Engineering Mechanics: Statics  ·  Chapter 10

Moments of Inertia

CHAPTER 10  ·  Engineering Mechanics: Statics

The design of structural members — beams, columns, and shafts — depends critically on how their cross-sectional areas resist bending and rotation. The moment of inertia of an area is the geometric property that governs this resistance, appearing throughout fluid mechanics, mechanics of materials, and structural design.

10.1 Definition of Moments of Inertia for Areas

When a distributed load acts perpendicular to an area with intensity varying linearly with depth, the moment of that loading about an axis involves an integral of the form \(\int y^2\,dA\). This integral is called the moment of inertia of the area (also the second moment of the area), and it arises pervasively in structural and mechanical formulas.

Moment of Inertia of an Area

The moments of inertia of a differential area element \(dA\) about the \(x\) and \(y\) axes are \(dI_x = y^2\,dA\) and \(dI_y = x^2\,dA\). For the entire area \(A\), integration yields \(I_x\) and \(I_y\). These quantities are always positive because they involve distance squared times area, and carry units of length to the fourth power (m⁴, mm⁴, ft⁴, in⁴).

Area Moments of Inertia
\[I_x = \int_A y^2\,dA \qquad I_y = \int_A x^2\,dA\]

Polar Moment of Inertia

The polar moment of inertia \(J_O\) is defined about the pole (z-axis) passing through point O. Each area element contributes \(r^2\,dA\), where \(r\) is the perpendicular distance from the z-axis to \(dA\). Because \(r^2 = x^2 + y^2\), the polar moment equals the sum of the two rectangular moments of inertia.

Polar Moment of Inertia
\[J_O = \int_A r^2\,dA = I_x + I_y\]
y x x y dA O r
Area element \(dA\) located at coordinates \((x,y)\) from origin O
Fig. 10–2

10.2 Parallel-Axis Theorem for an Area

The parallel-axis theorem provides a way to find the moment of inertia about any axis, given the centroidal moment of inertia and the perpendicular distance between the two axes.

Parallel-Axis Theorem

If \(\bar{I}\) is the moment of inertia of an area about a centroidal axis, then the moment of inertia \(I\) about any parallel axis equals \(\bar{I}\) plus the product of the total area \(A\) and the square of the distance \(d\) between the two parallel axes. The second integral \(\int y'\,dA\) vanishes because the primed axis passes through the centroid, where \(\bar{y}=0\).

Parallel-Axis Theorem
\[I_x = \bar{I}_{x'} + A\,d_y^2 \qquad I_y = \bar{I}_{y'} + A\,d_x^2 \qquad J_O = \bar{J}_C + A\,d^2\]
x' x d_y C dA
Distance \(d_y\) between the reference \(x\)-axis and the centroidal \(x'\)-axis
Fig. 10–3

10.3 Radius of Gyration of an Area

Radius of Gyration

The radius of gyration \(k\) is a length that condenses the moment of inertia and area of a cross-section into a single dimension. It is widely used in column design in structural mechanics. The analogy is direct: just as \(dI_x = y^2\,dA\) for a differential element, the integrated result gives \(I_x = k_x^2 A\).

Radii of Gyration
\[k_x = \sqrt{\frac{I_x}{A}} \qquad k_y = \sqrt{\frac{I_y}{A}} \qquad k_O = \sqrt{\frac{J_O}{A}}\]

Key Points

  • Moment of inertia is a geometric property of an area, not a physical mass property. It measures resistance to bending or the location of a resultant pressure force.
  • Because it involves distance squared times area, \(I_x\), \(I_y\), and \(J_O\) are always positive.
  • Units are always length to the fourth power: m⁴, mm⁴, ft⁴, or in⁴.
  • If the centroidal moment of inertia is known, the parallel-axis theorem gives the moment about any parallel axis immediately.

Procedure for Analysis — Single Integration

  1. Express the boundary curve as \(y = f(x)\) and select a rectangular differential element with finite length and differential width.
  2. Case 1 (element parallel to axis): Orient the element so its length is parallel to the axis of interest. Every point of the element is at the same moment-arm distance. Apply \(I = \int r^2\,dA\) directly.
  3. Case 2 (element perpendicular to axis): Not all points share the same moment arm, so first determine the centroidal moment of inertia of the element, then transfer to the reference axis using the parallel-axis theorem.

10.4 Moments of Inertia for Composite Areas

Composite Area Method

Any cross-section that can be decomposed into rectangles, triangles, circles, and similar standard shapes is a composite area. Because moment of inertia is an integral (hence additive), the total moment of inertia equals the algebraic sum of the contributions from all parts. Areas representing holes or cutouts are subtracted.

