copyright joseph greene 2003 all rights reserved 1 cm 197 mechanics of materials chap 8: moments of...

Copyright Joseph Greene 2003 All Rights Reserved 1

CM 197Mechanics of Materials

Chap 8: Moments of Inertia

Professor Joe Greene

CSU, CHICO

Reference: Statics and Strength of Materials, 2nd ed., Fa-Hwa Cheng, Glencoe/McGraw Hill, Westerville, OH (1997)

CM 197


Chap 8: Moments of Inertia• Objectives

– Introduction– Moments of Inertia and Radii of Gyration– Parallel Axis Theorem– Moments of Inertia of Composite Areas– Moments of Inertia of Built-up Sections


Automotive Applications ofPlastics and Composites

Composite Intensive Vehicles


Automotive Plastics and Composites UseAutomotive Plastics and Composites Use


Structural Applications• Products today under go stresses in daily use.

– Loads are applied to structure that cause the structure to move, bend, or twist.• The product moves, bends, ans/or twists (Inertia) relative to a center of the part (Centroid). Moment of

Inertia controls movement.

– Bridges• Wind, Snow, environment• Vehicles, earthquakes• Deflections, strain, stress, bending

– Buildings• Wind, Snow, environment• People, furniture, earthquakes• Deflections, strain, stress, bending

– Vehicles • Weight, movement, crash• Deflections, strain, stress, failure

– Moment of inertia is needed to calculate how much structure will move during an applied load


Introduction

• The concept of a moment of inertia is important in many design and analysis problems encountered in Civil Engineering.

• Used in the design of machines and bridges.– Used in calculations of beam deflections and stresses, torsion of shafts, and

buckling of columns.

• Computation of moments of inertia can often be cumbersome.

• Review (Chap 7)– Center of gravity- is the central point where weight of a solid is located

– Centroid- Location is the first moment of the area. xA

– Moment of inertia is the second moment of the area. x2A

• References– Web sites from yahoo.com or google.com

– http://www.mapleapps.com/categories/engineering/mechanical/html/momen1.html


Moments of Inertia and Radii of Gyration

• Moment of inertia of an area– Second moment of the area computed with respect to the axis– Figure 8.1

• Moments of inertia of an area element A about the x and y axes are Ix = y2A and Iy= x2A

• Moments of inertia for the entire area A about the x and y axes are Ix = y2A and Iy= x2A

• Note: the value of the moment of inertia is always positive, regardless of location of the axis.

• Note: Units are length to the 4th power, such as in4 or m4

Figure 8.1

x

z

AA

y

y

x

r

O


Moments of Inertia and Radii of Gyration• Polar Moment of inertia of an area

– Similarly, Second moment of the area computed with respect to the the pole O or the z axis.

– Figure 8.1• Moments of inertia for the entire area A about the z axes is

JO = r2A, where r is the distance from the pole to the element A since r2 = x2 + y2

JO = r2A = JO = (x2 + y2)A = x2 A + y2 A Therefore, JO = Ix + Iy

• Note: Moments of inertia are evaluated with respect to the axes on the plane of the area.

• Note: Polar moments of inertia are evaluated with respect to the pole or the axes perpendicular to the plane of the area.

x

z

AA

y

y

x

r

O


Moments of Inertia and Radii of Gyration• Radius of Gyration of an area with respect to the x axis

– The distance at which the entire area could be located from the x-axis to produce the same moment of inertia of the area with respect to the x-axis. Radius of gyration, rx,

Ix = A rx2 or rx = (Ix /A)½

Similarly, Iy = A ry2 or ry = (Iy /A)½

Centroidal axis The coordinate axis with the origin located at the centroid of an area.

Properties of Areas of Common Shapes Table 8.1 (Page 300) Note: Centroidal axes passing through the centroid C of the area are

denoted with a line bar over the x and y. Note: the properties are also denoted with a bar for Ix and Iy, J, and rx.


Parallel Axis Theorem• The parallel axis theorem

• Moment of inertia of an area with respect to a noncentroidal axis is equal to the moment of inertia of the area with respect to the parallel centroidal axis plus the product of the area and the square of the perpendicular distance between two axes.

