Moments of Inertia - Theory & Concepts

Learning Objectives

  • Define the moment of inertia and product of inertia for an area.
  • Apply integration techniques to determine moments of inertia.
  • Understand the concept of principal moments of inertia and how to find them using Mohr's Circle.
  • Utilize the Parallel-Axis Theorem to calculate the moment of inertia for composite areas.
  • Define and apply the concepts of polar moment of inertia and radius of gyration.

Introduction

While the centroid defines the "center" of a cross-section, the Moment of Inertia (Area Moment of Inertia) is a geometric property that quantifies how that area is distributed relative to an axis. It is a fundamental property used to calculate a structural member's resistance to bending and buckling.

Definition of Moment of Inertia

Moment of Inertia of an Area

The Moment of Inertia of an area AA with respect to the xx and yy axes is defined by its distance squared integrals.

Moment of Inertia Integrals

The basic integral equations for calculating the area moment of inertia.

Ix=Ay2dAI_x = \int_A y^2 \, dAIy=Ax2dAI_y = \int_A x^2 \, dA

Variables

SymbolDescriptionUnit
IxI_xMoment of inertia about the x-axism4\text{m}^4
IyI_yMoment of inertia about the y-axism4\text{m}^4
dAdAInfinitesimal element of aream2\text{m}^2
x,yx, yPerpendicular distances from the respective axes to the element dAm

Always Positive

Because the distance is squared (x2,y2x^2, y^2), the moment of inertia is always positive and has units of length to the fourth power (e.g., m4,mm4,in4\text{m}^4, \text{mm}^4, \text{in}^4). The further the area is located from the axis, the larger the moment of inertia.

Product of Inertia

The product of inertia of an area AA with respect to the xx and yy axes is defined by the integral of xydAxy \, dA.

Product of Inertia Integral

The basic integral equation for calculating the product of inertia.

Ixy=AxydAI_{xy} = \int_A xy \, dA

Variables

SymbolDescriptionUnit
IxyI_{xy}Product of inertia with respect to the x and y axesm4\text{m}^4
x,yx, yCoordinates of the differential area elementm
dAdAInfinitesimal element of aream2\text{m}^2

Product of Inertia Characteristics

Unlike IxI_x and IyI_y, the product of inertia can be positive, negative, or zero. It is zero if either the xx or yy axis is an axis of symmetry for the area. This property is crucial in determining the principal axes of inertia, which correspond to the maximum and minimum moments of inertia.

Mass Moment of Inertia

While the area moment of inertia relates to the cross-section's resistance to bending, the Mass Moment of Inertia relates to a solid body's resistance to rotational acceleration (Dynamics).

Mass Moment of Inertia Integral

The basic integral equation for calculating the mass moment of inertia.

I=mr2dmI = \int_m r^2 \, dm

Variables

SymbolDescriptionUnit
IIMass moment of inertiakgm2\text{kg}\cdot\text{m}^2
rrPerpendicular distance from the axis of rotation to the infinitesimal massm
dmdmInfinitesimal element of masskg

Integration Techniques for II

When calculating the Moment of Inertia for a shape defined by a mathematical function (e.g., y=f(x)y = f(x)), direct integration is necessary.

  • Horizontal Strip Method: Best for calculating IxI_x. Use a differential area dA=xdydA = x \, dy located at a uniform distance yy from the x-axis. Ix=y2(xdy)I_x = \int y^2 \, (x \, dy)
  • Vertical Strip Method: Best for calculating IyI_y. Use a differential area dA=ydxdA = y \, dx located at a uniform distance xx from the y-axis. Iy=x2(ydx)I_y = \int x^2 \, (y \, dx)

Principal Moments of Inertia

Principal Axes

The moments of inertia (Ix,IyI_x, I_y) and the product of inertia (IxyI_{xy}) of an area depend on the orientation of the xx and yy axes. As the axes are rotated about the origin, these values change continuously.

There exists a specific orientation of the axes where the product of inertia (IxyI_{xy}) is exactly zero. The axes at this orientation are called the Principal Axes of Inertia. The moments of inertia about these axes are called the Principal Moments of Inertia, and they represent the maximum (ImaxI_{\text{max}}) and minimum (IminI_{\text{min}}) possible moments of inertia for the area about that origin.

