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MM222Strength of MaterialsLecture 04Spring 2015Hafiz Kabeer RazaResearch AssociateFaculty of Materials Science and Engineering, GIK InstituteContact: Office G13, Faculty Lobby raza@giki.edu.pk, hkabeerraza@gmail.com, 03344025392Revision Last WeekCLOsCourse contentsGrading policyBooksBasic concepts

Spring 2015By Hafiz Kabeer RazaMM222 Strength of MaterialsCLOsCLO-1: Demonstrate the fundamental concepts of stress and strain and strain-stress equations in order to solve problems for simple three-dimensional elastic solids. [PLO1]CLO-2: Analyze the problems relating to torsional deformation of bars and other simple tri-dimensional structures.[PLO2]CLO-3: Solve problems relating to pure and non-uniform bending of beams, columns and other simple structures.[PLO2]CLO-4: Evaluate the stresses from the strain energies data in elastic solids.[PLO2]PLO: Program Learning OutcomeSpring 2015By Hafiz Kabeer RazaMM222 Strength of MaterialsPLOsEngineering KnowledgeProblem AnalysisDesign/Development of SolutionsInvestigationModern Tool UsageThe Engineer and SocietyEnvironment and SustainabilityEthicsIndividual and Team WorkCommunicationProject ManagementLifelong LearningSpring 2015By Hafiz Kabeer RazaMM222 Strength of MaterialsBooksMechanics of Materials by Beer and Johnston, 6th ed. (Course textbook) Mechanics of Materials by Vable, 2nd ed. Strength of Materials and Structures by JOHN CASE, 4th Ed.Mechanics of Solid Materials by J. Lemaitre and J. L. ChabocheMechanical Metallurgy by George E. Dieter, SI Metric Ed. Applied Mechanics of Solids by Allan F. Bower (available online at http://solidmechanics.org/contents.htm)Spring 2015By Hafiz Kabeer RazaMM222 Strength of MaterialsBasic ConceptsEngineering and True stress and strainsNormal and Shear stressesStress tensor (9 components)Sign conventionsConditions of equilibriumFree body diagramHookes LawGeneralized Hookes LawSpring 2015By Hafiz Kabeer RazaMM222 Strength of MaterialsHookes LawFor springExtension in spring is proportional to the applied forceSpring constantFor elastic solidsStress is proportional to strainElastic modulii (E and G)Normal and shear stress-strain relations = E OR = GNormal and shear strains

Spring 2015By Hafiz Kabeer RazaMM222 Strength of MaterialsGeneralized form of Hookes LawSpring 2015By Hafiz Kabeer RazaMM222 Strength of MaterialsGeneralized form of Hookes LawAssumption: applied normal stresses do not produce shear strains on x, y or z planes while the applied shear stresses do not produce normal stresses on the x, y or z planesTotal strains in x, y and z direction is found by superposition of the components; Shear strains are simple related to the applied shear stresses. i.e.

Spring 2015By Hafiz Kabeer RazaMM222 Strength of MaterialsElastic Constants

Summation of the equations of Hookes Law

Spring 2015By Hafiz Kabeer RazaMM222 Strength of MaterialsRelation b/w engineering and true stress and strainsSpring 2015By Hafiz Kabeer RazaMM222 Strength of MaterialsMM222Strength of MaterialsLecture 05Spring 2015Hafiz Kabeer RazaResearch AssociateFaculty of Materials Science and Engineering, GIK InstituteContact: Office G13, Faculty Lobby raza@giki.edu.pk, hkabeerraza@gmail.com, 03344025392Example Equilibrium & Free body diagramsThe structure is designed to support a 30 kN load

The structure consists two componentsPerform a static analysis to determine the internal force in each structural member and the reaction forces at the supports

Spring 2015By Hafiz Kabeer RazaMM222 Strength of MaterialsStructure Free Body DiagramStructure is detached from supports andthe loads and reaction forces are indicatedAy and Cy can not be determined from these equations

Conditions for static equilibrium:

Important: The direction of Reactions in free body diagram may be taken along any axis.Spring 2015By Hafiz Kabeer RazaMM222 Strength of MaterialsComponent Free-Body DiagramIn addition to the complete structure, each component must satisfy the conditions for static equilibrium

Consider a free-body diagram for the boom:

substitute into the structure equilibrium equationResults:

Reaction forces are directed along boom and rod

Spring 2015By Hafiz Kabeer RazaMM222 Strength of Materials15Method of JointsThe boom and rod are 2-force members, i.e., the members are subjected to only two forces which are applied at member endsFor equilibrium, the forces must be parallel to to an axis between the force application points, equal in magnitude, and in opposite directions

Joints must satisfy the conditions for static equilibrium which may be expressed in the form of a force triangle:

