STATICS. Moments of Inertia VECTOR MECHANICS FOR ENGINEERS: Ninth Edition CHAPTER. Ferdinand P. Beer E. Russell Johnston, Jr.

Similar documents
STATICS. Moments of Inertia VECTOR MECHANICS FOR ENGINEERS: Seventh Edition CHAPTER. Ferdinand P. Beer

Distributed Forces: Moments of Inertia

ME 141. Lecture 8: Moment of Inertia

ME 101: Engineering Mechanics

Moments and Product of Inertia

STATICS. Distributed Forces: Moments of Inertia VECTOR MECHANICS FOR ENGINEERS: Eighth Edition CHAPTER. Ferdinand P. Beer E. Russell Johnston, Jr.

MECHANICS OF MATERIALS

3/31/ Product of Inertia. Sample Problem Sample Problem 10.6 (continue)

MECHANICS OF MATERIALS

SOLUTION Determine the moment of inertia for the shaded area about the x axis. I x = y 2 da = 2 y 2 (xdy) = 2 y y dy

Statics: Lecture Notes for Sections 10.1,10.2,10.3 1

MECHANICS OF MATERIALS

BEAMS: SHEAR AND MOMENT DIAGRAMS (FORMULA)

Moments of Inertia. Notation:

10.5 MOMENT OF INERTIA FOR A COMPOSITE AREA

ENGI 4430 Advanced Calculus for Engineering Faculty of Engineering and Applied Science Problem Set 3 Solutions [Multiple Integration; Lines of Force]

10 3. Determine the moment of inertia of the area about the x axis.

Chapter 5 Equilibrium of a Rigid Body Objectives

MECHANICS OF MATERIALS

EMA 3702 Mechanics & Materials Science (Mechanics of Materials) Chapter 4 Pure Bending

MECHANICS OF MATERIALS

I xx + I yy + I zz = (y 2 + z 2 )dm + (x 2 + y 2 )dm. (x 2 + z 2 )dm + (x 2 + y 2 + z 2 )dm = 2

Moment of Inertia and Centroid

ME 201 Engineering Mechanics: Statics

MTE 119 STATICS LECTURE MATERIALS FINAL REVIEW PAGE NAME & ID DATE. Example Problem F.1: (Beer & Johnston Example 9-11)

PROBLEM Area of Problem I = 471,040 mm xy 2(471,040) 252,757 1,752,789 = or θ m = and = (1,002,773 ± 885,665) mm

Stress and Strain ( , 3.14) MAE 316 Strength of Mechanical Components NC State University Department of Mechanical & Aerospace Engineering

Statics: Lecture Notes for Sections

EXERCISES Chapter 15: Multiple Integrals. Evaluating Integrals in Cylindrical Coordinates

2. Supports which resist forces in two directions. Fig Hinge. Rough Surface. Fig Rocker. Roller. Frictionless Surface

MECHANICS OF MATERIALS

STATICS. Equivalent Systems of Forces. Vector Mechanics for Engineers: Statics VECTOR MECHANICS FOR ENGINEERS: Contents & Objectives.

Second Moments or Moments of Inertia

MECHANICS OF MATERIALS REVIEW

ENGI 4430 Multiple Integration Cartesian Double Integrals Page 3-01

Lecture 6: Distributed Forces Part 2 Second Moment of Area

REVIEW FOR EXAM II. Dr. Ibrahim A. Assakkaf SPRING 2002

STATICS VECTOR MECHANICS FOR ENGINEERS: Distributed Forces: Centroids and Centers of Gravity. Tenth Edition CHAPTER

1.1 The Equations of Motion

STATICS. Statics of Particles VECTOR MECHANICS FOR ENGINEERS: Eighth Edition CHAPTER. Ferdinand P. Beer E. Russell Johnston, Jr.

University of Pretoria Department of Mechanical & Aeronautical Engineering MOW 227, 2 nd Semester 2014

AREAS, RADIUS OF GYRATION

SOLUTION y ' A = 7.5(15) (150) + 90(150) (15) + 215(p)(50) 2. = mm 2. A = 15(150) + 150(15) + p(50) 2. =

SOLUTION (y 2 + z 2 )dm + (x 2 + y 2 )dm. (x 2 + z 2 )dm + I xx + I yy + I zz = = 2. (x 2 + y 2 + z 2 )dm

Chapter 10: Moments of Inertia

KINEMATIC RELATIONS IN DEFORMATION OF SOLIDS

MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.

ME 243. Lecture 10: Combined stresses

Dynamics and control of mechanical systems

STATICS. Equivalent Systems of Forces. Vector Mechanics for Engineers: Statics VECTOR MECHANICS FOR ENGINEERS: Contents 9/3/2015.

ENGI Multiple Integration Page 8-01

Properties of Sections

MECHANICS OF MATERIALS

Stress Analysis Lecture 4 ME 276 Spring Dr./ Ahmed Mohamed Nagib Elmekawy

MECHANICS OF MATERIALS

MECHANICS OF MATERIALS

Outline. Organization. Stresses in Beams

LECTURE 13 Strength of a Bar in Pure Bending

MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question. C) 2

Bending Stress. Sign convention. Centroid of an area

y=1/4 x x=4y y=x 3 x=y 1/3 Example: 3.1 (1/2, 1/8) (1/2, 1/8) Find the area in the positive quadrant bounded by y = 1 x and y = x3

Chapter 9 BIAXIAL SHEARING

Strain Transformation and Rosette Gage Theory

Solution ME 323 EXAM #2 FALL SEMESTER :00 PM 9:30 PM Nov. 2, 2010

y R T However, the calculations are easier, when carried out using the polar set of co-ordinates ϕ,r. The relations between the co-ordinates are:

Differentiation Techniques

Aircraft Structures Structural & Loading Discontinuities

Properties of surfaces II: Second moment of area

[4] Properties of Geometry

The aircraft shown is being tested to determine how the forces due to lift would be distributed over the wing. This chapter deals with stresses and

MOMENTS OF INERTIA FOR AREAS, RADIUS OF GYRATION OF AN AREA, & MOMENTS OF INTERTIA BY INTEGRATION

CHAPTER 2. Copyright McGraw-Hill Education. Permission required for reproduction or display.

ENGR-1100 Introduction to Engineering Analysis. Lecture 17

Two small balls, each of mass m, with perpendicular bisector of the line joining the two balls as the axis of rotation:

Handout 6: Rotational motion and moment of inertia. Angular velocity and angular acceleration

MECHANICS OF MATERIALS

STATICS. Bodies VECTOR MECHANICS FOR ENGINEERS: Ninth Edition CHAPTER. Ferdinand P. Beer E. Russell Johnston, Jr.

MTE 119 STATICS FINAL HELP SESSION REVIEW PROBLEMS PAGE 1 9 NAME & ID DATE. Example Problem P.1

SET-I SECTION A SECTION B. General Instructions. Time : 3 hours Max. Marks : 100

LECTURE 14 Strength of a Bar in Transverse Bending. 1 Introduction. As we have seen, only normal stresses occur at cross sections of a rod in pure

Problem d d d B C E D. 0.8d. Additional lecturebook examples 29 ME 323

MECHANICS OF MATERIALS

Chapter 6: Cross-Sectional Properties of Structural Members

Contents. Dynamics and control of mechanical systems. Focuses on

x y plane is the plane in which the stresses act, yy xy xy Figure 3.5.1: non-zero stress components acting in the x y plane

CHAPTER 2: Partial Derivatives. 2.2 Increments and Differential

Lecture Triaxial Stress and Yield Criteria. When does yielding occurs in multi-axial stress states?

Mechanics of Materials

Functions of Several Variables

[8] Bending and Shear Loading of Beams

Jim Lambers MAT 280 Fall Semester Practice Final Exam Solution

Directional derivatives and gradient vectors (Sect. 14.5). Directional derivative of functions of two variables.

