7.3 Centroids, Center of Gravity & Moment of Inertia

6.3 Centroids, Center of Gravity & Moment of Inertia

Cross-sectional properties dictate the structural load capacity of mine shaft headframes, haulage rail tracks, dragline booms, and hydraulic cylinder casings. In the Coal India MT exam, evaluating centroids ($\bar{x}, \bar{y}$), section modulus ($Z$), polar moment of inertia ($J$), and radius of gyration ($k$) forms an essential mathematical foundation across Mechanics of Materials and Machine Design.

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1. Centroid vs. Center of Mass vs. Center of Gravity

It is vital to distinguish between these three closely related physical concepts:

Crucial Equivalence: For a homogeneous body ($\rho = \text{constant}$) situated in a uniform gravitational field ($g = \text{constant}$), the Centroid, COM, and COG coincide at the exact same spatial coordinates.

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2. Centroids of Standard Geometric Shapes

   Rectangle              Right Triangle               Semicircle
┌──────────────┐ b             ▲                          ┌──────────────┐
│      •       │ h             │ \                        │      • ȳ     │ r
└──────────────┘               │  \  h                    └──────┴───────┘
  x̄=b/2, ȳ=h/2                 └───►                      x̄ = r, ȳ = 4r/(3π)
                                b  x̄=b/3, ȳ=h/3

Reference Formulas for Standard Planar Profiles

First Moment of Area & Composite Centroid Formulation

For an assembly of $N$ simple sub-shapes with areas $A_i$ and centroidal coordinates $(\bar{x}_i, \bar{y}_i)$:

$$\bar{x} = \frac{\sum_{i=1}^{N} A_i \bar{x}_i}{\sum_{i=1}^{N} A_i}, \qquad \bar{y} = \frac{\sum_{i=1}^{N} A_i \bar{y}_i}{\sum_{i=1}^{N} A_i}$$

For cutout/subtracted regions (e.g., holes), treat the removed area as negative area ($-A_{\text{hole}}$).

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3. Theorems of Pappus-Guldinus

The Pappus-Guldinus theorems provide elegant geometric shortcuts for determining the surface areas and volumes of revolution.

Theorem 1: Surface Area of Revolution

The surface area $A_s$ generated by revolving a plane curve of length $L$ about a non-intersecting coplanar axis equals the product of the curve length $L$ and the distance traveled by its centroid during the revolution.

$$A_s = \theta \cdot \bar{r} \cdot L$$

For a complete $360^{\circ}$ revolution ($\theta = 2\pi$ radians):

$$A_s = 2\pi \bar{r} L$$

where $\bar{r}$ is the perpendicular distance from the curve centroid to the axis of revolution.

Theorem 2: Volume of Revolution

The volume $V$ generated by revolving a plane laminar area $A$ about a non-intersecting coplanar axis equals the product of the area $A$ and the distance traveled by its area centroid during the revolution.

$$V = \theta \cdot \bar{r} \cdot A$$

For a complete $360^{\circ}$ revolution ($\theta = 2\pi$ radians):

$$V = 2\pi \bar{r} A$$

Direct Application: Torus Properties

A torus is generated by revolving a circle of radius $r$ about an axis at distance $R$ ($R > r$):

• Surface Area: $A_s = (2\pi R) \times (2\pi r) = 4\pi^2 R r$
• Volume: $V = (2\pi R) \times (\pi r^2) = 2\pi^2 R r^2$

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4. Second Moment of Area (Area Moment of Inertia)

The Second Moment of Area quantifies the geometric resistance of a cross-section to bending deformation and deflection.

Definitions

• About $x$-axis: $I_{xx} = \int y^2 \, dA$
• About $y$-axis: $I_{yy} = \int x^2 \, dA$
• Product of Inertia: $I_{xy} = \int x y \, dA$ (Measures section asymmetry; $I_{xy} = 0$ if at least one axis is an axis of symmetry).

