Chapter 5 Engineering Mechanics: Statics

Centroids, Center of Gravity & Distributed Loads

Center of Gravity, Centroids of 2D/3D Composite Bodies, Integration for Curved Boundaries, and Distributed Loads

5.1 Centroid Locations: 2-D Objects (Common Shapes)

5.1.1 Centroid Location of 2-D Objects (with Common Shapes)

Figure 5.1 illustrates a flat plate whose surface area $A$ (in the $x-y$ plane) is much larger than its uniform thickness $t$ along the $z$-direction. In this configuration, the total weight $W$ of the plate acts at its center of gravity $G(\bar{x}, \bar{y})$. The fundamental objective of this section is to develop and apply the formulas required to compute the coordinates of the centroid and center of gravity.

Figure 5.1: Flat Plate with Center of Gravity — Figure 5.1: Flat plate with surface area A and thickness t, showing total weight W acting at centroid G(x̄, ȳ).

Let $\rho$ denote the uniform material density of the plate. We can conceptualize this plate as composed of thousands of discrete particles, where each particle $i$ has its own weight $w_i = \rho \cdot t \cdot A_i$ and coordinates $(x_i, y_i)$. The resultant weight $W$ of the entire body is the sum of these elemental weights:

$$W = \sum_{i=1}^n w_i$$

(5.1)

Applying the principle of moments (Varignon\'s Theorem), the moment of the resultant weight about any coordinate axis equals the sum of the moments of the individual particle weights. Taking moments about the $x$-axis and $y$-axis yields:

$$W \bar{y} = \sum_{i=1}^n w_i y_i, \quad W \bar{x} = \sum_{i=1}^n w_i x_i$$

(5.2)

Substituting $w_i = \rho \cdot t \cdot A_i$ and $W = \rho \cdot t \cdot A$, the density $\rho$ and uniform thickness $t$ cancel out, giving the definitive formulas for the centroid of area of a composite 2-D body:

$$\bar{x} = \frac{\sum A_i x_i}{\sum A_i} = \frac{\sum A_i x_i}{A}, \quad \bar{y} = \frac{\sum A_i y_i}{\sum A_i} = \frac{\sum A_i y_i}{A}$$

(5.3)

Before analyzing composite bodies, let us review the centroid locations of fundamental 2-D geometric shapes.

Properties of Triangular Centroids:

Consider a general triangle $ABC$ with vertices and opposite side midpoints $M, N, P$. The three median lines $AM, BN, CP$ intersect at a single point $G$, which is the centroid of the triangle (Figure 5.2). The centroid divides each median into segments with the ratio $2:1$:

$$AG = \frac{2}{3} AM, \quad BG = \frac{2}{3} BN, \quad CG = \frac{2}{3} CP$$

(5.4)

$$MG = \frac{1}{3} AM, \quad NG = \frac{1}{3} BN, \quad PG = \frac{1}{3} CP$$

(5.5)

Figure 5.2: Triangular Geometry with Medians — Figure 5.2: Triangular geometry showing medians AM, BN, CP intersecting at centroid G.

Important Geometric Rules for Triangular Centroids:

Rule 1 (Ratio Rule): The centroid $G$ is always located at a distance of $\frac{2}{3}$ of the median length from any vertex, or $\frac{1}{3}$ of the median length from the opposite midpoint.
Rule 2 (Median Line): Connect a vertex to the midpoint of the opposite side.
Rule 3 (Division): Divide the median into 3 equal segments; $G$ is 2 segments away from the vertex and 1 segment away from the base midpoint.
Rule 4 (Similar Triangles): Construct similar triangles along the median line to find horizontal and vertical coordinates.
Rule 5 (Equal Ratios): Equate coordinate ratios to establish centroid distances, as shown in Figures 5.3 through 5.5.

Figure 5.3: Similar Triangles Option A — Figure 5.3: Similar triangles formulation (Option a).

Figure 5.4: Similar Triangles Option B — Figure 5.4: Similar triangles formulation (Option b).

Figure 5.5: Similar Triangles Option C — Figure 5.5: Similar triangles formulation (Option c).

Figure 5.6 summarizes the centroid locations for standard 2-D shapes commonly encountered in engineering problems, including rectangles, right triangles, semicircles, circles, and circular sectors:

Figure 5.6: Standard Geometric Centroids — Figure 5.6: Centroid locations for standard geometric shapes (rectangles, triangles, semicircles, and sectors).

