Properties of MaterialsCompetenciesSTRUCTURE OF MATTERCategories of chemical bondsSlide 5STATES OF MATTERNUCLEATION OF GRAINSSlide 8STRENGTH PROPERTIESProblemsSlide 11Slide 12Slide 13Slide 14Slide 15ProblemSlide 17Slide 18Slide 19Slide 20Slide 21Slide 22SURFACE PROPERTIESSlide 24Slide 25Slide 26Slide 27Properties of Material (Iron and Steel)Properties of MaterialTYPES OF STEEL MAKING FURNACESSlide 31Slide 32Slide 33Slide 34Slide 35Chapter 2 IT208 1Properties of MaterialsChapter 2Chapter 2IT208 2CompetenciesDefine Stress, Strain, True Stress and Engineering Stress, Yield Strength, and CompressionCalculate Stress, Strain, True Stress and Engineering Stress, Yield Strength, Safety Factor and CompressionList and describe the 4 categories of chemical bonds.Define material fatigue and creep List materials used to produce iron leading to steel.Chapter 2IT208 3STRUCTURE OF MATTERAll properties of materials are a function of their structure. If the atomic structure, bonding structure, crystal structure, and the imperfections in the material are known, the properties of the material can be determined.Matter is composed of atoms, which are the smallest units of individual elements. Atoms are composed of proton, neutrons, and electrons. Atoms can combine to form molecules, which are the smallest units of chemical compounds. The atoms are held together by chemical “bonds.”Chapter 2IT208 4Categories of chemical bondsIn chemical bonds, atoms can either transfer or share their valence electronsionic – In the extreme case where one or more atoms lose electrons and other atoms gain them in order to produce a noble gas electron configuration, the bond is called an ionic bond.covalent - Covalent chemical bonds involve the sharing of a pair of valence electrons by two atoms, in contrast to the transfer of electrons in ionic bonds. Such bonds lead to stable molecules if they share electrons in such a way as to create a noble gas configuration for each atom. metallic - van der waal -Chapter 2 IT208 5Chapter 2IT208 6STATES OF MATTERGaseous State – individual atoms or molecules have little or not attraction to each other. They are in constant motion and are continuously bouncing off one other.Boiling Point – The temperature at which gaseous particles begin to bond to each other. To continue into the liquid state the heat of vaporization must be removed or to move from liquid to gas the heat must be added.Liquid State – having bonds of varying lengths relating to the viscosity of a materialSolid State – has a definite structureMelting point – the temperature at which enough energy to break one bond of a crystal. All true solids have a definite melting point.Chapter 2IT208 7NUCLEATION OF GRAINSThe phenomenon when the temperature of molten material is lowered to the melting point, little crystals or nuclei are formed at many points in the liquid. After the grains have been nucleated and grown together to form a solid, the process of grain growth occurs. Slow cooling to room temperature allows for larger grains to form, while rapid cooling only allows for small grains to form.Chapter 2IT208 8NUCLEATION OF GRAINSAtoms or particles align themselves into planes within each crystal, there is a uniform distance between particles. These plains can slide over each other, the more ductile the material becomes, the more ways slip can occur. A materials density, ductility, and malleability are a factor or crystalline structure resulting in planes for slip to occur.Chapter 2IT208 9STRENGTH PROPERTIESStress - defined as the load per unit cross section of area. CompressionTorsional Tension – forces pulling an object in opposite directions. If the load or force pulling on the material is divided by the cross-sectional area of the bar, the result is the tensile stress applied to the sample AREA:Width x HeightPi r2 Stress generally given in psi (english) or Pascal (metric))/()/(minareakglbloadStress APS Chapter 2IT208 10Problems1. If a tensile force of 500 lb is placed on a 0.75-in. diameter bar, what is the stress on the bar?  22/1130375.14.3500inlb2rloadS1130 lb/in^2Chapter 2IT208 11Problems2. What is the tensile strength of a metal if a 0.505 in.-diameter bar withstands a load of 15,000 lb before breaking? 2rloadS 22/750002525.14.315000inlb75,000 lbs/ in^2Chapter 2IT208 12Problems3. A cable in a motor hoist must lift a 700-lb engine. The steel cable is 0.375 in. in diameter. What is the stress in the cable? APS 6338 lb/in29.6337)2375.(.7002lbChapter 2IT208 13STRENGTH PROPERTIESStrain - the elongation of a specimen per unit of original length lengt horiginalelon gatio nstrain  lengthorigina llengthoriginallenthexte n d edstrainoozz-ze Chapter 2IT208 14STRENGTH PROPERTIESElastic limit - The maximum applied stress that metals and other materials can be stretch and rebound in much the same manner as a rubber band also called proportional limit. The rest of the curve, to the right of the elastic limit, is the plastic region.Chapter 2IT208 15STRENGTH PROPERTIESTensile strength – or ultimate strength is the maximum stress that a bar will withstand before failing and is “e” shown as point T on the curve. Rupture strength - or breaking strength is the stress at which at a bar breaks, point R on Figure 2-16. Yield strength - the engineering design strength of the material •The point intersection determined by measuring a distance of 0.002 inch/inch on the strain axis, then drawing a straight line parallel to the straight-line portion of the curve. (Figure 2-17).Chapter 2IT208 16Problem4. If a steel cable is rated to take 800 lb and the steel has a yield strength of 90,000 psi, what is the diameter of the cable? (Ignore safety factor.)D = 0.11 in. StressLoadD *2inpsilbsD 1063.14.3000,90800*2 Chapter 2IT208 17STRENGTH PROPERTIESModulus of Elasticity (Young’s modulus) is the change in stress divided by the change in strain while the material is in the elastic region. StrainStresslengthorigianlelongationareaload//oozzzAP/)(/Chapter 2IT208 18Problem5. If a tensile part in a machine is designed to hold 25,000 lb and the part is made from a material having yield strength of 75,000 psi, what diameter must the part have? 2rloadSsloadr 