Distributed Forces, Centroids, and Centers of Gravity - Theory & Concepts

Distributed Forces, Centroids, and Centers of Gravity - Theory & Concepts

In previous sections, we idealized forces as being concentrated at a single point. In reality, loads are often distributed over a line, an area, or a volume. To analyze these bodies effectively, we must find a single equivalent resultant force and determine its line of action.

Center of Gravity (CG)

The Earth exerts a gravitational force on every particle of a body. These forces form a parallel force system.

Center of Gravity

The Center of Gravity (CG) is a specific point through which the resultant weight of a body acts. The total weight

W

is the sum of the weights of all individual particles:

W = \int dW

To find the coordinates (

\bar{x}, \bar{y}, \bar{z}

) of the CG, we equate the moment of the resultant weight about any axis to the sum of the moments of the weights of all the particles about that same axis (Varignon's Theorem).

Coordinates of the Center of Gravity

Calculates the CG based on differential weight elements.

Center of Mass and Centroid

If a body exists in a uniform gravitational field (where

g

is constant everywhere), the Center of Gravity coincides with the Center of Mass.

Center of Mass Coordinate

Calculates the center of mass based on differential mass elements.

If the body is made of a homogeneous material (constant density

\rho

), the Center of Mass coincides with the geometric center of the body, called the Centroid.

Centroids of Lines, Areas, and Volumes

The centroid is a geometric property representing the "average" position of the figure's geometry.

Centroid of a Volume ( $V$ ): $\bar{x} = \frac{\int x \, dV}{\int dV}$
Centroid of an Area ( $A$ ): $\bar{x} = \frac{\int x \, dA}{\int dA}$
Centroid of a Line ( $L$ ): $\bar{x} = \frac{\int x \, dL}{\int dL}$

Note: (Similar equations apply for $\bar{y}$ and $\bar{z}$ .)

Important

Symmetry: If an area, volume, or line has an axis of symmetry, its centroid must lie somewhere on that axis. If it has two axes of symmetry, the centroid is located at their intersection. This significantly reduces calculation effort.

Composite Bodies

Many engineering structures are composed of simpler geometric shapes (rectangles, triangles, circles). The centroid of a composite body can be found without integration by breaking it down into these simple parts.

Centroid of a Composite Area

Calculates the centroid coordinates for a shape made of multiple simpler parts.

Variables

Symbol	Description	Unit
$\bar{X}, \bar{Y}$	Coordinates of the centroid of the entire composite shape	m
$\bar{x}_i, \bar{y}_i$	Centroid coordinates of each individual simple part	m
$A_i$	Area of each individual part	m^2

Note

Note: If a part is a "hole," its area $A_i$ is treated as negative in the summation.

Centroids of Composite Areas

A composite area consists of a series of simpler, standard shapes (e.g., rectangles, triangles, circles) that are connected together. By breaking down a complex area into these simple parts, we can find the overall centroid without using calculus.

Procedure

STEP-BY-STEP

Divide the Shape: Break the complex shape into simpler parts whose centroids are known from standard tables.
Establish Axes: Define an $x$ and $y$ coordinate system. The choice of origin is arbitrary, but placing it at a corner or along an axis of symmetry simplifies calculations.
Tabulate Properties: For each simple part $i$ , determine its individual area ( $A_i$ ) and the coordinates of its local centroid ( $\tilde{x}_i, \tilde{y}_i$ ) relative to the global origin. Create a table with columns for: Part, $A_i$ , $\tilde{x}_i$ , $\tilde{y}_i$ , $\tilde{x}_i A_i$ , and $\tilde{y}_i A_i$ .
Handle Holes (Cutouts):: If the composite shape has a hole or cutout, treat it as a part with a negative area. Its moment ( $A_i \tilde{x}_i$ ) will also be negative.
Calculate Resultants: Sum the individual areas to find the total area ( $\Sigma A$ ). Sum the individual moments to find the total moments ( $\Sigma \tilde{x}_i A_i$ and $\Sigma \tilde{y}_i A_i$ ).
Find Global Centroid: Divide the total moment by the total area to find the coordinates of the centroid for the entire composite shape: $\bar{x} = \frac{\Sigma \tilde{x}_i A_i}{\Sigma A}$ and $\bar{y} = \frac{\Sigma \tilde{y}_i A_i}{\Sigma A}$ .

Theorems of Pappus-Guldinus

The Theorems of Pappus and Guldinus provide a simple way to calculate the surface area and volume of a body of revolution (a shape created by rotating a curve or an area around an axis). These theorems relate the area/volume directly to the centroid of the generating curve/area.

First Theorem (Surface Area)

The surface area

A

of a surface of revolution is equal to the product of the length

L

of the generating curve and the distance traveled by the centroid of the curve during the rotation.

