Distributed Forces, Centroids, and Centers of Gravity - Theory & Concepts
Learning Objectives
- Understand the definitions and differences between the Center of Gravity, Center of Mass, and Centroid.
- Calculate the centroid of lines, areas, and volumes using integration.
- Determine the centroid of composite areas and bodies using simple geometric shapes.
- Apply the Theorems of Pappus-Guldinus to find the surface area and volume of bodies of revolution.
- Convert distributed loads on beams into equivalent concentrated resultant forces.
- Calculate hydrostatic pressure and resultant forces on flat and curved submerged surfaces.
Introduction to Distributed Forces
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
The Center of Gravity (CG) is a specific point through which the resultant weight of a body acts. The total weight is the sum of the weights of all individual particles: .
Center of Gravity (CG)
The Earth exerts a gravitational force on every particle of a body. These forces form a parallel force system.
To find the coordinates () 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.
Variables
| Symbol | Description | Unit |
|---|---|---|
| Coordinates of the center of gravity | m | |
| Coordinates of the differential weight element | m | |
| Differential weight element | N |
Center of Mass
If a body exists in a uniform gravitational field (where 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.
Variables
| Symbol | Description | Unit |
|---|---|---|
| x-coordinate of the center of mass | m | |
| x-coordinate of the differential mass element | m | |
| Differential mass element | kg |
Centroid
If the body is made of a homogeneous material (constant density ), the Center of Mass coincides with the geometric center of the body, called the Centroid. The centroid is a geometric property representing the "average" position of the figure's geometry.
Centroids of Lines, Areas, and Volumes
The formulas for the centroid of a line, area, or volume are derived similarly to the center of mass, substituting the appropriate differential geometric property for mass.
Note: (Similar equations apply for and .)
Centroids of Geometric Properties
Calculates the centroid based on differential volume, area, or line elements.
Variables
| Symbol | Description | Unit |
|---|---|---|
| x-coordinate of the centroid for volume, area, and line | m | |
| x-coordinate of the differential element | m | |
| Differential volume element | ||
| Differential area element | ||
| Differential line element | m |
Symmetry
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 |
|---|---|---|
| Coordinates of the centroid of the entire composite shape | ||
| Centroid coordinates of each individual simple part | ||
| Area of each individual part |
Holes in Composite Areas
Note: If a part is a "hole," its area 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.
Calculating Centroids of Composite Areas
- Divide the Shape: Break the complex shape into simpler parts whose centroids are known from standard tables.
- Establish Axes: Define an and 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 , determine its individual area () and the coordinates of its local centroid () relative to the global origin. Create a table with columns for: Part, , , , , and .
- Handle Holes (Cutouts): If the composite shape has a hole or cutout, treat it as a part with a negative area. Its moment () will also be negative.
- Calculate Resultants: Sum the individual areas to find the total area (). Sum the individual moments to find the total moments ( and ).
- Find Global Centroid: Divide the total moment by the total area to find the coordinates of the centroid for the entire composite shape: and .
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 of a surface of revolution is equal to the product of the length of the generating curve and the distance traveled by the centroid of the curve during the rotation.
First Theorem of Pappus-Guldinus
Calculates the surface area of a surface of revolution.
Variables
| Symbol | Description | Unit |
|---|---|---|
| Surface area of the surface of revolution | ||
| Angle of revolution | ||
| Perpendicular distance from the axis of revolution to the centroid of the generating curve | m | |
| Length of the generating curve | m |
Second Theorem (Volume)
The volume of a solid of revolution is equal to the product of the generating area and the distance traveled by the centroid of the area during the rotation.
Second Theorem of Pappus-Guldinus
Calculates the volume of a solid of revolution.
Variables
| Symbol | Description | Unit |
|---|---|---|
| Volume of the solid of revolution | ||
| Angle of revolution | ||
| Perpendicular distance from the axis of revolution to the centroid of the generating area | m | |
| Generating area |
Axis Intersection Caution
The generating area must not cross the axis of revolution when applying the Second Theorem of Pappus-Guldinus.
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 (units: Force/Length, e.g., ).
To solve equilibrium problems, this distributed load must be replaced by a single equivalent concentrated resultant force ().
Magnitude: The magnitude of is equal to the total area under the load distribution diagram.
Equivalent Point Load Magnitude
Integrates the distributed load over the length.
Variables
| Symbol | Description | Unit |
|---|---|---|
| Equivalent concentrated resultant force | N | |
| Length of the beam over which the load is distributed | m | |
| Distributed load function | ||
| Differential length element | m |
Equivalent Point Load Location
Location: The line of action of passes through the centroid () of the area under the load diagram.
Equivalent Point Load Location
Finds the centroid of the distributed load area.
Variables
| Symbol | Description | Unit |
|---|---|---|
| Location of the equivalent concentrated resultant force | m | |
| Distance from the origin | m | |
| Length of the beam over which the load is distributed | m | |
| Distributed load function | ||
| Differential length element | m |
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 at a depth in a fluid with specific weight is calculated by integrating the pressure over the area.
The pressure increases linearly with depth, forming a triangular or trapezoidal load distribution on vertical or inclined submerged flat plates. The point where this resultant force acts is called the Center of Pressure (). It is always located below the centroid because the pressure is higher at the bottom of the plate.
Hydrostatic Pressure and Force Equations
Calculates pressure, resultant force, and center of pressure for flat submerged surfaces.
Variables
| Symbol | Description | Unit |
|---|---|---|
| Hydrostatic pressure at depth z | ||
| Specific weight of the fluid | ||
| Depth in the fluid | m | |
| Resultant hydrostatic force | N | |
| Depth to the centroid of the submerged area | m | |
| Submerged area | ||
| Center of pressure depth | m | |
| 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 (): 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 (): 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.
Interactive Simulation
Use this simulation to visualize how changing the dimensions of a composite T-section affects the position of its centroid.
Composite Centroid Simulation
Ȳ = (Σ y_i * A_i) / Σ A_i
- 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: .
- 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.