– Expressed as a moment of inertia with respect to a parallel centroidal axis• Figure 8-2

– x axis is a centroidal axis through the centroid C of the area and– x axis is a parallel noncentroidal axis located at a perpendicular distance d

from the centroidal axis– y is the coordinate of an area element A from the centroidal x axis– Moment of inertia of the area about the noncentroidal x axis is

• This equation is the Parallel-axis theorem Ix = (y + d)2 A = (y2 + 2 dy + d2) A Eqn 8-7

Ix = y2 A + 2 dy A + d2 A

Note: This equation only to two parallel axes and one axis must be a centroidal axis.

Example 8-1, Example 8-2

x

* AA

d

y xC


Moment of Inertia of Composite Areas• Moments of Inertia of Composite Areas

– Most shapes are not simple and are a combination of rectangular, circular shapes, tubular, triangular, or other shapes.

• The moment of inertia of a composite area about an axis is the sum of the moments of inertia of the component parts about the same axis.

• Note: Before adding the moments of inertia, the parallel-axis theorem must be used to transfer the moment of inertia of each area to the desired axis.

– The moment of inertia of a composite area about the x-axis may: Ix = [I + Ay2] Eqn

8-8 Note: I is the moment of inertia about the centroidal axis and y is the

distance from the centroid of a component area to the x-axis For an engineering application where a structure is undergoing bending, twisting,

pulling, or contracting, the resulting strength and deflection of the structure is calculated by FIRST finding the moments of inertia of a composite section about the centroidal axis

Ix = [I + A ( y-y)2] Eqn 8-9

Note: I is the moment of inertia about the centroidal axis and y is the distance from the centroid of a component area to the x-axis and y is the distance from the centroid of the entire section to the reference x axis.


Moment of Inertia of Composite Areas• Steps for solving Moments of Inertia of Composite Areas

1. Select a horizontal reference axis (e.g., bottom of the structure)2. Subdivide the section into rectangular sections3. Calculate the location of the centroid of an area for uniform thickness

(Eqn 7-6) and the corresponding centroidal axis.4. Calculate the moment of inertia of each component about its own

horizontal centroidal axis (Table 8.1)5. Calculate the moment of inertia of a section with respect to the centroidal

axis (Eqn 8.9) Ix = [I + A ( y-y)2]

– Hints: • Use a table to organize calculations.• For odd shaped structures, break it up into rectangular and/or circular

sections• For shapes with holes or notches, treat the values of their areas and their

moments of inertia as negative (example 8-4)• For symmetric shapes, the centroidal axes (x and y) are located at the

middle of the section.

– Example 8-3, Example 8-4, Example 8-5

A

Ayy

A

Axx

;


Moment of Inertia of Built-up Sections• Structural Steel Shapes

– Rolled into a wide variety of shapes and sizes.• Designated by letters that specify their sizes

– W21 x 83 : Wide flanged steel beam having nominal depth of 21in and weighing 83 lb per linear foot.

– S20 x 75 : American standard steel I beam 20in deep and weighing 75 lb per linear foot.

– C12 x 30 : American standard steel channel section 12in deep and weighing 30 lb per linear foot.

– L6 x 6 x ½ : Equal-leg steel angle of legs 6in long and 1/2in thick

• American Institute of Steel Construction (AISC) manual provides detailed information for structural steel shapes.

– Selected shapes are in the appendix for US and SI units.

» Tables list dimensions, areas, locations of centroidal axes, centroidal moments of inertia, centroidal radii of gyration, etc. for selected steel shapes


•


Moment of Inertia of Built-up Sections• Built-up Structural Steel Sections

– Made up of several shapes that are welded, riveted, or bolted together to form a single member.• Beams: To calculate strengths and deflections for the steel sections

– Need to calculate centroidal moments of inertia• Columns: To calculate strengths and deflections for the steel sections

– Need to calculate radius of gyration

– Calculations: Use same procedure as for composite area from before.

• Example 8-6– Section is symmetric, thus centroidal axis is in the middle of part.

• Select properties of I beam from appendix to calculate d and I1.

• Select the distance d for the steel I beam not including the cover plates.• Select the distance y as between the center of the cover plate to the center of the I

beam.

– Calculate moment of inertia Use Eqn 8.8

• Example 8-7– Determine the moment of inertia from built up section similar to Example 8-3

copyright joseph greene 2003 all rights reserved 1 cm 197 mechanics of materials chap 8: moments of...

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