Moment of Inertia of Standard Cross-Sections

      Rectangle                  Triangle                   Circle
┌──────────────────┐ b             ▲                        ┌─────────┐
│                  │               │ \                      │    ┼    │ d
│────────G─────────│ h             │──G── h                 └─────────┘
│                  │               └──┴──►                      I_G = π*d⁴/64
└──────────────────┘                 b
   I_G = b*h³/12                    I_G = b*h³/36
   I_base = b*h³/3                  I_base = b*h³/12
                                    I_apex = b*h³/4

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5. Fundamental Moment of Inertia Theorems

1. Parallel Axis Theorem (Steiner's Theorem)

The moment of inertia of any planar area about an arbitrary axis $AB$ is equal to the moment of inertia about a parallel axis passing through its area centroid $G$ ($I_G$), PLUS the product of the total area $A$ and the square of the perpendicular distance $d$ separating the two parallel axes.

$$I_{AB} = I_G + A d^2$$

Mandatory Rule: The Parallel Axis Theorem can ONLY be applied directly from the centroidal axis $G$. If transferring between two non-centroidal parallel axes $1$ and $2$: $I_2 = (I_1 - A d_1^2) + A d_2^2 = I_G + A d_2^2$

2. Perpendicular Axis Theorem

For any flat two-dimensional planar laminar figure lying in the $xy$-plane, the moment of inertia about the perpendicular $z$-axis ($I_{zz}$) passing through the origin is equal to the sum of the moments of inertia about the two orthogonal coplanar axes $x$ and $y$ ($I_{xx}$ and $I_{yy}$).

$$I_{zz} = I_{xx} + I_{yy}$$

Restriction: Strictly valid only for 2D thin planar plates/areas, never for 3D volumetric bodies.

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6. Polar Moment of Inertia ($J$) & Radius of Gyration ($k$)

Polar Moment of Inertia ($J$ or $I_p$)

The polar moment of inertia characterizes a cross-section's resistance to torsional twisting and angular deflection under torque $T$:

$$J = I_{zz} = \int r^2 \, dA = I_{xx} + I_{yy}$$

• Solid Circular Shaft: $J = \frac{\pi d^4}{32} = \frac{\pi r^4}{2}$
• Hollow Circular Shaft: $J = \frac{\pi (D^4 - d^4)}{32}$

Radius of Gyration ($k$)

The radius of gyration represents the radial distance from the reference axis at which the entire area $A$ could be concentrated as a thin strip without changing its moment of inertia:

$$I = A k^2 \implies k = \sqrt{\frac{I}{A}}$$

• Rectangle ($b \times h$ about centroidal $x$-axis): $k_{xx} = \sqrt{\frac{b h^3 / 12}{b h}} = \frac{h}{\sqrt{12}} = \frac{h}{2\sqrt{3}}$
• Solid Circle (Radius $r$): $k = \sqrt{\frac{\pi r^4 / 4}{\pi r^2}} = \frac{r}{2} = \frac{d}{4}$

Application in Column Buckling: In Euler's column formula ($P_{\text{cr}} = \frac{\pi^2 E I}{L_e^2} = \frac{\pi^2 E A}{\lambda^2}$), the slenderness ratio is $\lambda = \frac{L_e}{k_{\text{min}}}$, where $k_{\text{min}} = \sqrt{I_{\text{min}} / A}$. Buckling always occurs about the axis of minimum moment of inertia.

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7. Step-by-Step Worked Engineering Calculations

Worked Example 6.3.1: Centroid and Moment of Inertia of an Asymmetric T-Section

Problem: A structural steel T-section consists of a top horizontal flange of $160\text{ mm} \times 20\text{ mm}$ and a vertical web of $20\text{ mm} \times 140\text{ mm}$ (Total height $H = 160\text{ mm}$). Calculate:

1. The location of the centroidal axis $\bar{y}$ measured from the bottom base.
2. The centroidal second moment of area $I_{xx,G}$.

