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Chapter ISSUES TO ADDRESS... Stress and strain: What are they and stress-strain curve. Elastic behavior: When loads are small, how much deformation occurs? What materials deform least? Plastic behavior: At what point does permanent deformation occur? What materials are most resistant to permanent deformation? Ductility, Toughness, Resilience, Hardness Chapter 6: Mechanical Properties

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Chapter 6 - CONCEPTS OF STRESS AND STRAIN If a load is static or changes relatively slowly with time and is applied uniformly over a cross section or surface of a member, the mechanical behavior may be ascertained by a simple stress–strain test; There are four principal ways in which a load may be applied: namely, tension, compression, shear, and torsion 2

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Chapter 6 - 3

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4 Elastic means reversible Time independent! Elastic Deformation 1. Initial2. Small load3. Unload F bonds stretch return to initial F Linear- elastic Non-Linear- elastic

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Chapter Plastic means permanent! Plastic Deformation (Metals) F linear elastic linear elastic plastic 1. Initial2. Small load3. Unload planes still sheared F elastic + plastic bonds stretch & planes shear plastic

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Chapter Stress has units: N/m 2 or lb f /in 2 Engineering Stress Shear stress, : Area, A F t F t F s F F F s = F s A o Tensile stress, : original area before loading Area, A F t F t = F t A o 2 f 2 m N or in lb =

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Chapter Simple tension: cable Common States of Stress A o = cross sectional area (when unloaded) FF o F A o M R I M M A o 2R2R F s A c Torsion (a form of shear): drive shaft Ski lift

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Chapter (photo courtesy P.M. Anderson) Canyon Bridge, Los Alamos, NM o F A Simple compression: Note: compressive structure member ( < 0 here). (photo courtesy P.M. Anderson) OTHER COMMON STRESS STATES (1) A o Balanced Rock, Arches National Park

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Chapter Bi-axial tension: Hydrostatic compression: Pressurized tank < 0 h (photo courtesy P.M. Anderson) (photo courtesy P.M. Anderson) OTHER COMMON STRESS STATES (2) Fish under water z > 0

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Chapter Tensile strain: Lateral strain: Shear strain: Strain is always dimensionless. Engineering Strain 90º 90º - y xx = x/y = tan L o L L w o /2 L L o w o L=L-L 0

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Chapter Stress-Strain Testing Typical tensile test machine specimen extensometer Typical tensile specimen gauge length We can obtained many important properties of materials from stress-strain diagram Load measured Strain measured

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Chapter (linear)Elastic Properties of materials Modulus of Elasticity, E : (also known as Young's modulus) Hooke's Law: = E Linear- elastic E F F simple tension test

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Chapter Poisson's ratio, Poisson's ratio, : Units: E: [GPa] or [psi] : dimensionless L metals: ~ 0.33 ceramics: ~ 0.25 polymers: ~ 0.40

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Chapter Mechanical Properties Slope of stress strain plot (which is proportional to the elastic modulus) depends on bond strength of metal The magnitude of E is measured of the resistance to separation of adjacent atoms, that is, the interatomic bonding forces So E α (dF/dr)

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Chapter 6 - T – E diagram 15 Plot of modulus of elasticity versus temperature for tungsten, steel, and aluminum.

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Chapter Elastic Shear modulus, G: G = G Other Elastic Properties simple torsion test M M Special relations for isotropic materials: 2(1 ) E G 3(1 2 ) E K Elastic Bulk modulus, K: pressure test: Init. vol =V o. Vol chg. = V P PP P = -K V V o P V K V o stress strain

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Chapter Metals Alloys Graphite Ceramics Semicond Polymers Composites /fibers E(GPa) Composite data based on reinforced epoxy with 60 vol% of aligned carbon (CFRE), aramid (AFRE), or glass (GFRE) fibers. Young’s Moduli: Comparison 10 9 Pa

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Chapter Simple tension: FLFL o EA o L Fw o EA o Material, geometric, and loading parameters all contribute to deflection. Larger elastic moduli minimize elastic deflection. Useful Linear Elastic Relationships F A o /2 L LoLo w o Simple torsion: 2ML o r o 4 G M = moment = angle of twist 2ro2ro LoLo

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Chapter

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Chapter

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Chapter (at lower temperatures, i.e. T < T melt /3) Plastic (Permanent) Deformation Simple tension test: engineering stress, engineering strain, Elastic+Plastic at larger stress permanent (plastic) after load is removed pp plastic strain Elastic initially y Yielding stress

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Chapter 6 - