated the centroid can. Moment inertia for section ④ I1 = 1 x (2)3 0.667 in4. 12 similarly, I 2 = 2 x (1) 0.167 inu 12 3 . Iz = 1 & (21) 12 110.917 444. I4 3x (173 12 0.25 int Is = 2 x (11)3 221.833 14 12 and so, since section is segonmetric, I6= I4 = 0.25 in 4. I I 110.917 in 4 If = 0.167 in 4 Iz = Tg = Is = 0.667 in 4.

So, the Mome ut of înentia of the whole section would be. Iz (12+ Izt Ist Ih) x2 + Ic I 2 = (0.667 + 0.167 + 210.917+ 0.25 ) x2 + 221:833 I 224.002 + 221.833 Iz 445.835 inu. Ansi