Rectangle (base $b$, height $h$): $A = b \cdot h$, $\bar{x} = \frac{b}{2}$, $\bar{y} = \frac{h}{2}$
Right Triangle (base $b$, height $h$): $A = \frac{1}{2} b \cdot h$, $\bar{x} = \frac{b}{3}$, $\bar{y} = \frac{h}{3}$ (measured from the right-angle corner)
Semicircle (radius $r$): $A = \frac{\pi r^2}{2}$, $\bar{y} = \frac{4r}{3\pi}$ from the diameter base
Quarter-Circle (radius $r$): $A = \frac{\pi r^2}{4}$, $\bar{x} = \frac{4r}{3\pi}$, $\bar{y} = \frac{4r}{3\pi}$ from the center corner
Circular Sector (radius $r$, total angle $2\alpha$): $A = \alpha r^2$, $\bar{x} = \frac{2r\sin\alpha}{3\alpha}$, $\bar{y} = 0$ (with $\alpha$ in radians)

Example 5.1

Find the centroid location $(\bar{x}, \bar{y})$ for the 2-D composite object shown in Figure 5.7, which has 3 components:

Rectangle $OACE$, with centroid location $G_1(\bar{x}_1, \bar{y}_1)$
Half-circle $ABC$, with centroid location $G_2(\bar{x}_2, \bar{y}_2)$
Triangle $ECD$, with centroid location $G_3(\bar{x}_3, \bar{y}_3)$

Given dimensions: Rectangle base 6 in, height 12 in; Half-circle radius $r = 3\text{ in}$; Triangle base 6 in, height 12 in.

Figure 5.7: Composite 2-D Shape for Example 5.1 — Figure 5.7: 2-D composite object for Example 5.1 consisting of rectangle, half-circle, and triangle.

Solution

To find the centroid coordinates of the composite body, we compute the area and centroid location for each simple component:

1. Rectangle $OACE$: Base $b = 6\text{ in}$, height $h = 12\text{ in}$.

$A_1 = (6\text{ in}) \times (12\text{ in}) = 72\text{ in}^2$
$\bar{x}_1 = \frac{6}{2} = 3\text{ in}, \quad \bar{y}_1 = \frac{12}{2} = 6\text{ in}$

2. Half-Circle $ABC$: Radius $r = 3\text{ in}$, base on $y = 12\text{ in}$.

$A_2 = \frac{\pi (3\text{ in})^2}{2} \approx 14.14\text{ in}^2$
$\bar{x}_2 = 3\text{ in}, \quad \bar{y}_2 = 12 + \frac{4(3)}{3\pi} = 12 + \frac{4}{\pi} \approx 13.27\text{ in}$

3. Triangle $ECD$: Base $b = 6\text{ in}$, height $h = 12\text{ in}$.

$A_3 = \frac{1}{2}(6\text{ in}) \times (12\text{ in}) = 36\text{ in}^2$
$\bar{x}_3 = 6 + \frac{2}{3}(3) = 8\text{ in}, \quad \bar{y}_3 = \frac{1}{3}(12) = 4\text{ in}$

Table 5.1: Centroid Calculation for 2-D Composite Shape (Example 5.1)

Component #	$A_i$ ($\text{in}^2$)	$\bar{x}_i$ ($\text{in}$)	$\bar{y}_i$ ($\text{in}$)	$A_i \bar{x}_i$ ($\text{in}^3$)	$A_i \bar{y}_i$ ($\text{in}^3$)
1. Rectangle $OACE$	$72.00$	$3.00$	$6.00$	$216.00$	$432.00$
2. Half-circle $ABC$	$14.14$	$3.00$	$13.27$	$42.42$	$187.64$
3. Triangle $ECD$	$36.00$	$8.00$	$4.00$	$288.00$	$144.00$
Total ($\sum$)	$A = 122.14$	—	—	$\sum A_i \bar{x}_i = 546.42$	$\sum A_i \bar{y}_i = 763.64$

Applying Equation (5.3) to compute the composite centroid coordinates:

$\bar{x} = \frac{\sum A_i \bar{x}_i}{\sum A_i} = \frac{546.42\text{ in}^3}{122.14\text{ in}^2} \approx 4.47\text{ in}$
$\bar{y} = \frac{\sum A_i \bar{y}_i}{\sum A_i} = \frac{763.64\text{ in}^3}{122.14\text{ in}^2} \approx 6.25\text{ in}$

Answers: $\bar{x} = 4.47\text{ in}$ (or $5.47\text{ in}$), $\bar{y} = 6.25\text{ in}$.

Example 5.2

Find the centroid location $(\bar{x}, \bar{y})$ for the 2-D composite object shown in Figure 5.8, which has the same exterior dimensions as Example 5.1, but where the semicircle $ABC$ is now a circular cutout (hole) rather than added solid material.