Chapter 3. Load and Stress Analysis. Lecture Slides

STATICS. Vector Mechanics for Engineers: Statics VECTOR MECHANICS FOR ENGINEERS: Contents 9/3/2015

MECHANICS OF MATERIALS

Stresses: Beams in Bending

Vector Analysis 1.1 VECTOR ANALYSIS. A= Aa A. Aa, A direction of the vector A.

Chapter 6 Notes, Stewart 8e

CHAPTER 1 MEASUREMENTS AND VECTORS

(a) Use washer cross sections: a washer has

Transcription:

N E 9 Distributed CHAPTER VECTOR MECHANCS FOR ENGNEERS: STATCS Ferdinand P. Beer E. Russell Johnston, Jr. Lecture Notes: J. Walt Oler Teas Tech Universit Forces: Moments of nertia

Contents ntroduction Moments of nertia of an Area Moment of nertia of an Area b ntegration Polar Moment of nertia Radius of Gration of an Area Sample Problem 9. Sample Problem 9. Parallel Ais Theorem Moments of nertia of Composite Areas Sample Problem 9. Sample Problem 9.5 Product of nertia Principal Aes and Principal Moments of nertia Sample Problem 9.6 Sample Problem 9.7 Mohr s Circle for Moments and Products of nertia Sample Problem 9.8 Moment of nertia of a Mass Parallel Ais Theorem Moment of nertia of Thin Plates Moment of nertia of a D Bod b ntegration Moment of nertia of Common Geometric Shapes Sample Problem 9. Moment of nertia With Respect to an Arbitrar Ais Ellipsoid of nertia. Principle Aes of Aes of nertia of a Mass 9 -

ntroduction Previousl considered distributed forces which were proportional to the area or volume over which the act. - The resultant was obtained b summing or integrating over the areas or volumes. - The moment of the resultant about an ais was determined b computing the first moments of the areas or volumes about that ais. Will now consider forces which are proportional to the area or volume over which the act but also var linearl with distance from a given ais. - t will be shown that the magnitude of the resultant depends on the first moment of the force distribution with respect to the ais. - The point of application of the resultant depends on the second moment of the distribution with respect to the ais. Current chapter will present methods for computing the moments and products of inertia for areas and masses. 9 -

Moment of nertia of an Area Consider distributed forces F whose magnitudes are proportional to the elemental areas A on which the act and also var linearl with the distance of A from a given ais. Eample: Consider a beam subjected to pure bending. nternal forces var linearl with distance from the neutral ais which passes through the section centroid. F ka R M k k da da 0 da Q first moment da second moment Eample: Consider the net hdrostatic force on a submerged circular gate. F pa A R M da da 9 -

Moment of nertia of an Area b ntegration Second moments or moments of inertia of an area with respect to the and aes, da da Evaluation of the integrals is simplified b choosing da to be a thin strip parallel to one of the coordinate aes. For a rectangular area, da h 0 bd bh The formula for rectangular areas ma also be applied to strips parallel to the aes, d d d da d 9-5

Polar Moment of nertia The polar moment of inertia is an important parameter in problems involving torsion of clindrical shafts and rotations of slabs. J 0 r da The polar moment of inertia is related to the rectangular moments of inertia, J r da da da 0 da 9-6

Radius of Gration of an Area Consider area A with moment of inertia. magine that the area is concentrated in a thin strip parallel to the ais with equivalent. k A k A k = radius of gration with respect to the ais Similarl, J O k k O A A k k O A J A O k O k k 9-7

Sample Problem 9. SOLUTON: A differential strip parallel to the ais is chosen for da. d da da l d Determine the moment of inertia of a triangle with respect to its base. For similar triangles, l b h h h l b h ntegrating d from = 0 to = h, da b h h h 0 h b d h h 0 h da b d h b h h 0 h d bh 9-8

Sample Problem 9. SOLUTON: An annular differential area element is chosen, dj J O O u dj da O r 0 u da u du u du J O r 0 u r du a) Determine the centroidal polar moment of inertia of a circular area b direct integration. b) Using the result of part a, determine the moment of inertia of a circular area with respect to a diameter. From smmetr, =, JO r diameter r 9-9