Equations for Principal Moments

Mohr's Circle equations for finding principal moments of inertia and their orientation.

Imax,Imin=Ix+Iy2±(IxIy2)2+Ixy2I_{\text{max}}, I_{\text{min}} = \frac{I_x + I_y}{2} \pm \sqrt{\left(\frac{I_x - I_y}{2}\right)^2 + I_{xy}^2}tan(2θp)=2IxyIxIy\tan(2\theta_p) = \frac{-2I_{xy}}{I_x - I_y}

Variables

SymbolDescriptionUnit
Imax,IminI_{\text{max}}, I_{\text{min}}Maximum and minimum principal moments of inertiam4\text{m}^4
Ix,IyI_x, I_yMoments of inertia about the x and y axesm4\text{m}^4
IxyI_{xy}Product of inertiam4\text{m}^4
θp\theta_pAngle of orientation for the principal axesrad\text{rad}

Symmetry in Principal Axes

If an area has an axis of symmetry, that axis and any axis perpendicular to it are principal axes, meaning Ixy=0I_{xy} = 0 for those axes. Finding the principal moments of inertia is critical for analyzing asymmetric bending in beams and determining the critical buckling axis for columns.

Polar Moment of Inertia

Polar Moment Overview

While IxI_x and IyI_y relate to bending resistance (flexure), the Polar Moment of Inertia (JOJ_O) relates to a member's resistance to twisting (torsion).

It is defined as the integral of the area multiplied by the square of the radial distance rr from the pole (origin OO):

Polar Moment of Inertia Definition

Integral definition of polar moment of inertia.

JO=Ar2dAJ_O = \int_A r^2 \, dA

Variables

SymbolDescriptionUnit
JOJ_OPolar moment of inertia about the originm4\text{m}^4
rrRadial distance from the origin to the element dAm
dAdAInfinitesimal element of aream2\text{m}^2

Relation to Rectangular Moments

Since r2=x2+y2r^2 = x^2 + y^2, the polar moment of inertia relates to the rectangular moments of inertia by the sum of IxI_x and IyI_y.

Polar Moment of Inertia Sum

Relation of polar moment of inertia to rectangular components.

JO=A(x2+y2)dA=Ix+IyJ_O = \int_A (x^2 + y^2) \, dA = I_x + I_y

Variables

SymbolDescriptionUnit
JOJ_OPolar moment of inertiam4\text{m}^4
IxI_xMoment of inertia about the x-axism4\text{m}^4
IyI_yMoment of inertia about the y-axism4\text{m}^4

Parallel-Axis Theorem

Theorem Introduction

This is perhaps the most frequently used theorem in structural analysis when calculating the moment of inertia of complex, composite shapes.

Parallel-Axis Theorem

If the moment of inertia of an area about its own centroidal axis (Iˉ\bar{I}) is known, the moment of inertia (II) about any other parallel axis can be found using the parallel-axis theorem.

Parallel-Axis Theorem

Calculates the moment of inertia about any axis parallel to the centroidal axis.

I=Iˉ+Ad2I = \bar{I} + Ad^2

Variables

SymbolDescriptionUnit
IIMoment of inertia about the parallel axism4\text{m}^4
Iˉ\bar{I}Moment of inertia about the centroidal axism4\text{m}^4
AATotal area of the shapem2\text{m}^2
ddPerpendicular distance between the centroidal axis and the parallel axism

Parallel-Axis Theorem for Product of Inertia

Just as the parallel-axis theorem applies to moments of inertia, it also applies to the product of inertia.

Parallel-Axis Theorem for Product of Inertia

Calculates the product of inertia for any set of parallel axes.

Ixy=Iˉxy+AxˉyˉI_{xy} = \bar{I}_{xy} + A \bar{x} \bar{y}

Variables

SymbolDescriptionUnit
IxyI_{xy}Product of inertia about the parallel x and y axesm4\text{m}^4
Iˉxy\bar{I}_{xy}Product of inertia about the centroidal axes parallel to the x and y axesm4\text{m}^4
AATotal area of the shapem2\text{m}^2
xˉ,yˉ\bar{x}, \bar{y}Perpendicular distances from the y and x axes to the centroid of the area (with signs)m

Radius of Gyration

Radius of Gyration Applications

The radius of gyration of an area is a geometric property used frequently in the design of columns in structural mechanics to prevent buckling. It describes the distribution of the cross-sectional area around its centroidal axis.