Solution

When an object contains a void or cutout, we treat the cutout as a negative area. Its area and moments of area are subtracted from the totals:

For the semicircular hole of radius $r = 3\text{ in}$ whose diameter lies along $y = 12\text{ in}$ (opening downward into the rectangle):

$A_2 = -\frac{\pi r^2}{2} = -\frac{\pi (3\text{ in})^2}{2} \approx -14.14\text{ in}^2$
$\bar{x}_2 = 3\text{ in}$
$\bar{y}_2 = 12 - \frac{4(3)}{3\pi} = 12 - \frac{4}{\pi} \approx 12 - 1.27 = 10.73\text{ in}$

Table 5.2: Centroid Calculation for 2-D Shape with Circular Cutout (Example 5.2)

Component #	$A_i$ ($\text{in}^2$)	$\bar{x}_i$ ($\text{in}$)	$\bar{y}_i$ ($\text{in}$)	$A_i \bar{x}_i$ ($\text{in}^3$)	$A_i \bar{y}_i$ ($\text{in}^3$)
1. Rectangle $OACE$	$72.00$	$3.00$	$6.00$	$216.00$	$432.00$
2. Half-circle Hole $ABC$	$-14.14$	$3.00$	$10.73$	$-42.42$	$-151.72$
3. Triangle $ECD$	$36.00$	$8.00$	$4.00$	$288.00$	$144.00$
Total ($\sum$)	$A = 93.86$	—	—	$\sum A_i \bar{x}_i = 461.58$	$\sum A_i \bar{y}_i = 424.28$

Applying Equation (5.3) for the composite centroid with cutout:

$\bar{x} = \frac{\sum A_i \bar{x}_i}{\sum A_i} = \frac{461.58\text{ in}^3}{93.86\text{ in}^2} \approx 4.92\text{ in}$
$\bar{y} = \frac{\sum A_i \bar{y}_i}{\sum A_i} = \frac{424.28\text{ in}^3}{93.86\text{ in}^2} \approx 4.52\text{ in}$

Answers: $\bar{x} = 4.92\text{ in}$, $\bar{y} = 4.52\text{ in}$.

5.2 Centroid Locations: 2-D Objects with Curved Boundaries

5.2.1 Integration Method for Curved Boundaries

When the boundary of a 2-D area is defined by a mathematical curve $y = f(x)$ or $x = g(y)$, the centroid cannot be computed simply from standard geometric formulas. Instead, we divide the area into infinitesimal differential elements $dA$ and perform calculus integration (Figure 5.9):

Figure 5.9: Differential Strip Element for Integration — Figure 5.9: Differential strip element dA for integration along curved boundary y = f(x).

1. Vertical Strip Element:

A vertical strip has width $dx$ and height $y$. The differential area element and its elemental centroid $(\bar{x}_{el}, \bar{y}_{el})$ are:

$$dA = y \, dx, \quad \bar{x}_{el} = x, \quad \bar{y}_{el} = \frac{y}{2}$$

(5.6)

2. Horizontal Strip Element:

A horizontal strip has height $dy$ and length $x$. The differential area element and its elemental centroid are:

$$dA = x \, dy, \quad \bar{x}_{el} = \frac{x}{2}, \quad \bar{y}_{el} = y$$

(5.7)

Integrating over the entire region gives the centroid coordinates:

$$\bar{x} = \frac{\int x \, dA}{\int dA} = \frac{\int x y \, dx}{\int y \, dx}, \quad \bar{y} = \frac{\int \bar{y}_{el} \, dA}{\int dA} = \frac{\frac{1}{2} \int y^2 \, dx}{\int y \, dx}$$

(5.8)

Example 5.3

Find the centroid coordinates $(\bar{x}, \bar{y})$ for the shaded 2-D composite shape shown in Figure 5.10, which consists of three simple components:

Component #1: Area under the parabolic curve $y = x^2$ from $x = 0$ to $x = 6\text{ in}$
Component #2: Rectangle $ACFB$ of dimensions $3\text{ in} \times 36\text{ in}$
Component #3: Quarter-circle cutout (hole) of radius $r = 1\text{ in}$

Figure 5.10: Shaded Composite Area with Parabolic Boundary — Figure 5.10: Shaded composite area with parabolic boundary for Example 5.3.

Solution

First, determine the area and centroid of the parabolic region under $y = x^2$ from $x = 0$ to $x = 6\text{ in}$ via integration:

$A_1 = \int_0^6 x^2 \, dx = \left[ \frac{x^3}{3} \right]_0^6 = \frac{216}{3} = 72\text{ in}^2$
$A_1 \bar{x}_1 = \int_0^6 x (x^2) \, dx = \left[ \frac{x^4}{4} \right]_0^6 = \frac{1296}{4} = 324\text{ in}^3 \implies \bar{x}_1 = \frac{324}{72} = 4.50\text{ in}$
$A_1 \bar{y}_1 = \int_0^6 \frac{x^2}{2} (x^2 \, dx) = \frac{1}{2} \left[ \frac{x^5}{5} \right]_0^6 = \frac{7776}{10} = 777.60\text{ in}^3 \implies \bar{y}_1 = \frac{777.60}{72} = 10.80\text{ in}$

Next, compute components #2 and #3, and assemble the results into Table 5.3:

Table 5.3: Centroid Calculation for Area with Curved Boundary (Example 5.3)

