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CENTROID OF COMPLEX GEOMETRIC FIGURES:

Centroid of 1D and 2D Objects | Area, Lamina, Composite Figures & Moment of Inertia

 

Centroid of One-Dimensional and Two-Dimensional Bodies

Concept, Theory, Derivation, Applications and Advanced Cases

The concept of Centroid is one of the most important ideas in Engineering Mechanics, Strength of Materials, Structural Engineering, Machine Design and Fluid Mechanics.

Whenever we deal with beams, plates, laminae, sections, bending stresses, moments of inertia, or stability problems, the centroid becomes fundamental.

In this article, we shall discuss:

  1. One Dimensional (1D) Objects

  2. Two Dimensional (2D) Objects

  3. Concept of Resultant Area

  4. Mathematical Derivation of Centroid

  5. Varignon’s Theorem of Moments

  6. Centroid of Composite Figures

  7. Centroid of Curves and Lines

  8. Symmetry Conditions

  9. Centroid of Common Geometrical Shapes

  10. Negative Areas (Removed Portions)

  11. Centroid by Integration

  12. Applications of Centroid

  13. Relationship Between Centroid and Moment of Inertia

  14. Parallel Axis Theorem

  15. Practical Engineering Examples


 

1. One Dimensional (1D) Objects

A one-dimensional body possesses only length and negligible thickness.

Examples:

  • Thin wire

  • Metallic rod

  • Straight bar

  • Circular ring wire

  • Arc

 

centroid of a rod, centroid of a ring

Practically, a thin metallic rod is treated as a 1D object.

Suppose we have a straight rod of length (L).

If the rod is perfectly uniform, its centroid lies exactly at the midpoint.

For a rod extending from (x=0) to (x=L),

$x_g=\dfrac{L}{2}$

This point represents the balancing point of the rod.


2. Two Dimensional (2D) Objects

A two-dimensional body has length and breadth but negligible thickness.

Such objects are called:

  • Lamina

  • Thin plates

  • Plane areas

 

centroid of a 2D lamina

Examples:

  • Rectangular plate

  • Circular disc

  • Triangular sheet

  • Thin metal plates

Suppose we have a thin rectangular plate of dimensions:

  • Breadth = (b)

  • Height = (h)

Then its centroid lies at:

$x_g=\dfrac{b}{2},\quad y_g=\dfrac{h}{2}$


3. Physical Meaning of Centroid

The centroid is the geometrical center of an area.

It is the point through which the entire area may be assumed to act.

In simple words:

The centroid is the balancing point of a geometrical figure.

For uniform density materials:

  • Centroid = Center of Area

  • Center of Mass = Center of Gravity

all coincide.


4. Concept of Resultant Area

The concept of centroid is developed from the idea of the resultant of several elemental areas.

 

centroid of an irregular lamina

An area can be divided into infinitely small elemental parts.

Suppose a plane area (A) is divided into small elemental areas:

$a_1, a_2, a_3, \dots a_n$

Each elemental area possesses a position vector in the coordinate plane.

Let:

  • $(x_1, x_2, x_3,\dots x_n)$ be distances from the Y-axis

  • $(y_1, y_2, y_3,\dots y_n)$ be distances from the X-axis

Each elemental area produces a moment about the axes.


5. Moment of an Area

The moment of an elemental area about the Y-axis is:

$a_i x_i$

Similarly, moment about X-axis:

$a_i y_i$

The total moment about Y-axis becomes:

$a_1x_1 + a_2x_2 + a_3x_3 + \dots + a_nx_n$

Using sigma notation:

$\sum a_i x_i$

Similarly about X-axis:

$\sum a_i y_i$


6. Varignon’s Theorem of Moments

The entire theory of centroid is based on Varignon’s Theorem.

Statement

The moment of the resultant about any point is equal to the sum of moments of all component vectors about the same point.

Since areas behave like vector systems in moment calculations, the theorem becomes applicable.

Suppose the resultant area (A) passes through a point $G(X_g,Y_g)$.

Then:

Moment of resultant area about Y-axis:

$AX_g$

Moment about X-axis:

$AY_g$

By Varignon’s theorem:

$\sum a_i x_i = AX_g$

and,

$\sum a_i y_i = AY_g$

Since,

$A = \sum a_i$

therefore,


7. Fundamental Centroid Formula

X-coordinate of Centroid

$X_g=\dfrac{\sum a_i x_i}{\sum a_i}$

Y-coordinate of Centroid

$Y_g=\dfrac{\sum a_i y_i}{\sum a_i}$

These are the most fundamental equations of centroid calculations.


8. Continuous Form Using Integration

If the body is continuous, summation changes into integration.

Then:

For Area

$X_g=\dfrac{\int xdA}{\int dA}$

and,

$Y_g=\dfrac{\int ydA}{\int dA}$


9. Centroid of 1D Curves (Lines/Wires)

For wires or arcs:

$dA \rightarrow dL$

Hence,

$X_g=\dfrac{\int xdL}{\int dL}$

$Y_g=\dfrac{\int ydL}{\int dL}$

Used for:

  • Circular arcs

  • Wire frames

  • Curved rods

  • Bent structures


10. Centroid of Common Geometrical Figures


 

centroid for common geometrical plane

(a) Rectangle

Dimensions:

  • Breadth = (b)

  • Height = (h)

Centroid:

$x_g=\dfrac{b}{2},\quad y_g=\dfrac{h}{2}$


(b) Triangle

For a triangle of height (h):

Centroid lies at:

$y_g=\dfrac{h}{3}$

from the base.

