MECHANICS OF MATERIALSFourth EditionFerdinand P. BeerE. Russell Johnston, Jr.John T. DeWolfLecture Notes:J. Walt OlerTexas Tech UniversityCHAPTER© 2006 The McGraw-Hill Companies, Inc. All rights reserved.Introduction –Concept of Stress© 2006 The McGraw-Hill Companies, Inc. All rights reserved. MECHANICS OF MATERIALSFourthEditionBeer • Johnston • DeWolf1- 2ContentsConcept of StressReview of StaticsStructure Free-Body DiagramComponent Free-Body DiagramMethod of JointsStress AnalysisDesignAxial Loading: Normal StressCentric & Eccentric LoadingShearing StressShearing Stress ExamplesBearing Stress in ConnectionsStress Analysis & Design ExampleRod & Boom Normal StressesPin Shearing StressesPin Bearing StressesStress in Two Force MembersStress on an Oblique PlaneMaximum StressesStress Under General LoadingsState of StressFactor of Safety© 2006 The McGraw-Hill Companies, Inc. All rights reserved. MECHANICS OF MATERIALSFourthEditionBeer • Johnston • DeWolf1- 3Concept of Stress• The main objective of the study of the mechanics of materials is to provide the future engineer with the means of analyzing and designing various machines and load bearing structures.• Both the analysis and design of a given structure involve the determination of stresses and deformations. This chapter is devoted to the concept of stress.© 2006 The McGraw-Hill Companies, Inc. All rights reserved. MECHANICS OF MATERIALSFourthEditionBeer • Johnston • DeWolf1- 4Review of Statics• The structure is designed to support a 30 kN load• Perform a static analysis to determine the internal force in each structural member and the reaction forces at the supports• The structure consists of a boom and rod joined by pins (zero moment connections) at the junctions and supports© 2006 The McGraw-Hill Companies, Inc. All rights reserved. MECHANICS OF MATERIALSFourthEditionBeer • Johnston • DeWolf1- 5Structure Free-Body Diagram• Structure is detached from supports and the loads and reaction forces are indicated• Ayand Cycan not be determined from these equations()()()kN300kN300kN400kN40m8.0kN30m6.00=+=−+==−=−=+===−==∑∑∑yyyyyxxxxxxxCCACAFACCAFAAM• Conditions for static equilibrium:© 2006 The McGraw-Hill Companies, Inc. All rights reserved. MECHANICS OF MATERIALSFourthEditionBeer • Johnston • DeWolf1- 6Component Free-Body Diagram• In addition to the complete structure, each component must satisfy the conditions for static equilibrium• Results:↑=←=→= kN30kN40kN40yxCCAReaction forces are directed along boom and rod()0m8.00=−==∑yyBAAM• Consider a free-body diagram for the boom:kN30=yCsubstitute into the structure equilibrium equation© 2006 The McGraw-Hill Companies, Inc. All rights reserved. MECHANICS OF MATERIALSFourthEditionBeer • Johnston • DeWolf1- 7Method of Joints• The boom and rod are 2-force members, i.e., the members are subjected to only two forces which are applied at member endskN50kN403kN30540=====∑BCABBCABBFFFFFr• Joints must satisfy the conditions for static equilibrium which may be expressed in the form of a force triangle:• For equilibrium, the forces must be parallel to to an axis between the force application points, equal in magnitude, and in opposite directions© 2006 The McGraw-Hill Companies, Inc. All rights reserved. MECHANICS OF MATERIALSFourthEditionBeer • Johnston • DeWolf1- 8Stress Analysis• Conclusion: the strength of member BC is adequateMPa165all=σ• From the material properties for steel, the allowable stress isCan the structure safely support the 30 kN load? MPa159m10314N105026-3=××==APBCσ• At any section through member BC, the internal force is 50 kN with a force intensity or stressofdBC= 20 mm• From a statics analysisFAB= 40 kN (compression) FBC= 50 kN (tension)© 2006 The McGraw-Hill Companies, Inc. All rights reserved. MECHANICS OF MATERIALSFourthEditionBeer • Johnston • DeWolf1- 9Design• Design of new structures requires selection of appropriate materials and component dimensions to meet performance requirements• For reasons based on cost, weight, availability, etc., the choice is made to construct the rod from aluminum (σall= 100 MPa). What is an appropriate choice for the rod diameter?()mm2.25m1052.2m10500444m10500Pa10100N105022622663=×=×===×=××===−−−πππσσAddAPAAPallall• An aluminum rod 26 mm or more in diameter is adequate© 2006 The McGraw-Hill Companies, Inc. All rights reserved. MECHANICS OF MATERIALSFourthEditionBeer • Johnston • DeWolf1- 10Axial Loading: Normal Stress• The normal stress at a particular point may not be equal to the average stress but the resultant of the stress distribution must satisfy∫∫===AavedAdFAPσσ• The resultant of the internal forces for an axially loaded member is normal to a section cut perpendicular to the member axis.APAFaveA=∆∆=→∆σσ0lim• The force intensity on that section is defined as the normal stress.• The detailed distribution of stress is statically indeterminate, i.e., can not be found from statics alone.© 2006 The McGraw-Hill Companies, Inc. All rights reserved. MECHANICS OF MATERIALSFourthEditionBeer • Johnston • DeWolf1- 11• If a two-force member is eccentrically loaded, then the resultant of the stress distribution in a section must yield an axial force and a moment.Centric & Eccentric Loading• The stress distributions in eccentrically loaded members cannot be uniform or symmetric.• A uniform distribution of stress in a section infers that the line of action for the resultant of the internal forces passes through the centroid of the section. • A uniform distribution of stress is only possible if the concentrated loads on the end sections of two-force members are applied at the section centroids. This is referred to as centric loading.© 2006 The McGraw-Hill Companies, Inc. All rights reserved. MECHANICS OF MATERIALSFourthEditionBeer • Johnston • DeWolf1- 12Shearing Stress• Forces P and P’ are applied transversely to the member AB.AP=aveτ• The corresponding average shear stress is,• The resultant of the internal shear force distribution is defined as the shear of the section and is equal to the load P.• Corresponding internal forces act in the plane of section C and are called shearing forces.• Shear stress distribution varies from zero at the member surfaces to maximum values that may be much larger than the average value. • The shear