Procedure for Composite Areas

  1. Sketch and decompose: Divide the cross-section into simple parts. Mark the centroid location of each part and the perpendicular distance to the reference axis.
  2. Apply the parallel-axis theorem: For each part whose centroidal axis does not coincide with the reference axis, compute \(I = \bar{I} + A\,d^2\). Use tabulated \(\bar{I}\) values.
  3. Sum algebraically: Add contributions from all solid parts and subtract contributions from any holes.

10.5 Product of Inertia for an Area

Product of Inertia

The product of inertia \(I_{xy}\) is needed to find the axes about which the moment of inertia is maximum or minimum. Unlike \(I_x\) and \(I_y\), the product of inertia can be positive, negative, or zero. It is zero whenever either the \(x\) or the \(y\) axis is an axis of symmetry of the area, because symmetric element pairs contribute equal and opposite products that cancel.

Product of Inertia
\[I_{xy} = \int_A xy\,dA\]
Parallel-Axis Theorem for Product of Inertia
\[I_{xy} = \bar{I}_{x'y'} + A\,d_x\,d_y\]

Here \(d_x\) and \(d_y\) are the signed distances from the centroid to the reference \(y\)- and \(x\)-axes respectively. Maintaining the correct algebraic signs for \(d_x\) and \(d_y\) is essential.

10.6 Moments of Inertia for an Area about Inclined Axes

When axes \(u\) and \(v\) are rotated by an angle \(\theta\) from the \(x\)–\(y\) axes, the moments and product of inertia transform via the following relations. These are derived by substituting the rotation of coordinates \(u = x\cos\theta + y\sin\theta\), \(v = y\cos\theta - x\sin\theta\) into the definitions and applying double-angle identities.

Transformation Equations for Moments of Inertia
\[I_u = \frac{I_x+I_y}{2} + \frac{I_x-I_y}{2}\cos 2\theta - I_{xy}\sin 2\theta\] \[I_v = \frac{I_x+I_y}{2} - \frac{I_x-I_y}{2}\cos 2\theta + I_{xy}\sin 2\theta\] \[I_{uv} = \frac{I_x-I_y}{2}\sin 2\theta + I_{xy}\cos 2\theta\]

Note that \(I_u + I_v = I_x + I_y = J_O\) regardless of the angle \(\theta\) — the polar moment of inertia is invariant under rotation.

Principal Moments of Inertia

The axes about which \(I_u\) reaches its maximum and minimum values are called the principal axes of inertia. They are found by setting \(dI_u/d\theta = 0\). The orientation angle \(\theta_p\) of the principal axes and the corresponding extreme values are given below. Crucially, the product of inertia is zero about the principal axes — so any axis of symmetry is automatically a principal axis.

Principal Axes and Principal Moments of Inertia
\[\tan 2\theta_p = \frac{-I_{xy}}{(I_x - I_y)/2}\] \[I_{\max/\min} = \frac{I_x+I_y}{2} \pm \sqrt{\left(\frac{I_x-I_y}{2}\right)^2 + I_{xy}^2}\]

10.7 Mohr's Circle for Moments of Inertia

Mohr's Circle

The transformation equations for \(I_u\) and \(I_{uv}\) can be combined algebraically to yield the equation of a circle in the \((I, I_{xy})\) plane. This circle — named after the German engineer Otto Mohr (1835–1918) — provides a graphical method for finding principal moments and axes. The center lies at \(((I_x+I_y)/2,\,0)\) and the radius is \(R = \sqrt{((I_x-I_y)/2)^2 + I_{xy}^2}\).

Mohr's Circle — Center and Radius
\[\text{Center: } \left(\frac{I_x+I_y}{2},\;0\right) \qquad R = \sqrt{\left(\frac{I_x-I_y}{2}\right)^2 + I_{xy}^2}\] \[I_{\max} = \frac{I_x+I_y}{2} + R \qquad I_{\min} = \frac{I_x+I_y}{2} - R\]

Procedure — Mohr's Circle

  1. Compute \(I_x\), \(I_y\), and \(I_{xy}\) for the area about the chosen origin.
  2. Draw axes: horizontal axis for \(I\), vertical axis for \(I_{xy}\).
  3. Plot the center \(O\) at \((I_x+I_y)/2\) on the \(I\)-axis and the reference point \(A(I_x, I_{xy})\).
  4. Draw the radius \(OA\) and construct the circle. The intersections with the \(I\)-axis give \(I_{\max}\) and \(I_{\min}\).
  5. The angle \(2\theta_{p_1}\) from \(OA\) to the positive \(I\)-axis (measured in the same rotational sense) gives the principal axis orientation \(\theta_{p_1}\).
I I_xy O A(I_x, I_xy) I_min I_max 2θ_p1 R
Mohr's Circle for moments of inertia. The horizontal axis intercepts give the principal moments \(I_{\min}\) and \(I_{\max}\).
Fig. 10–19

10.8 Mass Moment of Inertia

Mass Moment of Inertia

While the area moment of inertia is a purely geometric property, the mass moment of inertia is a physical property of a body that quantifies its resistance to angular acceleration. It appears in Newton's rotational law (\(\sum M = I\alpha\)) and is fundamental to dynamics. Units are kg·m² or slug·ft².

