Moments of Inertia

Moments of Inertia • Definitions of Moments of Inertia for Areas • Parallel-Axis Theorem for an Area • Radius of Gyration of an Area • Moments of Inertia for Composite Areas • Product of Inertia for an Area • Moments of Inertia for an Area about Inclined Axes • Mohr’s Circle for Moments of Inertia • Mass Moment of Inertia

1. Definition of Moments of Inertia for Areas • Centroid for an area is determined by the first moment of an area about an axis • Second moment of an area is referred as the moment of inertia • Moment of inertia of an area originates whenever one relates the normal stress σ or force per unit area

1. Definition of Moments of Inertia for Areas Moment of Inertia • Consider area A lying in the x-y plane • Be definition, moments of inertia of the differential plane area dA about the x and y axes • For entire area, moments of inertia are given by

1. Definition of Moments of Inertia for Areas Moment of Inertia • Formulate the second moment of dA about the pole O or z axis • This is known as the polar axis where r is perpendicular from the pole (z axis) to the element dA • Polar moment of inertia for entire area,

2. Parallel Axis Theorem for an Area • For moment of inertia of an area known about an axis passing through its centroid, determine the moment of inertia of area about a corresponding parallel axis using the parallel axis theorem • Consider moment of inertia of the shaded area • A differential element dA is located at an arbitrary distance y’from the centroidal x’ axis

2. Parallel Axis Theorem for an Area • The fixed distance between the parallel x and x’ axes is defined as dy • For moment of inertia of dA about x axis • For entire area • First integral represent the moment of inertia of the area about the centroidal axis

2. Parallel Axis Theorem for an Area • Second integral = 0 since x’ passes through the area’s centroid C • Third integral represents the total area A • Similarly • For polar moment of inertia about an axis perpendicular to the x-y plane and passing through pole O (z axis)

3. Radius of Gyration of an Area • Radius of gyration of a planar area has units of length and is a quantity used in the design of columns in structural mechanics • For radii of gyration • Similar to finding moment of inertia of a differential area about an axis

Example 1 Determine the moment of inertia for the rectangular area with respect to (a) the centroidal x’ axis, (b) the axis xb passing through the base of the rectangular, and (c) the pole or z’ axis perpendicular to the x’-y’ plane and passing through the centroid C.

Solution Part (a) Differential element chosen, distance y’ from x’ axis. Since dA = b dy’, Part (b) By applying parallel axis theorem,

Solution Part (c) For polar moment of inertia about point C,

4. Moments of Inertia for Composite Areas • Composite area consist of a series of connected simpler parts or shapes • Moment of inertia of the composite area = algebraic sum of the moments of inertia of all its parts Procedure for Analysis Composite Parts • Divide area into its composite parts and indicate the centroid of each part to the reference axis Parallel Axis Theorem • Moment of inertia of each part is determined about its centroidal axis

4. Moments of Inertia for Composite Areas Procedure for Analysis Parallel Axis Theorem • When centroidal axis does not coincide with the reference axis, the parallel axis theorem is used Summation • Moment of inertia of the entire area about the reference axis is determined by summing the results of its composite parts

Example 2 Compute the moment of inertia of the composite area about the x axis.

Solution Composite Parts Composite area obtained by subtracting the circle form the rectangle. Centroid of each area is located in the figure below.

Solution Parallel Axis Theorem Circle Rectangle

Solution Summation For moment of inertia for the composite area,

5. Product of Inertia for an Area • Moment of inertia for an area is different for every axis about which it is computed • First, compute the product of the inertia for the area as well as its moments of inertia for given x, y axes • Product of inertia for an element of area dA located at a point (x, y) is defined as dIxy = xydA • Thus for product of inertia,

5. Product of Inertia for an Area Parallel Axis Theorem • For the product of inertia of dA with respect to the x and y axes • For the entire area, • Forth integral represent the total area A,

Example 3 Determine the product Ixy of the triangle.

Solution Differential element has thickness dx and area dA = y dx Using parallel axis theorem, locates centroid of the element or origin of x’, y’ axes

Solution Due to symmetry, Integrating we have

Solution Differential element has thickness dy and area dA = (b - x) dy. For centroid, For product of inertia of element

6. Moments of Inertia for an Area about Inclined Axes • In structural and mechanical design, necessary to calculate Iu, Iv and Iuv for an area with respect to a set of inclined u and v axes when the values of θ, Ix, Iy and Ixy are known • Use transformation equations which relate the x, y and u, v coordinates

6. Moments of Inertia for an Area about Inclined Axes • Integrating, • Simplifying using trigonometric identities,

6. Moments of Inertia for an Area about Inclined Axes • We can simplify to • Polar moment of inertia about the z axis passing through point O is,

6. Moments of Inertia for an Area about Inclined Axes Principal Moments of Inertia • Iu, Iv and Iuv depend on the angle of inclination θ of the u, v axes • The angle θ = θp defines the orientation of the principal axes for the area

6. Moments of Inertia for an Area about Inclined Axes Principal Moments of Inertia • Substituting each of the sine and cosine ratios, we have • Result can gives the max or min moment of inertia for the area • It can be shown that Iuv = 0, that is, the product of inertia with respect to the principal axes is zero • Any symmetric axis represent a principal axis of inertia for the area

Example 4 Determine the principal moments of inertia for the beam’s cross-sectional area with respect to an axis passing through the centroid.

