6 Stress-Strain Diagrams Nominal Stress: σ = P/A0Where A0 is original cross sectionNominal Strain: ε = δ/L0Where L0 is the original gauge length andδ is the change in gauge lengthNo two stress-strain diagrams for a particular material will be exactly the same since the results depend on:material’s compositionmicroscopic imperfections,way material is manufacturedrate of loadingtemperature
10 DuctilityThe extent of plastic deformation that a material undergoes before fracture.
11 Ductile MaterialA material that can be subjected to large strains before it ruptures.Ductility can be measured by percent elongation or percent reduction in area at the time of fracture.Mild Steel : 38%Mild Steel : 60%
12 Ductility Why use ductile materials? Capable of absorbing shock or energyWhen overloaded, usually exhibit large deformation before failing.
19 Hooke’s LawModulus of ElasticityE = Stress / Strain = E
20 Modulus of ElasticityModulus of Elasticity (E) indicates stiffness of a material.If material is stiff, E is large(for steel, E = 200 GPa)If material is spongy, E is small(for vulcanized rubber, E = 0.70 MPa)
21 Material Properties Strength Ductility Brittleness Stiffness ResilienceToughnessEnduranceRigidity
22 Strain HardeningStrain hardening is used to establish a higher yield point for a materialThe modulus of elasticity stays the same.The ductility decreases.
23 Chapter 3 Lecture Example 1 A tension test for a steel alloy results in the stress-strain diagram shown. Calculate the modulus of elasticity and the yield strength based on a 0.2% offset. Identify on the graph the ultimate stress and the fracture stress.
24 Strain EnergyEnergy stored in a material due to deformation
26 Chapter 3 Lecture Example 2 The stress-strain diagram for an aluminum alloy that is used for making aircraft parts is shown. If a specimen of this material is stressed to 600 MPa, determine the permanent strain that remains in the specimen when the load is released. Also, find the modulus of resilience both before and after the load application.
27 Toughness The area under the stress-strain curve The amount of energy per unit volume that the material dissipates prior to fracture.
32 French mathematician Simeon Denis Poisson Poisson’s Ratio French mathematician Simeon Denis PoissonWhen a deformable body is subjected to a tensile force, not only does it elongate, but it also contracts.Longitudinal is in the direction of the tensile force.
33 Poisson’s Ratio Value of is positive Value of same in tension and compressionRange of to 0.35 constant only in the elastic rangeMaximum possible value of is (Section 10.6)Only Longitudinal force is acting to cause the lateral strain.
36 Derivation of this equation in section 10.6 Modulus of RigidityDerivation of this equation in section 10.6
37 Chapter 3 Lecture Example 3 This is a titanium alloy. Determine the shear modulus, G, the proportional limit, and the ultimate shear stress. Find the maximum d where the material behaves elastically. What is the magnitude of V to cause d?
38 Failure of turbine blade due to creep Failure of steam pipe due to creep
39 CreepCreep is the time-related deformation of a material for which temperature and stress play an important role.Creep results from sustained loading below the measured yield point.Members are designed to resist the effects of creep based on their creep strength, which is the highest amount of stress a member can withstand during a specified amount of time without experiencing creep strain.
41 FatigueFatigue occurs in metals when stress or strain is cycled. It causes a brittle fracture to occur.Members are design to resist fatigue by ensuring that the stress in the member does not exceed its endurance limit.This the maximum stress member can resist when enduring a specified number of cycles.
42 Concept QuestionsHow would you explain strain to one of your engineering classmates?Strain is a linear deformation due to stressPercent increase in length under tension or compression
43 Concept QuestionsHow would you characterize a ductile material? Give several examples.A material that can be subjected to large strains before it ruptures.Steel, wood, natural rubber, brass, copper, gold, aluminum
44 Concept QuestionsHow would you characterize a brittle material? Give several examples.A material that exhibits little or no yielding before failure.Cast iron, concrete, glass, ceramics
45 Concept QuestionsWhat is the difference between engineering stress and true stress?Engineering stress assumes a constant cross sectional area during elongation.
46 A tale of two cities How can you explain Elasticity? Property of material by which it returns to original dimensions on unloading.How can you explain Plasticity?Characteristic of material by which it cannot return to original dimensions on unloading and undergoes permanent deformation.
47 Material PropertiesStrength – Capacity to resist loads – yield stress. Higher y, higher strength.Resilience – Measure of energy absorbed without permanent damage .Toughness – Measure of energy absorbed before fracturing.Ductility – Property of material which allows large deformation before fracture.Brittleness – Property of material which allows little or no yielding before fracture.Endurance – Ability to sustain cyclic loads.Stiffness – Mechanical property indicated by E. Higher E means stiff material. Steeper slope in the Stress – Strain diagram.Rigidity - Mechanical property of material indicated by G.
48 Some fun links Making Swords Balance of material properties – ductility and hardness.Knives
49 3.7 (a) A structural member in a nuclear reactor is made of Zirconium alloy. If an axial load of 4 kip is to be supported by the member, determine the required cross sectional area. Yield stress for Zirconium is 57.5 ksi. Factor of safety (b) What value of load would cause an elongation of 0.02 inch in this member if length of member is 3 ft. Ezr = 14 x 103 ksi.
50 3.26 A short cylindrical block of 2014-T6 Aluminum having an original diameter of 0.5 in and an original length of 1.5 in is placed in the smooth jaws of a vise and a compressive force of 800 lb is applied. Determine (a) The decrease in length. (b) The new diameter.