Part B Find the location x z of its mass Centre of Gravity

Part B)

Find the location (x?, ?, z ¯) of its mass Centre of Gravity.

The composite plate is made from both steel (A) and brass (B) segments. Take rho_st = 7.85 Mg/m^3 and rho_br = 8.74 Mg/m^3. Suppose that L = 185 mm.

Since the thickness is uniform, we can readily say that Ycg = 30/2 = 15 mm towards the negative y-axis. Or,

Ycg = -15 mm

Now area of triangle B = 1/2 * 225 * 150 = 16875 mm^2

Mass of triangle B = 8.74*10^6 * 16875*10^-6 = 147487.5 kg/m

Let\'s divide trapezium A into two parts, a rectangle of sides 185 mm and 225 mm. And a right triangle with sides 225*150.

Area of triangle portion of A = 1/2 * 225 * 150 = 16875 mm^2

Area of rectangle poriton = 185*225 = 41625 mm^2

Mass of triangle poriton of A = 7.85*10^6 * 16875*10^-6 = 132468.75 kg/m

Mass of rectangle poriton of A = 7.85*10^6 * 41625*10^-6 = 326756.25 kg/m

Now, Xcg of rectangle portion = 185/2 = 92.5 mm

Xcg of triangle portion of A = 185 + 150/3 = 235 mm

Xcg of triangle B = 185 + 150 - 150/3 = 285 mm

Therefore Xcg of total area = (326756.25*92.5 + 132468.75*235 + 147487.5*285) / (326756.25 + 132468.75 + 147487.5)

Xcg = 170.409 mm

Now, Zcg of rectangle portion = 225/2 = 112.5 mm

Zcg of triangle portion of A = 2/3 * 225 = 150 mm

Zcg of triangle B = 225/3 = 75 mm

Therefore Zcg of total area = (326756.25*112.5 + 132468.75*150 + 147487.5*75) / (326756.25 + 132468.75 + 147487.5)

Zcg = 111.572 mm