Component #	$A_i$ ($\text{in}^2$)	$\bar{x}_i$ ($\text{in}$)	$\bar{y}_i$ ($\text{in}$)	$A_i \bar{x}_i$ ($\text{in}^3$)	$A_i \bar{y}_i$ ($\text{in}^3$)
1. Parabolic Area $A_1$	$72.00$	$4.50$	$10.80$	$324.00$	$777.60$
2. Rectangle $ACFB$	$108.00$	$7.50$	$18.00$	$810.00$	$1944.00$
3. Quarter-Circle Hole	$-0.79$	$8.58$	$0.42$	$-6.74$	$-0.33$
Total ($\sum$)	$A = 179.21$	—	—	$\sum A_i \bar{x}_i = 1127.26$	$\sum A_i \bar{y}_i = 2721.27$

Applying Equation (5.8) to determine the overall centroid:

$\bar{x} = \frac{\sum A_i \bar{x}_i}{\sum A_i} = \frac{1127.26\text{ in}^3}{179.21\text{ in}^2} \approx 6.29\text{ in}$
$\bar{y} = \frac{\sum A_i \bar{y}_i}{\sum A_i} = \frac{2721.27\text{ in}^3}{179.21\text{ in}^2} \approx 15.19\text{ in}$

Answers: $\bar{x} \approx 6.29\text{ in}$ (or $5.89\text{ in}$), $\bar{y} \approx 15.19\text{ in}$.

Example 5.4

Find the centroid coordinates $(\bar{x}, \bar{y})$ for the shaded 2-D object $O-D-C-A-O$ as shown in Figure 5.11, which consists of three components evaluated by composite subtraction:

Component #1: Bounding rectangle $ODAB$ of dimensions $6\text{ in} \times 36\text{ in}$
Component #2: Area below parabola $y = x^2$ from $x = 0$ to $x = 6\text{ in}$ (hole)
Component #3: Semicircular cutout $DCA$ of radius $r = 3\text{ in}$ (hole)

Figure 5.11: 2-D Area Bounded by Parabola and Cutout — Figure 5.11: 2-D area bounded by parabolic curve and semicircular cutout for Example 5.4.

Solution

We solve this composite problem by subtracting the void below the curve and the semicircular hole from the bounding rectangle:

1. Bounding Rectangle $ODAB$: Dimensions $6\text{ in} \times 36\text{ in}$.

$A_1 = (6\text{ in}) \times (36\text{ in}) = 216\text{ in}^2, \quad \bar{x}_1 = 3\text{ in}, \quad \bar{y}_1 = 18\text{ in}$

2. Parabolic Void under $y = x^2$ from $x = 0$ to $x = 6\text{ in}$:

$A_2 = -\int_0^6 x^2 \, dx = -\left[ \frac{x^3}{3} \right]_0^6 = -72\text{ in}^2$
$A_2 \bar{x}_2 = -\int_0^6 x (x^2) \, dx = -\left[ \frac{x^4}{4} \right]_0^6 = -324\text{ in}^3 \implies \bar{x}_2 = \frac{-324}{-72} = 4.50\text{ in}$
$A_2 \bar{y}_2 = -\int_0^6 \frac{x^2}{2} (x^2) \, dx = -\left[ \frac{x^5}{10} \right]_0^6 = -777.60\text{ in}^3 \implies \bar{y}_2 = \frac{-777.60}{-72} = 10.80\text{ in}$

3. Semicircular Cutout $DCA$: Radius $r = 3\text{ in}$ at the top of the rectangle.

$A_3 = -\frac{\pi (3\text{ in})^2}{2} \approx -14.14\text{ in}^2$
$\bar{x}_3 = 3\text{ in}, \quad \bar{y}_3 = 36 - \frac{4(3)}{3\pi} \approx 34.73\text{ in}$

The tabulated values are compiled in Table 5.4:

Table 5.4: Centroid Calculation for Shaded Area under Curve (Example 5.4)

Component #	$A_i$ ($\text{in}^2$)	$\bar{x}_i$ ($\text{in}$)	$\bar{y}_i$ ($\text{in}$)	$A_i \bar{x}_i$ ($\text{in}^3$)	$A_i \bar{y}_i$ ($\text{in}^3$)
1. Rectangle $ODAB$	$216.00$	$3.00$	$18.00$	$648.00$	$3888.00$
2. Parabolic Void (Hole)	$-72.00$	$4.50$	$10.80$	$-324.00$	$-777.60$
3. Semicircular Cutout (Hole)	$-14.14$	$3.00$	$34.73$	$-42.42$	$-491.08$
Total Net Area	$A = 129.86$	—	—	$\sum A_i \bar{x}_i = 281.58$	$\sum A_i \bar{y}_i = 2619.32$

Applying the composite centroid formula:

$\bar{x} = \frac{\sum A_i \bar{x}_i}{A_{\text{net}}} = \frac{281.58\text{ in}^3}{129.86\text{ in}^2} \approx 2.17\text{ in}$
$\bar{y} = \frac{\sum A_i \bar{y}_i}{A_{\text{net}}} = \frac{2619.32\text{ in}^3}{129.86\text{ in}^2} \approx 20.17\text{ in}$

Answers: $\bar{x} = 2.17\text{ in}$, $\bar{y} = 20.17\text{ in}$.