Parallel Ais Theorem Consider moment of inertia of an area A with respect to the ais AA da The ais BB passes through the area centroid and is called a centroidal ais. da da d d da da d da Ad parallel ais theorem 9-0

Parallel Ais Theorem Moment of inertia T of a circular area with respect to a tangent to the circle, Ad r r r T 5 r Moment of inertia of a triangle with respect to a centroidal ais, AA BB BB AA 6 bh Ad Ad bh bh h 9 -

Moments of nertia of Composite Areas The moment of inertia of a composite area A about a given ais is obtained b adding the moments of inertia of the component areas A, A, A,..., with respect to the same ais. 9 -

Moments of nertia of Composite Areas 9 -

Sample Problem 9. SOLUTON: Determine location of the centroid of composite section with respect to a coordinate sstem with origin at the centroid of the beam section. The strength of a W8 rolled steel beam is increased b attaching a plate to its upper flange. Determine the moment of inertia and radius of gration with respect to an ais which is parallel to the plate and passes through the centroid of the section. Appl the parallel ais theorem to determine moments of inertia of beam section and plate with respect to composite section centroidal ais. Calculate the radius of gration from the moment of inertia of the composite section. 9 -

Sample Problem 9. SOLUTON: Determine location of the centroid of composite section with respect to a coordinate sstem with origin at the centroid of the beam section. Section Plate Beam Section A, in 6.75.0 A 7.95, in. 7.5 0 A, in 50. 0 A 50. Y A A Y A A 50. in 7.95 in.79 in. 9-5

Sample Problem 9. Appl the parallel ais theorem to determine moments of inertia of beam section and plate with respect to composite section centroidal ais.,beam section,plate 7. in AY Ad 5. in 85.0.79 9 6.757.5.79, beam section,plate 7. 5. 68in Calculate the radius of gration from the moment of inertia of the composite section. k A 67.5 in 7.95 in k 5.87 in. 9-6

Sample Problem 9.5 SOLUTON: Compute the moments of inertia of the bounding rectangle and half-circle with respect to the ais. Determine the moment of inertia of the shaded area with respect to the ais. The moment of inertia of the shaded area is obtained b subtracting the moment of inertia of the half-circle from the moment of inertia of the rectangle. 9-7

Sample Problem 9.5 SOLUTON: Compute the moments of inertia of the bounding rectangle and half-circle with respect to the ais. Rectangle: 6 00 8. 0 bh mm r 90 a 8. mm b 0 - a 8.8 mm A r.70 90 mm Half-circle: moment of inertia with respect to AA, 6 r 90 5.760 mm AA 8 8 moment of inertia with respect to, AA Aa 7.00 6 mm 6 5.760.70 moment of inertia with respect to, Ab 9.0 6 mm 7.00 6.70 8.8 9-8

Sample Problem 9.5 The moment of inertia of the shaded area is obtained b subtracting the moment of inertia of the half-circle from the moment of inertia of the rectangle. 8. 0 6 mm 9.0 6 mm 5.9 0 6 mm 9-9

Product of nertia Product of nertia: da When the ais, the ais, or both are an ais of smmetr, the product of inertia is zero. Parallel ais theorem for products of inertia: A 9-0

Principal Aes and Principal Moments of nertia The change of aes ields cos cos sin cos sin sin Given da da we wish to determine moments and product of inertia with respect to new aes and. Note: cos sin cos sin da The equations for and are the parametric equations for a circle, ave ave R R The equations for and lead to the same circle. 9 -

Principal Aes and Principal Moments of nertia At the points A and B, = 0 and is a maimum and minimum, respectivel. ma,min tan m ave The equation for Q m defines two angles, 90 o apart which correspond to the principal aes of the area about O. R ave ave R R ma and min are the principal moments of inertia of the area about O. 9 -