                 ◄────── 160 mm ──────►
                 ┌────────────────────┐ ▲ 20 mm (Flange)
                 └────────┬──┬────────┘ ▼
                          │  │
                          │  │ 140 mm (Web)
                          │  │
                          └──┴── ▲ 20 mm Width
                          Datum (Base)

Step-by-step Solution:

1. Divide into Sub-areas & Establish Datum at Bottom Base:

   • Web (Part 1): $b_1 = 20\text{ mm}, h_1 = 140\text{ mm}$
     • $A_1 = 20 \times 140 = 2800\text{ mm}^2$
     • $y_1 = \frac{140}{2} = 70\text{ mm}$
   • Flange (Part 2): $b_2 = 160\text{ mm}, h_2 = 20\text{ mm}$
     • $A_2 = 160 \times 20 = 3200\text{ mm}^2$
     • $y_2 = 140 + \frac{20}{2} = 150\text{ mm}$

2. Calculate Centroid Location $\bar{y}$:

   $$\sum A = A_1 + A_2 = 2800 + 3200 = 6000\text{ mm}^2$$ $$\sum A y = (2800 \times 70) + (3200 \times 150) = 196{,}000 + 480{,}000 = 676{,}000\text{ mm}^3$$ $$\bar{y} = \frac{\sum A y}{\sum A} = \frac{676{,}000}{6000} = 112.67\text{ mm} \quad \text{(from bottom base)}$$

3. Calculate Distances from Centroidal Axis ($d_i = \lvert y_i - \bar{y} \rvert$):

   • $d_1 = \lvert 70 - 112.67 \rvert = 42.67\text{ mm}$
   • $d_2 = \lvert 150 - 112.67 \rvert = 37.33\text{ mm}$

4. Apply Parallel Axis Theorem for Each Component:

   • Web:

     $$I_{G1} = \frac{b_1 h_1^3}{12} = \frac{20 \times 140^3}{12} = \frac{20 \times 2{,}744{,}000}{12} = 4{,}573{,}333.33\text{ mm}^4$$ $$A_1 d_1^2 = 2800 \times (42.67)^2 = 2800 \times 1820.73 = 5{,}098{,}044.44\text{ mm}^4$$ $$I_{xx1} = 4{,}573{,}333.33 + 5{,}098{,}044.44 = 9{,}671{,}377.77\text{ mm}^4$$

   • Flange:

     $$I_{G2} = \frac{b_2 h_2^3}{12} = \frac{160 \times 20^3}{12} = \frac{160 \times 8000}{12} = 106{,}666.67\text{ mm}^4$$ $$A_2 d_2^2 = 3200 \times (37.33)^2 = 3200 \times 1393.53 = 4{,}459{,}296.00\text{ mm}^4$$ $$I_{xx2} = 106{,}666.67 + 4{,}459{,}296.00 = 4{,}565{,}962.67\text{ mm}^4$$

5. Total Centroidal Moment of Inertia $I_{xx,G}$:

   $$I_{xx,G} = I_{xx1} + I_{xx2} = 9{,}671{,}377.77 + 4{,}565{,}962.67 = 14{,}237{,}340.44\text{ mm}^4 \approx 14.24 \times 10^6\text{ mm}^4$$

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Worked Example 6.3.2: Volume and Surface Area via Pappus-Guldinus Theorem

Problem: A solid steel flywheel rim is formed by revolving a triangular cross-section (base $b = 60\text{ mm}$ parallel to the rotation axis, axial height $h = 40\text{ mm}$) through a full revolution of $360^{\circ}$ about the flywheel center shaft. The base of the triangle is located at a radial distance of $r_0 = 200\text{ mm}$ from the shaft centerline, with the apex pointing radially outward. Determine the exact volume of the rim.

Step-by-step Solution:

1. Calculate Area of Cross-Section:

   $$A = \frac{1}{2} b h = \frac{1}{2} \times 60 \times 40 = 1200\text{ mm}^2$$

2. Locate Radial Centroid $\bar{r}$ from Axis of Revolution: For a triangle, the centroid is at a distance of $h/3$ from its base:

   $$\bar{r} = r_0 + \frac{h}{3} = 200 + \frac{40}{3} = 200 + 13.33 = 213.33\text{ mm}$$

3. Apply Pappus-Guldinus Second Theorem ($V = 2\pi \bar{r} A$):

   $$V = 2\pi \times 213.333 \times 1200 = 2\pi \times 256{,}000 = 512{,}000\pi \approx 1{,}608{,}495.44\text{ mm}^3 \approx 1.6085 \times 10^{-3}\text{ m}^3$$