Spring 2015By Hafiz Kabeer RazaMM222 Strength of Materials16Another methodMake the sum of forces zero (in the entire structure) 1st condition of equilibriumCut the joint B from rest of the structureNow we have three forces, say 30 kN, F1 and F2Apply the condition of static equilibrium and solveIn the same way solve for the other two joints

Important for this method: the direction of force is always taken outward, if the force in component is unknownIf the force is known: tensile is taken outward while compression is taken inward

Spring 2015By Hafiz Kabeer RazaMM222 Strength of MaterialsConclusions of exampleTwo force memberthe members subjected to only two forces which are applied at member endsThe directions of forces at supports can be determined from the orientation and shape of component attachedConcept of Internal force

Spring 2015By Hafiz Kabeer RazaMM222 Strength of MaterialsConcept of StressDetermination of forces at supports [1st]Determination of internal forces in members [2nd]These two are the initial steps of problem solving in mechanics.However this does not tell about whether the structure (and the individual members) can safely bear the applied load!!!Intensity of force the stressFor every material in application, a limiting stress value is defined; it cannot be used beyond this stress valueUnits of stress (Pa, psi; 1.0 psi = 6895 Pa)

Spring 2015By Hafiz Kabeer RazaMM222 Strength of MaterialsStress analysisThe next step of problem solving in mechanics is stress analysis.The internal forces in the last example were +50 kN (FBC) and -40 kN (FAB). The cross sections are AB=3050 mm2 and BC=20 mm dia.For example the allowable stress (all) for steel is 165 MPa. Check whether the structure will work safely!!!For aluminum all = 100 MPa, check the safety and if not safe, calculate the required minimum cross sections

Spring 2015By Hafiz Kabeer RazaMM222 Strength of MaterialsAxial LoadingAxial Loading Normal stressWhen the applied load is perpendicular to the cross-section of interestExample: last problem solved, lifting a bucket

Spring 2015By Hafiz Kabeer RazaMM222 Strength of MaterialsMM222Strength of MaterialsLecture 06Spring 2015Hafiz Kabeer RazaResearch AssociateFaculty of Materials Science and Engineering, GIK InstituteContact: Office G13, Faculty Lobby raza@giki.edu.pk, hkabeerraza@gmail.com, 03344025392Shear LoadingShear Shear StressWhen the applied load is parallel to the cross-section of interestExample: walking on ground, the force between tire and road, force for twisting a rod, shearing (cutting) by a scissor, a bold joining two plates under axial loadingDefinition of shear stress: Force/AreaSingle shear, double shearShearing force is taken only from one side of the cross-section

Spring 2015By Hafiz Kabeer RazaMM222 Strength of MaterialsBearing stressAs a reaction of applied shear, the bolts/pins/rivets exert an opposite force on the contact surface of the sheets/plates (or any thing to which they are connected)That force divided by the area is named as bearing stressHowever the distribution of the force is quite complicated so the area in bearing stress is taken as product of plate/sheet thickness and diameter of the bolt/pin/rivet.Remember: thickness taken will be along the axis of bolt/pin/rivet.

Spring 2015By Hafiz Kabeer RazaMM222 Strength of MaterialsCalculation of Shearing and Bearing stressesRecall the last exampleNote (all dimensions in mm)Rod BC: Dia = 20 & Flat ends = 2040)Boom AB: Cross section = 5030At C: a simple pin of dia 25At A: pin (25) fitted into double bracket (25, 25)At B: the members connected by a pin (25 dia) & Load is hanged through a U-shaped bracket. BC is connected through flat end while AB fitted with clevis (branch thickness of 20).

Spring 2015By Hafiz Kabeer RazaMM222 Strength of MaterialsCalculation of Shearing and Bearing stressesShearing stressesAt CSingle shearP = 50 kNDia = 25mm = 102 MPa At ADouble shearP = F/2Dia = 25mm = 40.7 MPa

Spring 2015By Hafiz Kabeer RazaMM222 Strength of MaterialsCalculation of Shearing and Bearing stressesShearing stressesAt B5 segmentsShearing sections = 4Forces from one sideDE (At E)P = 15 kNDEG (At G)P = 25 kN

Spring 2015By Hafiz Kabeer RazaMM222 Strength of MaterialsCalculation of Shearing and Bearing stressesBearing stressesAt C:In Member: d =25, t=20, P = 50 kNb = 100 MPa Same in supportAt A:In member: d=25, t=30, P=40 kN b = 53.3 MPaIn bracket: d=25, t=50, P=40 kN b = 32 MpaHalf on each branch

Spring 2015By Hafiz Kabeer RazaMM222 Strength of Materials28Maximum Normal StressesMaximum normal stress occurs where the cross section is minimumThe portions near holesCross-section at hole at C20(40-25)= 166.7 MPaCross-section at hole at B30(50-25)= 53.3 MPaCompare with the average stress values (159 MPa in BC and 26.7 MPa in AB)

Spring 2015By Hafiz Kabeer RazaMM222 Strength of Materials