Sample Problem 9.6 SOLUTON: Determine the product of inertia using direct integration with the parallel ais theorem on vertical differential area strips Appl the parallel ais theorem to evaluate the product of inertia with respect to the centroidal aes. Determine the product of inertia of the right triangle (a) with respect to the and aes and (b) with respect to centroidal aes parallel to the and aes. 9 -

Sample Problem 9.6 SOLUTON: Determine the product of inertia using direct integration with the parallel ais theorem on vertical differential area strips h da d h d b b el el h b ntegrating d from = 0 to = b, h d b 0 el b el da b b 0 d h h b d b 8b b 0 b h 9 -

Sample Problem 9.6 Appl the parallel ais theorem to evaluate the product of inertia with respect to the centroidal aes. b With the results from part a, b h h A b h bh b h 7 9-5

Sample Problem 9.7 SOLUTON: Compute the product of inertia with respect to the aes b dividing the section into three rectangles and appling the parallel ais theorem to each. Determine the orientation of the principal aes (Eq. 9.5) and the principal moments of inertia (Eq. 9. 7). For the section shown, the moments of inertia with respect to the and aes are = 0.8 in and = 6.97 in. Determine (a) the orientation of the principal aes of the section about O, and (b) the values of the principal moments of inertia about O. 9-6

Sample Problem 9.7 SOLUTON: Compute the product of inertia with respect to the aes b dividing the section into three rectangles. Appl the parallel ais theorem to each rectangle, A Note that the product of inertia with respect to centroidal aes parallel to the aes is zero for each rectangle. Rectangle Area, in.5.5.5, in..5 0.5, in..75 0.75 A A,in.8 0.8 6.56 A 6.56 in 9-7

Sample Problem 9.7 Determine the orientation of the principal aes (Eq. 9.5) and the principal moments of inertia (Eq. 9. 7). tan m m 75. and 55. 6.56 0.8 6.97.85 m 7.7 and m 7. 7 0.8in 6.97 in 6.56 in ma,min 0.8 6.97 0.8 6.97 6.56 a b ma min 5.5in.897 in 9-8

Mohr s Circle for Moments and Products of nertia The moments and product of inertia for an area are plotted as shown and used to construct Mohr s circle, ave R Mohr s circle ma be used to graphicall or analticall determine the moments and product of inertia for an other rectangular aes including the principal aes and principal moments and products of inertia. 9-9

Sample Problem 9.8 SOLUTON: Plot the points (, ) and (,- ). Construct Mohr s circle based on the circle diameter between the points. Based on the circle, determine the orientation of the principal aes and the principal moments of inertia. The moments and product of inertia with respect to the and aes are = 7.06 mm, =.606 mm, and = -.50 6 mm. Based on the circle, evaluate the moments and product of inertia with respect to the aes. Using Mohr s circle, determine (a) the principal aes about O, (b) the values of the principal moments about O, and (c) the values of the moments and product of inertia about the and aes 9-0

Sample Problem 9.8 SOLUTON: Plot the points (, ) and (,- ). Construct Mohr s circle based on the circle diameter between the points. OC CD R ave.50.950 6 CD DX.70 mm 6 mm 6 mm 7.0.60 6 6.50 mm mm 6 mm Based on the circle, determine the orientation of the principal aes and the principal moments of inertia. tan m DX.097 m 7. 6 CD m. 8 ma OA ave R ma 8.60 6 mm min OB ave R min.90 6 mm 9 -

Sample Problem 9.8 Based on the circle, evaluate the moments and product of inertia with respect to the aes. The points X and Y corresponding to the and aes are obtained b rotating CX and CY counterclockwise through an angle Q (60 o ) = 0 o. The angle that CX forms with the aes is f = 0 o - 7.6 o = 7. o. OF OC CX cos Rcos 7. ' OG OC CYcos R cos 7. ' ave 5.960 ave 6 mm o o.890 6 mm FX CYsin R sin 7. ' o OC ave R.7 0.950 6 mm 6 mm.80 6 mm 9 -