Solution Moment and product of inertia of the cross-sectional area, Using the angles of inclination of principal axes u and v,

Solution For principal of inertia with respect to the u and v axes

7. Mohr’s Circle for Moments of Inertia • It is found that • In a given problem, Iu and Iv are variables and Ix, Iy and Ixy are known constants • When this equation is plotted on a set of axes that represent the respective moment of inertia and the product of inertia, the resulting graph represents a circle

7. Mohr’s Circle for Moments of Inertia • The circle constructed is known as a Mohr’s circle with radius and center at (a, 0) where

7. Mohr’s Circle for Moments of Inertia Procedure for Analysis Determine Ix, Iy and Ixy • Establish the x, y axes for the area, with the origin located at point P of interest and determine Ix, Iy and Ixy Construct the Circle • Construct a rectangular coordinate system such that the abscissa represents the moment of inertia I and the ordinate represent the product of inertia Ixy

7. Mohr’s Circle for Moments of Inertia Construct the Circle • Determine center of the circle O, which is located at a distance (Ix + Iy)/2 from the origin, and plot the reference point a having coordinates (Ix, Ixy) • By definition, Ix is always positive, whereas Ixy will either be positive or negative • Connect the reference point A with the center of the circle and determine distance OA (radius of the circle) by trigonometry • Draw the circle

7. Mohr’s Circle for Moments of Inertia Principal of Moments of Inertia • Points where the circle intersects the abscissa give the values of the principle moments of inertia Imin and Imax • Product of inertia will be zero at these points Principle Axes • This angle represent twice the angle from the x axis to the area in question to the axis of maximum moment of inertia Imax • The axis for the minimum moment of inertia Imin is perpendicular to the axis for Imax

Example 5 Using Mohr’s circle, determine the principle moments of the beam’s cross-sectional area with respect to an axis passing through the centroid.

Solution Determine Ix, Iy and Ixy Moments of inertia and the product of inertia have been determined in previous examples Construct the Circle Center of circle, O, lies from the origin, at a distance

Solution Construct the Circle With reference point A (2.90, -3.00) connected to point O, radius OA is determined using Pythagorean theorem Principal Moments of Inertia Circle intersects I axis at points (7.54, 0) and (0.960, 0)

Solution Principal Axes Angle 2θp1 is determined from the circle by measuring CCW from OA to the direction of the positive I axis The principal axis for Imax = 7.54(109) mm4 is therefore orientated at an angle θp1 = 57.1°, measured CCW from the positive x axisto the positive u axis. v axis is perpendicular to this axis.

8. Mass Moment of Inertia • Mass moment of inertia is defined as the integral of the second moment about an axis of all the elements of mass dm which compose the body • For body’s moment of inertia about the z axis, • The axis that is generally chosen for analysis, passes through the body’s mass center G

8. Mass Moment of Inertia • If the body consists of material having a variable density ρ = ρ(x, y, z), the element mass dm of the body may be expressed as dm = ρ dV • Using volume element for integration, • When ρ being a constant,

8. Mass Moment of Inertia Procedure for Analysis Shell Element • For a shell element having height z, radius y and thickness dy, volume dV = (2πy)(z)dy Disk Element • For disk element having radius y, thickness dz, volume dV = (πy2) dz

Example 6 Determine the mass moment of inertia of the cylinder about the z axis. The density of the material is constant.

Solution Shell Element For volume of the element, For mass, Since the entire element lies at the same distance r from the z axis, for the moment of inertia of the element,

Solution Integrating over entire region of the cylinder, For the mass of the cylinder So that

8. Mass Moment of Inertia Parallel Axis Theorem • If the moment of inertia of the body about an axis passing through the body’s mass center is known, the moment of inertia about any other parallel axis may be determined by using parallel axis theorem • Using Pythagorean theorem, r 2 = (d + x’)2 + y’2 • For moment of inertia of body about the z axis,

8. Mass Moment of Inertia Parallel Axis Theorem • For moment of inertia about the z axis, I = IG + md2 Radius of Gyration • For moment of inertia expressed using k, radius of gyration,

Example 7 If the plate has a density of 8000kg/m3 and a thickness of 10mm, determine its mass moment of inertia about an axis perpendicular to the page and passing through point O.

Solution The plate consists of 2 composite parts, the 250mm radius disk minus the 125mm radius disk. Disk For moment of inertia of a disk, Mass center of the disk is located 0.25m from point O

Moments of Inertia