5.3 Centroid Locations: 3-D and 1-D Objects

5.3.1 Centroids of 3-D Volumes and 1-D Wires

The centroid concept extends naturally to 3-D volume bodies and 1-D wire frameworks:

Centroid of 3-D Volumes:

For a 3-D body of uniform material density $\rho$ and total volume $V$, the coordinates of the centroid are given by the ratio of the first moments of volume to the total volume:

$$\bar{x} = \frac{\sum V_i x_i}{V}, \quad \bar{y} = \frac{\sum V_i y_i}{V}, \quad \bar{z} = \frac{\sum V_i z_i}{V}$$

(5.9)

Centroid of 1-D Wires:

When a body consists of a slender wire whose cross-sectional area is small compared to its length $L$, the centroid coordinates are given by:

$$\bar{x} = \frac{\sum L_i x_i}{L}, \quad \bar{y} = \frac{\sum L_i y_i}{L}, \quad \bar{z} = \frac{\sum L_i z_i}{L}$$

(5.10)

Example 5.5

A 1-D bent wire $ABCD$ is in equilibrium as shown in Figure 5.12. Using point $B$ as the coordinate origin, find the centroid location $\bar{x}$ of this composite wire.

Figure 5.12: 1-D Bent Wire Assembly — Figure 5.12: 1-D bent wire assembly ABCD for Example 5.5.

Solution

The composite wire consists of three straight segments:

$L_1 = AB = \sqrt{(60\text{ cm})^2 + (80\text{ cm})^2} = 100\text{ cm}, \quad \bar{x}_1 = \frac{80}{2} = 40\text{ cm}$
$L_2 = BC = 80\text{ cm}, \quad \bar{x}_2 = \frac{80}{2} = 40\text{ cm}$
$L_3 = CD = 120\text{ cm}, \quad \bar{x}_3 = 80 + \frac{120}{2} = 140\text{ cm}$
$L = L_1 + L_2 + L_3 = 100 + 80 + 120 = 300\text{ cm}$

The segment properties are summarized in Table 5.5:

Table 5.5: Centroid Calculation for 1-D Bent Wire Assembly (Example 5.5)

Wire Segment	$L_i$ ($\text{cm}$)	$\bar{x}_i$ ($\text{cm}$)	$L_i \bar{x}_i$ ($\text{cm}^2$)
Segment $AB$	$100.00$	$40.00$	$4000.00$
Segment $BC$	$80.00$	$40.00$	$3200.00$
Segment $CD$	$120.00$	$140.00$	$16800.00$
Total ($\sum$)	$L = 300.00$	—	$\sum L_i \bar{x}_i = 24000.00$

Applying Equation (5.10):

$\bar{x} = \frac{\sum L_i \bar{x}_i}{\sum L_i} = \frac{24000\text{ cm}^2}{300\text{ cm}} = 80.00\text{ cm}\text{ (measured from point } B\text{)}$

Answer: $\bar{x} = 80.00\text{ cm}$ from point $B$ (coincides with point $C$).

Example 5.6

Referring to Figure 5.12, suppose the length $L_3 = CD$ is unknown. What must be the length $L_3$ so that the centroid of the entire wire assembly is located exactly at point $C$ ($\bar{x} = 80\text{ cm}$), allowing segment $CD$ to remain horizontal in equilibrium?

Solution

If the centroid of the entire wire is at point $C$, then $\bar{x} = 80\text{ cm}$ measured from point $B$. Taking moments of length about point $B$ via Equation (5.10):

$(L_1 + L_2 + L_3) \bar{x} = L_1 \bar{x}_1 + L_2 \bar{x}_2 + L_3 \bar{x}_3$
$(100 + 80 + L_3)(80) = (100)(40) + (80)(40) + L_3\left(80 + 0.5 L_3\right)$
$80 L_3 + 14400 = 4000 + 3200 + 80 L_3 + 0.5 L_3^2$
$80 L_3 + 14400 = 7200 + 80 L_3 + 0.5 L_3^2$
$0.5 L_3^2 - 7200 = 0 \implies L_3^2 = 14400 \implies L_3 = 120.00\text{ cm}$

Answer: $L_3 = 120.00\text{ cm}$.

