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MM222Strength of MaterialsLecture 01Spring 2015Hafiz Kabeer RazaResearch AssociateFaculty of Materials Science and Engineering, GIK InstituteContact: Office G13, Faculty Lobby raza@giki.edu.pk, hkabeerraza@gmail.com, 03344025392Course Learning Outcomes (CLOs)An outcome is the criterion on which the assessment of students will be based.A statement which tells about what the students are able to do after studying this course.In strength of materials (SOM) MM222 the main CLO isApply the concepts of mechanics in engineering applications

Spring 2015By Hafiz Kabeer RazaMM222 Strength of MaterialsCourse contentsTheory of solid mechanics, Elastic and plastic strains, Engineering vs. True stress and strains, the equilibrium and constitutive equations - (05 Lectures)Shear Force and Bending Moment Diagram, Free-body Diagram - (06 Lectures)Factor of Safety, Stress concentration (02 Lectures)Deflection of beams and columns (06 Lectures)Torsional loading of solid and hollow circular shafts (06 Lectures)Pressure in thin and thick walled cylinders / vessels (05 Lectures)Thermal Stresses, Combined Stresses (05 Lectures)Principal stresses and Mohr's Circle of stress and strain (05 Lectures)Energy methods, Computational methods and Yielding Criteria (05 Lectures)Spring 2015By Hafiz Kabeer RazaMM222 Strength of MaterialsGrading PolicyProject (10%)Quizzes 20% Mid-semester Exam. 30%Final Exam. 40%Spring 2015By Hafiz Kabeer RazaMM222 Strength of MaterialsBooksMechanics of Materials by Beer and Johnston, 6th ed. (Course textbook) Strength of Materials and Structures by JOHN CASE, 4th Ed.Mechanics of Solid Materials by J. Lemaitre and J. L. ChabocheMechanical Behavior of Materials by William F. HosfordMechanical Metallurgy by Georgy E. Dieter, SI Metric Ed. Applied Mechanics of Solids by Allan F. Bower (available online at http://solidmechanics.org/contents.htm)ASM Volumes (Related and special topics)E-JournalsSpring 2015By Hafiz Kabeer RazaMM222 Strength of MaterialsLab ExperimentsComparison of yield strength, ultimate tensile strength and fracture strength of different metallic materials.Impact, creep, and fatigue tests of engineering materials.

Spring 2015By Hafiz Kabeer RazaMM222 Strength of MaterialsBasic ConceptsStressStrain Stress strain curvesElastic and Plastic DeformationYieldingFractureEngineering and True stress and strains

Spring 2015By Hafiz Kabeer RazaMM222 Strength of MaterialsMM222Strength of MaterialsLecture 02Spring 2015Hafiz Kabeer RazaResearch AssociateFaculty of Materials Science and Engineering, GIK InstituteContact: Office G13, Faculty Lobby raza@giki.edu.pk, hkabeerraza@gmail.com, 03344025392Basic 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 MaterialsNormal and Shear componentsF is along S (the total stress)Fn = Fsin(90-) = FcosFs = Fcos(90-) = Fsin = Fcos/A = Fsin/Ax = Fsinsin/Ay = Fsincos/A

yxS

nSpring 2015By Hafiz Kabeer RazaMM222 Strength of Materials9 components of stressSuppose n number of forces are acting on a bodyThe total effect of all the forces can be reduced to 9 components of stress as shownSince the body is in equilibrium; there are 3 components of shear symmetric to other 3.So, finally we can reduce the calculations to only 6 components

Spring 2015By Hafiz Kabeer RazaMM222 Strength of MaterialsSign conventionsNormal StressTypeSignTensile+veCompression-veShear StressPlaneDirectionSign+++--++---+-

yxz(001)(010)+y-y-x+x(100)Spring 2015By Hafiz Kabeer RazaMM222 Strength of MaterialsConditions of equilibriumF = 0Fx = 0Fy = 0 = 0anticlock = clock

Spring 2015By Hafiz Kabeer RazaMM222 Strength of MaterialsFree body diagramA free body diagram, sometimes called a force diagram, is a pictorial device, often a rough working sketch, used by engineers and physicists to analyze the forces and moments acting on a body.ExampleSpring 2015By Hafiz Kabeer RazaMM222 Strength of MaterialsExample Equilibrium & Free body diagramsThe structure is designed to support a 30 kN load

The structure consists of a boom and rod joined by pins (zero moment connections) at the junctions and supports

Perform a static analysis to determine the internal force in each structural member and the reaction forces at the supports

Solutions by two methods

Spring 2015By Hafiz Kabeer RazaMM222 Strength of MaterialsMM222Strength of MaterialsLecture 03Spring 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

Solutions by two methods

Suppose that one of components is your arm what kind of stress would you be feeling?Suppose for each componentSpring 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 Materials19Method 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 Materials20Another methodMake the sum of forces zero (in the entire structure)Separate 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 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 ConstantsSpring 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 Materials