### Centroid of a Composite Area

• Calculate centroid of composite shapes, with or without cavities, using formula for centroid of primitive shapes.

### Solved Example:

#### 26-2-01

The centre of gravity of a triangle is at the point where three:

Solution:
A centroid of a triangle is the point where the three medians of the triangle meet. A median of a triangle is a line segment from one vertex to the mid point on the opposite side of the triangle.

### Solved Example:

#### 26-2-02

If an area has one line of symmetry the centroid will:

Solution:
The centroid of a 2D surface is a point that corresponds to the center of gravity of a very thin homogeneous plate of the same area and shape. If the area (or section or body) has one line of symmetry, the centroid will lie somewhere along the line of symmetry.

### Solved Example:

#### 26-2-03

Calculate centroid of the figure shown. Solution:

• For quarter circle Area = 314.16 $cm^2$

x = $\dfrac{4R}{3\pi}$ = 8.4883 cm
y = $\dfrac{4R}{3\pi}$ = 8.4883 cm

• For semi circle Area = 157.08 $cm^2$

x = $\dfrac{20}{2}$ = 10 cm
y = $\dfrac{4R}{3\pi}$ = 4.2441 cm

$\bar{x} = \dfrac{\Sigma A_x}{\Sigma A} = \dfrac{1095.87}{157.08} = 6.98\ cm$ $\bar{y} = \dfrac{\Sigma A_y}{\Sigma A} = \dfrac{2000}{157.08} = 12.73\ cm$

### Solved Example:

#### 26-2-04

Considering bottom left corner point as origin, what is the y coordinate of the centroid of the object shown? Solution:
Let part 1 be the rectangle and part 2 be the semicircle. $A_1 = 6 \times 3 = 18\ cm^2$ $y_1 = 1.5$ $A_2 = \dfrac{(\pi)(4)}{2} = 6.28\ cm^2$ \begin{equation*} \begin{split} y_2 & = 3 - \dfrac{4R}{3\pi} \\ & = 3 - \dfrac{8}{3 \pi} \\ & = 2.1511\ cm \end{split} \end{equation*} \begin{equation*} \begin{split} \bar{y} & = \dfrac{A_1 y_1 - A_2 y_2}{A_1 -A_2} \\ & = \dfrac{18 \times 1.5 - 6.28 \times 2.1511}{18 - 6.28} \\ & = \dfrac{27 - 13.509}{11.72} \\ & = 1.15\ cm \end{split} \end{equation*}

### Solved Example:

#### 26-2-05

For the composite lamina shown in red in the figure, determine the coordinates of its centroid. (CBCGS Engineering Mechanics- Dec 2018) Solution:

Part I = Large Rectangle
Area = 90 $\times$ 60 = 5400 mm$^2$
$\bar{x}$ = -30, $\bar{y}$ = 5

Part II = Small Rectangle
Area = 50 $\times$ 40 = 2000 mm$^2$
$\bar{x}$ = 25, $\bar{y}$ = -20

Part III = Quarter Circle
Area = $\dfrac{\pi}{4}$ $\times$ 50$^2$ = 1963.5 mm$^2$
$\bar{x}$ = $\dfrac{4R}{3\pi}$=21.22
$\bar{y}$ = $\dfrac{4R}{3\pi}$=21.22

Part IV = Triangle (Removed part)
Area = $\dfrac{1}{2}$ $\times$ 75 $\times$ 90 = 3375 mm$^2$
$\bar{x}$ =-35, $\bar{y}$ =-10

\begin{align*} \bar{x} &= \dfrac{\Sigma A_i x_i}{\Sigma A_i}\\ &= \dfrac{5400(-30) + 2000 (25) +1963.5 (21.22) - 3375(-35)}{5400 + 2000 + 1963.5 - 3375}\\ &= \dfrac{47790.5}{5988.5} = 7.98\ mm \end{align*} \begin{align*} \bar{y}&= \dfrac{\Sigma A_i y_i}{\Sigma A_i}\\ &= \dfrac{5400(5) + 2000(-20) + 1963.5 (21.22) -3375(-10)}{5400 + 2000 + 1963.5 - 3375}\\ &= \dfrac{62415.5}{5988.5} = 10.42\ mm \end{align*}