Example 5.7

A 3-D composite body consisting of two rectangular plates is shown in Figure 5.13:

Plate #1 ($ABCD$): Dimensions $80\text{ cm} \times 60\text{ cm}$, uniform thickness $t = 4\text{ cm}$
Plate #2 ($AJMB$): Dimensions $80\text{ cm} \times 40\text{ cm}$, uniform thickness $t = 4\text{ cm}$

Determine the 3-D centroid coordinates $(\bar{x}, \bar{y}, \bar{z})$ of this composite object.

Figure 5.13: Three-Dimensional Composite Object — Figure 5.13: Three-dimensional composite object consisting of two rectangular plates for Example 5.7.

Solution

Calculate the volume and centroid coordinates of each plate:

1. Rectangular Plate #1 ($ABCD$):

$V_1 = (80\text{ cm}) \times (60\text{ cm}) \times (4\text{ cm}) = 19200\text{ cm}^3$
$\bar{x}_1 = \frac{80}{2} = 40\text{ cm}, \quad \bar{y}_1 = -\frac{4}{2} = -2\text{ cm}, \quad \bar{z}_1 = \frac{60}{2} + 4 = 34\text{ cm}$

2. Rectangular Plate #2 ($AJMB$):

$V_2 = (80\text{ cm}) \times (40\text{ cm}) \times (4\text{ cm}) = 12800\text{ cm}^3$
$\bar{x}_2 = \frac{80}{2} = 40\text{ cm}, \quad \bar{y}_2 = \frac{40}{2} = 20\text{ cm}, \quad \bar{z}_2 = 0\text{ cm}$

Table 5.6: Centroid Calculation for 3-D Composite Volume (Example 5.7)

Component	$V_i$ ($\text{cm}^3$)	$\bar{x}_i$ ($\text{cm}$)	$\bar{y}_i$ ($\text{cm}$)	$\bar{z}_i$ ($\text{cm}$)	$V_i \bar{x}_i$ ($\text{cm}^4$)	$V_i \bar{y}_i$ ($\text{cm}^4$)	$V_i \bar{z}_i$ ($\text{cm}^4$)
Plate #1 ($ABCD$)	$19200$	$40.00$	$-2.00$	$34.00$	$768000$	$-38400$	$652800$
Plate #2 ($AJMB$)	$12800$	$40.00$	$20.00$	$0.00$	$512000$	$256000$	$0$
Total ($\sum$)	$V = 32000$	—	—	—	$\sum V_i \bar{x}_i = 1280000$	$\sum V_i \bar{y}_i = 217600$	$\sum V_i \bar{z}_i = 652800$

Applying Equation (5.9) for the 3-D centroid coordinates:

$\bar{x} = \frac{\sum V_i \bar{x}_i}{\sum V_i} = \frac{1280000\text{ cm}^4}{32000\text{ cm}^3} = 40.00\text{ cm}$
$\bar{y} = \frac{\sum V_i \bar{y}_i}{\sum V_i} = \frac{217600\text{ cm}^4}{32000\text{ cm}^3} = 6.80\text{ cm}$
$\bar{z} = \frac{\sum V_i \bar{z}_i}{\sum V_i} = \frac{652800\text{ cm}^4}{32000\text{ cm}^3} = 20.40\text{ cm}$

Answers: $\bar{x} = 40.00\text{ cm}$, $\bar{y} = 6.80\text{ cm}$, $\bar{z} = 20.40\text{ cm}$.

5.4 User Friendly Application and Practice Problems

5.4.1 Description of User Friendly Application

Based on the analytical formulations presented in Chapter 5, a user-friendly application has been developed with an intuitive, menu-driven interface, interactive graphical displays of composite shapes, and automated analytical computation of centroid coordinates $(\bar{x}, \bar{y})$ and moments of inertia.

Input Data Requirements:

The user selects the analysis scenario (e.g., 2D Composite Shapes) and specifies the geometric parameters in the input table. The application dynamically adjusts input fields for rectangular, triangular, circular, and curved segments.

Figure 5.14: Input Tab of Application — Figure 5.14: Screenshot of the input tab of the user-friendly application for composite centroid analysis.

Output Data and Graphical Verification:

Once the "Calculate" button is clicked, the application computes the total area, first moments of area, and centroid location $G(\bar{x}, \bar{y})$. The output tab displays the results both numerically and graphically with visual verification (Figure 5.15):

Figure 5.15: Output Tab of Application — Figure 5.15: Screenshot of the output tab displaying computed centroid coordinates and visual verification plot.

5.4.2 Practice Problems

Solve the following practice problems manually using the tabular composite method, then verify your results using the user-friendly application.

Practice Problem 5.1

A 2-D composite shape consists of three components as shown in Figure 5.16:

Component #1: Area under parabolic curve with $L_1 = 6\text{ in}$, $H_4 = 3\text{ in}$
Component #2: Rectangle $ACFE$ of dimensions $3\text{ in} \times 33\text{ in}$
Component #3: Semicircular hole $ADC$ with diameter on $y = 33\text{ in}$

Find the centroid location $(\bar{x}, \bar{y})$ for this composite shape.

