mm222 lec 5
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MM222Strength of MaterialsLecture 05Spring 2015Hafiz Kabeer RazaResearch AssociateFaculty of Materials Science and Engineering, GIK InstituteContact: Office G13, Faculty Lobby [email protected], [email protected], 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 Materials4Method 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 Materials5Another 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 MaterialsShear 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 loadingSingle shear, double shearShearing force is taken only from one side of the cross-section
Spring 2015By Hafiz Kabeer RazaMM222 Strength of Materials