Mass Moment of Inertia
\[I = \int_m r^2\,dm \qquad \text{or, for constant density:} \quad I = \rho\int_V r^2\,dV\]

Integration Strategy for Axisymmetric Bodies

  1. Shell element (hollow cylinder of thickness \(dy\)): \(dV = 2\pi y\cdot z\,dy\). All points of the shell are at the same radius \(r = y\), so \(dI_z = r^2\,dm\) applies directly.
  2. Disk element (thin disk of thickness \(dz\)): \(dV = \pi y^2\,dz\). Points span a range of radii, so the disk's centroidal moment of inertia \(d\bar{I}_z = \tfrac{1}{2}(dm)y^2\) must be used, followed by integration.
Parallel-Axis Theorem for Mass
\[I = I_G + m\,d^2\]
Radius of Gyration (Mass)
\[I = mk^2 \quad \Rightarrow \quad k = \sqrt{\frac{I}{m}}\]

This section connects directly to Dynamics. The mass moment of inertia \(I_G\) is used in Newton's second law for rotation: \(\sum M_G = I_G\,\alpha\). Keep in mind: the parallel-axis theorem is your fastest friend for composite bodies — always find \(I_G\) for each part first, then transfer to the desired axis.

Chapter Review

Area Moment of Inertia — Key Facts

The area moment of inertia (second moment of area) measures how a cross-section's geometry resists bending and pressure. It is always positive and has units of length⁴. For irregular boundaries, integrate a rectangular differential element in the direction that simplifies the algebra.

Parallel-Axis Theorem

Transfer a known centroidal moment of inertia to any parallel axis by adding \(Ad^2\). Works for both area moments (\(I = \bar{I} + Ad^2\)) and mass moments (\(I = I_G + md^2\)). The same form extends to the product of inertia (\(I_{xy} = \bar{I}_{x'y'} + Ad_xd_y\)).

Principal Moments of Inertia

For any origin, there exist two mutually perpendicular axes (the principal axes) about which \(I\) reaches its maximum and minimum values and the product of inertia is zero. These can be found analytically or graphically via Mohr's circle.

Mass vs. Area Moment of Inertia

Both share the same mathematical structure (integral of \(r^2\) times an infinitesimal quantity), but area moment of inertia is purely geometric while mass moment of inertia depends on both geometry and material density. Area moment governs structural stiffness; mass moment governs rotational dynamics.

Learning objectives

  • Distinguish area, polar, and mass moments by their integrands and units.
  • Select centroidal, parallel-axis, composite-area, and principal-axis models.
  • Calculate section properties while retaining a stated reference axis.
  • Verify symmetry, signs, and orders of magnitude before using a result.

Physical meaning

Area moment weights material by squared distance from a bending axis: moving material outward is disproportionately effective. It is geometry, not strength; modulus, allowable stress, buckling criteria, and connections still govern design.

Problem-solving strategy

  1. Mark the requested axis and its units before choosing a table value.
  2. Locate each component centroid and treat holes as negative area.
  3. Transfer only between parallel axes, retaining signed offsets for products.
  4. Use symmetry before rotation formulas, then audit bounds and dimensions.
EXAMPLE 1Offset plate bending property
An 80 mm by 120 mm rectangular plate is used with its 80 mm bottom edge as the bending reference axis. Find its centroidal \(I_x\), edge-axis \(I_x\), and centroidal radius of gyration.

Recognition strategy

A standard rectangle is shifted to a parallel edge axis, so use its centroidal value and one parallel-axis transfer rather than integrate again.

Given information

  • \(b=80\text{ mm}\), \(h=120\text{ mm}\), \(A=9600\text{ mm}^2\), and centroid offset \(d=60\text{ mm}\).

Model and interpretation

The plate is a uniform planar area; \(x\) is horizontal and the requested edge axis is parallel to the centroidal horizontal axis.

Step-by-step solution
1. Identify. The centroidal rectangle formula is \(\bar I_x=bh^3/12\), appropriate because the axis is parallel to the width.
2. Calculate centroidal property. \(\bar I_x=80(120)^3/12=11.52\times10^6\text{ mm}^4\).
3. Transfer axes. \(I_{\rm edge}=\bar I_x+Ad^2=11.52\times10^6+9600(60)^2\text{ mm}^4\).
4. Condense the spread. \(k_x=\sqrt{\bar I_x/A}=\sqrt{11.52\times10^6/9600}=34.6\text{ mm}\).

Verification

Units: Both \(bh^3\) and \(Ad^2\) give \(\text{mm}^4\), while dividing by area leaves \(k\) in mm.

