Unit 3 - Notes

MEC107 6 min read

Unit 3: Centroid and Moment of Inertia

1. Center of Gravity

The Center of Gravity (CG) is the point through which the entire weight of a body acts, regardless of the orientation of the body in space.

It is a property of mass distribution.
For a 3D body of total weight $W$ , the coordinates of the center of gravity $(\bar{x}, \bar{y}, \bar{z})$ are found using the principle of moments:
$\bar{x} = \frac{\int x \, dW}{W}, \quad \bar{y} = \frac{\int y \, dW}{W}, \quad \bar{z} = \frac{\int z \, dW}{W}$
In uniform gravitational fields, the center of gravity coincides with the center of mass.

2. Centroids of Areas and Lines

The Centroid is the geometric center of a shape (line, area, or volume). If a body is homogeneous (constant density and thickness), its center of gravity coincides with its centroid.

2.1 Centroid of Lines (1D)

For a curve or wire of total length $L$ :
$\bar{x} = \frac{\int x \, dL}{L}, \quad \bar{y} = \frac{\int y \, dL}{L}, \quad \bar{z} = \frac{\int z \, dL}{L}$

2.2 Centroid of Areas (2D)

For a plane area of total area $A$ :
$\bar{x} = \frac{\int x \, dA}{A}, \quad \bar{y} = \frac{\int y \, dA}{A}$

Axis of Symmetry: If an area has an axis of symmetry, the centroid must lie on that axis. If it has two axes of symmetry, the centroid is at their intersection.

3. First Moments of Areas and Lines

The First Moment is the integral of distance with respect to a geometric quantity (area, line, volume). It represents the numerator in the centroid formulas.

3.1 First Moment of Area

About the y-axis: $Q_y = \int x \, dA = \bar{x}A$
About the x-axis: $Q_x = \int y \, dA = \bar{y}A$
Key Property: The first moment of an area about any centroidal axis is exactly zero.

3.2 First Moment of a Line

About the y-axis: $Q_y = \int x \, dL = \bar{x}L$
About the x-axis: $Q_x = \int y \, dL = \bar{y}L$

4. Centroids of Composite Plates and Wires

Complex shapes can be broken down into standard, simple geometric shapes (rectangles, triangles, circles) whose centroids are known.

Methodology for Composite Shapes:

Divide the complex body into $n$ simple parts.
Establish a coordinate system (reference axes).
Calculate the area ( $A_i$ ) or length ( $L_i$ ) of each part. Note: "Holes" or cut-outs are treated as negative areas/lengths.
Determine the centroid coordinates $(\bar{x}_i, \bar{y}_i)$ of each individual part with respect to the reference axes.
Apply the summation formulas:

TEXT

For Composite Areas:
X_bar = (Σ A_i * x_i) / (Σ A_i)
Y_bar = (Σ A_i * y_i) / (Σ A_i)

For Composite Lines:
X_bar = (Σ L_i * x_i) / (Σ L_i)
Y_bar = (Σ L_i * y_i) / (Σ L_i)

5. Moment of Inertia of Plane Sections

The Moment of Inertia (Second Moment of Area) measures a section's resistance to bending or buckling. It is crucial in structural engineering.

About the x-axis: $I_{xx} = \int y^2 \, dA$
About the y-axis: $I_{yy} = \int x^2 \, dA$
Polar Moment of Inertia ( $J$ ): Resistance to torsion (twisting).
$J = \int r^2 \, dA = I_{xx} + I_{yy}$ (where $r^2 = x^2 + y^2$ )
Radius of Gyration ( $k$ ): The distance from the axis at which the entire area could be concentrated to yield the same moment of inertia.
$k_x = \sqrt{\frac{I_{xx}}{A}}, \quad k_y = \sqrt{\frac{I_{yy}}{A}}$

6. Theorems of Moment of Inertia

6.1 Parallel Axis Theorem (Transfer Theorem)

The moment of inertia of an area about any axis is equal to the moment of inertia about a parallel centroidal axis plus the product of the area and the square of the perpendicular distance ( $d$ ) between the two axes.
$I = \bar{I} + Ad^2$

$I$ = Moment of inertia about the desired axis.
$\bar{I}$ = Moment of inertia about the centroidal axis parallel to the desired axis.
$A$ = Area of the section.
$d$ = Distance between the two axes.

6.2 Perpendicular Axis Theorem

For a planar body, the polar moment of inertia about an axis perpendicular to the plane (z-axis) is equal to the sum of the moments of inertia about two mutually perpendicular axes in the plane (x and y axes) intersecting at the same point.
$I_{zz} = I_{xx} + I_{yy}$

7. Moment of Inertia of Standard and Composite Sections

Standard Sections (Centroidal Axes)

Rectangle (width , depth ):
- $I_{xx} = \frac{bd^3}{12}$
- $I_{yy} = \frac{db^3}{12}$
Triangle (base , height ):
- $I_{xx}$ (about centroid parallel to base) = $\frac{bh^3}{36}$
- $I_{base} = \frac{bh^3}{12}$
Circle (radius , diameter ):
- $I_{xx} = I_{yy} = \frac{\pi R^4}{4} = \frac{\pi D^4}{64}$
- $J = \frac{\pi R^4}{2} = \frac{\pi D^4}{32}$
Semicircle (radius ):
- $I_{xx}$ (about horizontal centroidal axis) $\approx 0.11R^4$
- $I_{yy}$ (about vertical axis of symmetry) = $\frac{\pi R^4}{8}$

Composite Sections

To find the moment of inertia of a composite section (e.g., I-beams, T-sections):

Divide the section into standard shapes.
Locate the global centroid of the composite shape.
Calculate the local moment of inertia ( $\bar{I}$ ) for each standard shape about its own centroidal axis.
Use the Parallel Axis Theorem ( $I_i = \bar{I}_i + A_i d_i^2$ ) to transfer each local moment of inertia to the global centroidal axis.
Sum them up: $I_{total} = \Sigma (\bar{I}_i + A_i d_i^2)$ . Cut-outs are subtracted.

8. Mass Moment of Inertia of Thin Plates

The Mass Moment of Inertia ( $I_m$ ) measures a body's resistance to angular acceleration (rotational inertia).
$I_m = \int r^2 \, dm$

For a thin plate of uniform thickness $t$ and density $\rho$ :

Mass element: $dm = \rho \, dV = \rho \, t \, dA$
Mass moment of inertia becomes directly proportional to the area moment of inertia:
$I_{m, x} = \int y^2 (\rho \, t \, dA) = \rho \, t \int y^2 dA = \rho \, t \, I_{area, x}$
$I_{m, y} = \rho \, t \, I_{area, y}$
$I_{m, z} = \rho \, t \, J_{area}$
Where $m = \rho t A$ is the total mass of the plate.
For example, for a rectangular thin plate (mass , sides and ):
- $I_{m, x} = \frac{1}{12} m b^2$
- $I_{m, y} = \frac{1}{12} m a^2$
- $I_{m, z} = \frac{1}{12} m (a^2 + b^2)$ (using the perpendicular axis theorem for mass: $I_{m,z} = I_{m,x} + I_{m,y}$ )

Unit 2

Unit 4