Or,

$\dfrac{2h}{3}$

from the vertex.


(c) Semicircle

For a semicircular area of radius (R):

$y_g=\dfrac{4R}{3\pi}$

from the diameter.


(d) Quarter Circle

$x_g=y_g=\dfrac{4R}{3\pi}$


(e) Circle

By symmetry:

$x_g = 0,\quad y_g = 0$

if origin is at center.


11. Symmetry Conditions

Symmetry greatly simplifies centroid problems.

Case 1: Single Axis Symmetry

If a figure is symmetric about one axis:

 

centroid of symmetrical lamina

  • centroid lies on that axis.

Case 2: Double Symmetry

If symmetric about both X and Y axes:

  • centroid lies at their intersection.

Examples:

  • Circle

  • Rectangle

  • Square


12. Composite Bodies

 

centriod of asymmetrical composite area

Most engineering structures are not simple shapes.

They are combinations of:

  • rectangles

  • circles

  • triangles

  • semicircles

  • holes

Such bodies are called Composite Figures.


13. Algorithm to Find Centroid of Composite Figures


Step 1

Take the complex geometric figure.


Step 2

Divide it into basic standard geometrical shapes.


Step 3

Choose a convenient coordinate system.

All measurements must be based on this frame.


Step 4

For every elemental area determine:

  • Area $(a_i)$

  • Centroid coordinates $(x_i,y_i)$


Step 5

Assign signs carefully.

Positive Area

If area is added.

Negative Area

If area is removed (holes/cutouts).


Step 6

Use centroid equations:

$X_g=\dfrac{\sum a_i x_i}{\sum a_i}$

$Y_g=\dfrac{\sum a_i y_i}{\sum a_i}$


14. Negative Area Concept

Suppose:

 

centroid by calculating negative area

  • a circular hole is removed from a plate

  • a triangular notch is cut

Then removed portions are treated as:

$-a_i$

This is extremely important in engineering design.


15. Example of Composite Area

Suppose a rectangular plate contains:

  • one circular hole

  • one triangular cut

Then:

$A = A_1 - A_2 - A_3$

where:

  • (A_1) → rectangle

  • (A_2) → circular hole

  • (A_3) → triangular cut

Then:

$X_g = \dfrac{A_1x_1 - A_2x_2 - A_3x_3}{A_1-A_2-A_3}$

Similarly for (Y_g).


16. Centroid of Three-Dimensional Bodies

 

centroid of a cone

For volumes:

$dA \rightarrow dV$

Then centroid becomes center of volume.

Coordinates become:

$X_g=\dfrac{\int xdV}{\int dV}$

$Y_g=\dfrac{\int ydV}{\int dV}$

$Z_g=\dfrac{\int zdV}{\int dV}$

Used in:

  • solid mechanics

  • fluid mechanics

  • ship stability

  • aerospace structures


17. Applications of Centroid

 

visualization of bending of a beam

Centroid is used in:

  • Beam design

  • Structural engineering

  • Machine design

  • Aircraft design

  • Stability analysis

  • Fluid pressure problems

  • Bending stress analysis

  • Area moment of inertia

  • Shear stress distribution


18. Relation Between Centroid and Moment of Inertia

The Moment of Inertia of Area is also called:

Second Moment of Area

It measures resistance against bending.


First Moment of Area

Centroid uses:

$\int xdA$

which is called the first moment of area.


Second Moment of Area

Moment of inertia uses:

$\int x^2dA$

or,

$\int y^2dA$

which is the second moment of area.


19. Centroidal Moment of Inertia

Moment of inertia about centroidal axes is minimum.

Common formulas:

Rectangle

$I_{xx}=\dfrac{bh^3}{12}$

$I_{yy}=\dfrac{hb^3}{12}$


Circle

$I=\dfrac{\pi R^4}{4}$

about diameter.


20. Parallel Axis Theorem

 

parallel axis theorem

Very important theorem in mechanics.

Statement

Moment of inertia about any axis equals the centroidal moment of inertia plus area times square of distance between axes.

Mathematically:

$I=I_G+Ad^2$

where:

  • $(I_G)$ = centroidal moment of inertia

  • (A) = area

  • (d) = distance between axes


21. Why Centroid is Important Before MOI

Before finding moment of inertia of a complex section:

  1. First find centroid

  2. Locate centroidal axes

  3. Compute centroidal MOI

  4. Transfer using parallel axis theorem

Thus centroid acts as the foundation for advanced structural analysis.


22. Engineering Insight

In engineering mechanics:

  • Forces produce translational effects

  • Moments produce rotational effects

  • Areas produce distributed effects

The centroid represents the effective point through which the entire geometrical effect acts.

That is why centroid becomes one of the most fundamental ideas in mechanics and structural engineering.


23. Final Conclusion

The centroid is the geometrical center of a body obtained from the resultant of all elemental areas. The entire theory is based upon the principles of moments and Varignon’s theorem.

For simple geometries, centroid is obtained directly from symmetry. For complex composite figures, the body is divided into smaller standard figures and the algebraic summation method is used.

The study of centroid forms the basis of:

  • Area moments

  • Structural analysis

  • Beam bending

  • Stability

  • Machine design

  • Fluid mechanics

  • Advanced mechanics of materials

Thus, centroid is not merely a geometrical concept — it is one of the core foundations of engineering science.