Figure 5.16: Generic Composite Geometry — Figure 5.16: Generic composite geometry for Practice Problems 5.1 through 5.4.

Answers: $\bar{x} \approx 4.82\text{ in}, \quad \bar{y} \approx 14.25\text{ in}$

Practice Problem 5.2

For the composite geometry in Figure 5.16 with data: $L_1 = 6\text{ in}, L_2 = 3\text{ in}, L_3 = 3\text{ in}, L_4 = 3\text{ in}, L_5 = 3\text{ in}, H_3 = 3\text{ in}, H_4 = 3\text{ in}, H_5 = 48\text{ in}$:

Component #1: Area above the parabola
Component #2: Rectangle $ACFE$
Component #3: Semicircle $ABC$
Component #4: Triangle $CNF$

Find the centroid location $(\bar{x}, \bar{y})$.

Answers: $\bar{x} \approx 6.45\text{ in}, \quad \bar{y} \approx 22.10\text{ in}$

Practice Problem 5.3

For the composite geometry in Figure 5.16 with data: $L_4 = 3\text{ in}, L_5 = 3\text{ in}, L_6 = 3\text{ in}, L_7 = 3\text{ in}, H_1 = 3\text{ in}, H_2 = 3\text{ in}, H_3 = 3\text{ in}, H_4 = 3\text{ in}, H_5 = 33\text{ in}$:

Component #1: Triangle $JLN$
Component #2: Rectangle $HJNC$
Component #3: Quarter-circle hole $IKJ$
Component #4: Quarter-circle $NPM$

Find the centroid location $(\bar{x}, \bar{y})$.

Answers: $\bar{x} \approx 4.15\text{ in}, \quad \bar{y} \approx 16.80\text{ in}$

Practice Problem 5.4

Using the geometry from Figure 5.16 with the modified component set: Triangle $CNF$, Rectangle $HJNC$, Quarter-circle hole $IKJ$, and Quarter-circle $NPM$:

Compute the composite centroid location $(\bar{x}, \bar{y})$.

Answers: $\bar{x} \approx 3.85\text{ in}, \quad \bar{y} \approx 7.20\text{ in}$

Practice Problem 5.5

Find the centroid location $G(\bar{x}, \bar{y})$ associated with the shaded area $O-A-B-O$ bounded by the curve shown in Figure 5.17.

Figure 5.17: Area under Parabolic Curve — Figure 5.17: Shaded area under parabolic curve for Practice Problem 5.5.

Answers: $\bar{x} = 0.75\text{ ft}, \quad \bar{y} = 1.60\text{ ft}$

Practice Problem 5.6

Find the centroid location $G(\bar{x}, \bar{y})$ of the shaded area below the curve $y = x$ shown in Figure 5.18.

Figure 5.18: Area Bounded by Curve and Line — Figure 5.18: Area bounded between curve and straight line for Practice Problem 5.6.

Answers: $\bar{x} = 1.60\text{ ft}, \quad \bar{y} = 1.00\text{ ft}$

Practice Problem 5.7

Find the centroid location $G(\bar{x}, \bar{y})$ of the shaded parabolic region bounded by $O-A-B-O$ shown in Figure 5.19.

Figure 5.19: Parabolic Spandrel Area — Figure 5.19: Parabolic spandrel area for Practice Problem 5.7.

Answers: $\bar{x} = 1.00\text{ m}, \quad \bar{y} = 0.40\text{ m}$

Practice Problem 5.8

Find the centroid location $G(\bar{x}, \bar{y})$ of the shaded region bounded by $O-A-D-O$ as shown in Figure 5.20, by subtracting the spandrel area from the bounding rectangle.

Figure 5.20: Composite Area with Curve Cutout — Figure 5.20: Composite area with curve cutout for Practice Problem 5.8.

Answers: $\bar{x} \approx 2.45\text{ in}, \quad \bar{y} \approx 3.75\text{ in}$

Practice Problem 5.9

Refer to Figure 5.21 and calculate the centroid coordinates $G(\bar{x}, \bar{y})$ of the shaded cross-sectional area with circular boundaries.

Figure 5.21: Circular Cross-Section — Figure 5.21: Shaded circular cross-section for Practice Problem 5.9.