Radius of Gyration (kk)

The distance from the reference axis at which the entire area could be concentrated such that it would yield the exact same moment of inertia as the original distributed area.

Radius of Gyration Formulas

Equations to compute radius of gyration.

kx=IxAky=IyAkO=JOA\begin{aligned} k_x &= \sqrt{\frac{I_x}{A}} \\ k_y &= \sqrt{\frac{I_y}{A}} \\ k_O &= \sqrt{\frac{J_O}{A}} \end{aligned}

Variables

SymbolDescriptionUnit
kx,kyk_x, k_yRadius of gyration about the x and y axesm
kOk_OPolar radius of gyrationm
Ix,IyI_x, I_yMoments of inertia about the x and y axesm4\text{m}^4
JOJ_OPolar moment of inertiam4\text{m}^4
AATotal cross-sectional aream2\text{m}^2

Moment of Inertia of Composite Areas

Composite Shapes Overview

Just like centroids, the moment of inertia of a composite shape can be found by adding or subtracting the moments of inertia of its simple constituent parts.

Interactive Simulation

Interact with the simulation below to see how calculating properties works for basic geometric forms.

Parallel-Axis Theorem

Adjust the dimensions and distance of the rectangle to see how the moment of inertia changes relative to the reference axis.

I_bar (bh³/12):6666.7 cm⁴
Area (bh):200.0 cm²
Ad² term:45000.0 cm⁴
Total I:51666.7 cm⁴
Ref Xx'd

Steps to Calculate Composite $I$

  1. Divide the composite area into simple geometric shapes (rectangles, triangles, circles).
  2. Locate the centroid of each simple shape and determine the centroid of the entire composite area.
  3. Establish the reference axis (usually the centroidal axis of the composite shape).
  4. For each simple shape, calculate its moment of inertia about its own centroidal axis (Iˉ\bar{I}).
  5. Use the Parallel-Axis Theorem to transfer each Iˉ\bar{I} to the composite reference axis (I=Iˉ+Ad2I = \bar{I} + Ad^2).
  6. Sum the transferred moments of inertia: Itotal=Σ(Ipart)I_{\text{total}} = \Sigma (I_{\text{part}}).

Note: For a "hole", its II value must be subtracted from the total.

Mohr's Circle for Moments of Inertia

Mohr's Circle Graphical Method

Mohr's Circle provides a graphical method for determining principal moments of inertia and the orientation of the principal axes. It relies on the transformation equations for IxI_x, IyI_y, and IxyI_{xy}.

Constructing Mohr's Circle

  1. Plot the points (Ix,Ixy)(I_x, I_{xy}) and (Iy,Ixy)(I_y, -I_{xy}) on a coordinate system where the horizontal axis represents moments of inertia (II) and the vertical axis represents products of inertia (IxyI_{xy}).
  2. Connect the two points with a straight line. The intersection of this line with the horizontal axis is the center of the circle, located at Iavg=Ix+Iy2I_{\text{avg}} = \frac{I_x + I_y}{2}.
  3. The radius of the circle is R=(IxIy2)2+Ixy2R = \sqrt{\left(\frac{I_x - I_y}{2}\right)^2 + I_{xy}^2}.
  4. The principal moments of inertia are the points where the circle intersects the horizontal axis: Imax=Iavg+RI_{\text{max}} = I_{\text{avg}} + R and Imin=IavgRI_{\text{min}} = I_{\text{avg}} - R.
Key Takeaways
  • The Moment of Inertia (II) measures an area's distribution about an axis and indicates resistance to bending.
  • The Polar Moment of Inertia (JOJ_O) relates to resistance to torsion (twisting) and equals Ix+IyI_x + I_y.
  • The Parallel-Axis Theorem (I=Iˉ+Ad2I = \bar{I} + Ad^2) is essential for finding the moment of inertia about an axis parallel to a centroidal axis.
  • The Principal Moments of Inertia represent the maximum and minimum II values, occurring about axes where the product of inertia Ixy=0I_{xy} = 0.
  • To find II for a composite shape, divide it into simple parts, calculate Iˉ\bar{I} and Ad2Ad^2 for each part relative to the common axis, and sum them up.
  • The Radius of Gyration (kk) is a geometric property used primarily in column buckling analysis.