A = \theta \bar{r} L

Where:

$\theta$ is the angle of revolution in radians (for a full revolution, $\theta = 2\pi$ ).
$\bar{r}$ is the perpendicular distance from the axis of revolution to the centroid of the generating curve.
$L$ is the length of the generating curve.

Second Theorem (Volume)

The volume

V

of a solid of revolution is equal to the product of the generating area

A

and the distance traveled by the centroid of the area during the rotation.

V = \theta \bar{r} A

Where:

$\theta$ is the angle of revolution in radians.
$\bar{r}$ is the perpendicular distance from the axis of revolution to the centroid of the generating area.
$A$ is the generating area.

Note: The generating area must not cross the axis of revolution.

Distributed Loads on Beams

A common type of distributed force in structural analysis is a load distributed along the length of a beam, represented by a load function

w(x)

(units: Force/Length, e.g.,

\text{kN/m}

To solve equilibrium problems, this distributed load must be replaced by a single equivalent concentrated resultant force ( $F_R$ ).

Magnitude: The magnitude of

F_R

is equal to the total area under the load distribution diagram.

Equivalent Point Load Magnitude
Integrates the distributed load over the length.
Location: The line of action of $F_R$ passes through the centroid ( $\bar{x}$ ) of the area under the load diagram.
Equivalent Point Load Location
Finds the centroid of the distributed load area.

Fluid Statics (Distributed Loads on Surfaces)

A prime application of centroids and distributed loads in civil engineering is calculating the pressure exerted by a fluid at rest on submerged surfaces (e.g., dams, tanks).

Hydrostatic Pressure on Flat Surfaces

According to Pascal's Law, the pressure

p

at a depth

z

in a fluid with specific weight

\gamma

is:

p = \gamma z

The pressure increases linearly with depth, forming a triangular or trapezoidal load distribution on vertical or inclined submerged flat plates.

To find the resultant force

F_R

on a submerged area

A

F_R = \gamma \bar{z} A

Where

\bar{z}

is the depth to the centroid of the submerged area.

The point where this resultant force acts is called the Center of Pressure ( $z_p$ ). It is always located below the centroid because the pressure is higher at the bottom of the plate:

z_p = \bar{z} + \frac{\bar{I}_x}{\bar{z} A}

Where

\bar{I}_x

is the moment of inertia of the area about its centroidal axis.

Hydrostatic Pressure on Curved Surfaces

For curved surfaces, it is easier to resolve the resultant force into horizontal and vertical components:

Horizontal Component ( $F_H$ ): Equal to the resultant force on the vertical projection of the curved surface, acting at the center of pressure of that vertical projection.
Vertical Component ( $F_V$ ): Equal to the weight of the fluid volume directly above the curved surface extending up to the free surface. The line of action passes through the centroid of that fluid volume.

Composite Centroid Simulation

Use this simulation to visualize how changing the dimensions of a composite T-section affects the position of its centroid.

Composite Centroid Simulation

Flange Height ($h_1$)40 mm

Web Width ($w_2$)40 mm

Area 1 (Flange):8000 mm²

Area 2 (Web):6400 mm²

Total Area:14400 mm²

Notice how the centroid (Ȳ, red dot) shifts towards the part with the larger area.
Ȳ = (Σ y_i * A_i) / Σ A_i

Key Takeaways

The Center of Gravity is the point where the resultant weight acts. For uniform objects, it coincides with the Centroid (geometric center).
If a shape has an axis of symmetry, its centroid lies on that axis.
To find the centroid of a composite body, break it into simple shapes (rectangles, triangles, circles), calculate the area and centroid for each, and use the weighted average formula: $\bar{Y} = \frac{\Sigma \bar{y} A}{\Sigma A}$ .
Treat "holes" as negative areas when using the composite method.
A distributed load on a beam is replaced by an equivalent concentrated force whose magnitude equals the area under the load diagram, acting through the centroid of that area.

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Distributed Forces, Centroids, and Centers of Gravity - Theory & Concepts

Center of Gravity (CG)

Center of Gravity

Coordinates of the Center of Gravity

Center of Mass and Centroid

Center of Mass Coordinate

Centroids of Lines, Areas, and Volumes

Important

Composite Bodies

Centroid of a Composite Area

Note

Centroids of Composite Areas

Procedure

Theorems of Pappus-Guldinus

First Theorem (Surface Area)

Second Theorem (Volume)

Distributed Loads on Beams

Equivalent Point Load Magnitude

Equivalent Point Load Location

Fluid Statics (Distributed Loads on Surfaces)

Hydrostatic Pressure on Flat Surfaces

Hydrostatic Pressure on Curved Surfaces

Composite Centroid Simulation

Composite Centroid Simulation