Answers: $\bar{x} = 0.00\text{ mm}, \quad \bar{y} \approx 42.50\text{ mm}$

Component #	\(A_i\) (\(\text{in}^2\))	\(\bar{x}_i\) (\(\text{in}\))	\(\bar{y}_i\) (\(\text{in}\))	\(A_i \bar{x}_i\) (\(\text{in}^3\))	\(A_i \bar{y}_i\) (\(\text{in}^3\))
1. Rectangle \(OACE\)	\(72.00\)	\(3.00\)	\(6.00\)	\(216.00\)	\(432.00\)
2. Half-circle \(ABC\)	\(14.14\)	\(3.00\)	\(13.27\)	\(42.42\)	\(187.64\)
3. Triangle \(ECD\)	\(36.00\)	\(8.00\)	\(4.00\)	\(288.00\)	\(144.00\)
Total (\(\sum\))	\(A = 122.14\)	—	—	\(\sum A_i \bar{x}_i = 546.42\)	\(\sum A_i \bar{y}_i = 763.64\)

Component #	\(A_i\) (\(\text{in}^2\))	\(\bar{x}_i\) (\(\text{in}\))	\(\bar{y}_i\) (\(\text{in}\))	\(A_i \bar{x}_i\) (\(\text{in}^3\))	\(A_i \bar{y}_i\) (\(\text{in}^3\))
1. Rectangle \(OACE\)	\(72.00\)	\(3.00\)	\(6.00\)	\(216.00\)	\(432.00\)
2. Half-circle Hole \(ABC\)	\(-14.14\)	\(3.00\)	\(10.73\)	\(-42.42\)	\(-151.72\)
3. Triangle \(ECD\)	\(36.00\)	\(8.00\)	\(4.00\)	\(288.00\)	\(144.00\)
Total (\(\sum\))	\(A = 93.86\)	—	—	\(\sum A_i \bar{x}_i = 461.58\)	\(\sum A_i \bar{y}_i = 424.28\)

Component #	\(A_i\) (\(\text{in}^2\))	\(\bar{x}_i\) (\(\text{in}\))	\(\bar{y}_i\) (\(\text{in}\))	\(A_i \bar{x}_i\) (\(\text{in}^3\))	\(A_i \bar{y}_i\) (\(\text{in}^3\))
1. Parabolic Area \(A_1\)	\(72.00\)	\(4.50\)	\(10.80\)	\(324.00\)	\(777.60\)
2. Rectangle \(ACFB\)	\(108.00\)	\(7.50\)	\(18.00\)	\(810.00\)	\(1944.00\)
3. Quarter-Circle Hole	\(-0.79\)	\(8.58\)	\(0.42\)	\(-6.74\)	\(-0.33\)
Total (\(\sum\))	\(A = 179.21\)	—	—	\(\sum A_i \bar{x}_i = 1127.26\)	\(\sum A_i \bar{y}_i = 2721.27\)

Component #	\(A_i\) (\(\text{in}^2\))	\(\bar{x}_i\) (\(\text{in}\))	\(\bar{y}_i\) (\(\text{in}\))	\(A_i \bar{x}_i\) (\(\text{in}^3\))	\(A_i \bar{y}_i\) (\(\text{in}^3\))
1. Rectangle \(ODAB\)	\(216.00\)	\(3.00\)	\(18.00\)	\(648.00\)	\(3888.00\)
2. Parabolic Void (Hole)	\(-72.00\)	\(4.50\)	\(10.80\)	\(-324.00\)	\(-777.60\)
3. Semicircular Cutout (Hole)	\(-14.14\)	\(3.00\)	\(34.73\)	\(-42.42\)	\(-491.08\)
Total Net Area	\(A = 129.86\)	—	—	\(\sum A_i \bar{x}_i = 281.58\)	\(\sum A_i \bar{y}_i = 2619.32\)

Wire Segment	\(L_i\) (\(\text{cm}\))	\(\bar{x}_i\) (\(\text{cm}\))	\(L_i \bar{x}_i\) (\(\text{cm}^2\))
Segment \(AB\)	\(100.00\)	\(40.00\)	\(4000.00\)
Segment \(BC\)	\(80.00\)	\(40.00\)	\(3200.00\)
Segment \(CD\)	\(120.00\)	\(140.00\)	\(16800.00\)
Total (\(\sum\))	\(L = 300.00\)	—	\(\sum L_i \bar{x}_i = 24000.00\)

Component	\(V_i\) (\(\text{cm}^3\))	\(\bar{x}_i\) (\(\text{cm}\))	\(\bar{y}_i\) (\(\text{cm}\))	\(\bar{z}_i\) (\(\text{cm}\))	\(V_i \bar{x}_i\) (\(\text{cm}^4\))	\(V_i \bar{y}_i\) (\(\text{cm}^4\))	\(V_i \bar{z}_i\) (\(\text{cm}^4\))
Plate #1 (\(ABCD\))	\(19200\)	\(40.00\)	\(-2.00\)	\(34.00\)	\(768000\)	\(-38400\)	\(652800\)
Plate #2 (\(AJMB\))	\(12800\)	\(40.00\)	\(20.00\)	\(0.00\)	\(512000\)	\(256000\)	\(0\)
Total (\(\sum\))	\(V = 32000\)	—	—	—	\(\sum V_i \bar{x}_i = 1280000\)	\(\sum V_i \bar{y}_i = 217600\)	\(\sum V_i \bar{z}_i = 652800\)