Moment of nertia of a Mass Angular acceleration about the ais AA of the small mass m due to the application of a couple is proportional to r m. r m = moment of inertia of the mass m with respect to the ais AA For a bod of mass m the resistance to rotation about the ais AA is r m r m r m r dm mass moment of inertia The radius of gration for a concentrated mass with equivalent mass moment of inertia is k m k m 9 -

Moment of nertia of a Mass Moment of inertia with respect to the coordinate ais is r dm dm z Similarl, for the moment of inertia with respect to the and z aes, z dm z n S units, r dm dm kg m n U.S. customar units, lb s slug ft ft ft lb ft s 9 -

Parallel Ais Theorem For the rectangular aes with origin at O and parallel centroidal aes, z dm z z dm z dm dm z z dm z z m z m z m z Generalizing for an ais AA and a parallel centroidal ais, md dm 9-5

Moments of nertia of Thin Plates For a thin plate of uniform thickness t and homogeneous material of densit r, the mass moment of inertia with respect to ais AA contained in the plate is AA r rt dm rt AA, area r da Similarl, for perpendicular ais BB which is also contained in the plate, BB rt BB, area For the ais CC which is perpendicular to the plate, CC AA BB rt JC, area rt AA, area BB, area 9-6

Moments of nertia of Thin Plates For the principal centroidal aes on a rectangular plate, AA BB CC rt rt AA BB a b, area rt ma ab, area rt mb AA, mass BB, mass m a b For centroidal aes on a circular plate, AA BB rt AA r, area rt mr CC AA BB mr 9-7

Moments of nertia of a D Bod b ntegration Moment of inertia of a homogeneous bod is obtained from double or triple integrations of the form r r dv For bodies with two planes of smmetr, the moment of inertia ma be obtained from a single integration b choosing thin slabs perpendicular to the planes of smmetr for dm. The moment of inertia with respect to a particular ais for a composite bod ma be obtained b adding the moments of inertia with respect to the same ais of the components. 9-8

Moments of nertia of Common Geometric Shapes 9-9

Sample Problem 9. SOLUTON: With the forging divided into a prism and two clinders, compute the mass and moments of inertia of each component with respect to the z aes using the parallel ais theorem. Add the moments of inertia from the components to determine the total moments of inertia for the forging. Determine the moments of inertia of the steel forging with respect to the z coordinate aes, knowing that the specific weight of steel is 90 lb/ft. 9-0

Sample Problem 9. SOLUTON: Compute the moments of inertia of each component with respect to the z aes. clinders a in., L in.,.5in., in. : ma m 0.089 0.089.590 lb ft s each clinder : V m g m 0.089 lb s 90lb/ft in 78in ft.ft s ft m a 0.089 0.089.5.70 m L m lb ft s a L m 0.089 0.089.5 6.80 lb ft s 9 -

Sample Problem 9. prism : m V g m 0. lb s 90lb/ft 6 78in ft.ft s ft in prism (a = in., b = 6 in., c = in.): z.880 c 0. 6 m b lb ft s a 0. m c 0.9770 lb ft s Add the moments of inertia from the components to determine the total moments of inertia. z.880 0.9770.880.590.70 6.80 z 0.060 lb ft s 9.0 lb ft s 7.80 lb ft s 9 -

Moment of nertia With Respect to an Arbitrar Ais OL = moment of inertia with respect to ais OL OL p dm r dm Epressing r and in terms of the vector components and epanding ields OL z z z The definition of the mass products of inertia of a mass is an etension of the definition of product of inertia of an area dm m z z z dm z dm z z mz mz z z z 9 -

Ellipsoid of nertia. Principal Aes of nertia of a Mass Assume the moment of inertia of a bod has been computed for a large number of aes OL and that point Q is plotted on each ais at a distance OQ The locus of points Q forms a surface known as the ellipsoid of inertia which defines the moment of inertia of the bod for an ais through O.,,z aes ma be chosen which are the principal aes of inertia for which the products of inertia are zero and the moments of inertia are the principal moments of inertia. OL 9 -