Signs and direction: The offset is squared, so the edge-axis value must exceed the positive centroidal value regardless of which side is chosen.

Plausibility: The transfer term is \(34.56\times10^6\text{ mm}^4\), three times the centroidal term, reasonable for an axis 60 mm away.

Final result: \(\boxed{\bar I_x=11.52\times10^6\text{ mm}^4,\ I_{\rm edge}=46.08\times10^6\text{ mm}^4,\ k_x=34.6\text{ mm}}\)

Engineering interpretation

The edge reference sees four times the centroidal inertia, showing why the neutral-axis location must be known before a bending calculation.

EXAMPLE 2Panel with centered service opening
A 200 mm wide by 160 mm tall panel has an 80 mm by 60 mm rectangular opening centered 100 mm above its bottom edge. Find the net area moment of inertia about the bottom edge.

Recognition strategy

This is a composite area about a specified noncentroidal axis; subtract the opening after transferring each component to the same bottom axis.

Given information

  • Outer rectangle \(200\times160\text{ mm}\); opening \(80\times60\text{ mm}\) with centroid height 100 mm.

Model and interpretation

All dimensions use mm and the horizontal bottom edge is \(x\); the void has negative area because it carries no bending-area contribution.

Step-by-step solution
1. Outer area. About its base, \(I_o=bh^3/3=200(160)^3/3=273.067\times10^6\text{ mm}^4\).
2. Opening centroid value. \(\bar I_h=80(60)^3/12=1.440\times10^6\text{ mm}^4\).
3. Transfer the opening. \(I_h=\bar I_h+(80\cdot60)(100)^2=49.440\times10^6\text{ mm}^4\).
4. Superpose. \(I_x=I_o-I_h=(273.067-49.440)\times10^6\text{ mm}^4\).

Verification

Units: Every contribution is an area times a squared length, so the reported property retains \(\text{mm}^4\).

Signs and direction: The opening is explicitly subtracted; its positive offset squared does not change the required negative composite contribution.

Plausibility: Removing material reduces \(I_x\) from 273.067 to 223.627 million \(\text{mm}^4\), and the result remains positive.

Final result: \(\boxed{I_x=223.627\times10^6\text{ mm}^4}\)

Engineering interpretation

An opening away from the edge removes more inertia than its centroidal size alone suggests; reinforcement decisions also require stress and buckling review.

EXAMPLE 3Principal axes from section data
A measured unsymmetric section has centroidal \(I_x=8.0\times10^6\), \(I_y=20.0\times10^6\), and \(I_{xy}=-6.0\times10^6\text{ mm}^4\). Determine its principal moments and one principal-axis angle.

Recognition strategy

A nonzero product signals coupled axes, so rotate to principal axes instead of treating \(I_x\) or \(I_y\) as an extreme value.

Given information

  • All properties are centroidal and use the same \(\text{mm}^4\) unit; positive rotation is counterclockwise.

Model and interpretation

The reported signs follow \(I_{xy}=\int xy\,dA\); a different sign convention changes the angle label but not the two principal magnitudes.

Step-by-step solution
1. Find orientation. \(\tan2\theta_p=-2I_{xy}/(I_x-I_y)=12/(-12)=-1\), so one solution is \(\theta_p=-22.5^\circ\).
2. Form the average. \(I_{\rm avg}=(I_x+I_y)/2=14.0\times10^6\text{ mm}^4\).
3. Form the radius. \(R=\sqrt{[(I_x-I_y)/2]^2+I_{xy}^2}=8.485\times10^6\text{ mm}^4\).
4. Read extrema. \(I_{1,2}=I_{\rm avg}\pm R=22.485,\ 5.515\times10^6\text{ mm}^4\).

Verification

Units: The average and Mohr-circle radius combine only \(\text{mm}^4\) quantities, preserving \(\text{mm}^4\) in both results.

Signs and direction: The negative \(I_{xy}\) sets the selected clockwise angle; the perpendicular principal axis is \(67.5^\circ\) counterclockwise from it.

Plausibility: The extrema bracket both 8.0 and 20.0 million \(\text{mm}^4\), and their sum equals the invariant \(28.0\times10^6\text{ mm}^4\).

Final result: \(\boxed{I_1=22.485\times10^6,\ I_2=5.515\times10^6\text{ mm}^4;\ \theta_p=-22.5^\circ}\)

Engineering interpretation

Principal directions decouple bending measures for an unsymmetric section; fabrication tolerances and the full bending-stress model remain necessary for design release.

Chapter recap

  • Choose the axis first, then use integration or tabulated centroidal properties.
  • Transfer parallel axes and sum solids minus holes consistently.
  • Before moving on: verify \(\text{length}^4\), symmetry, and whether a principal-axis transformation is needed.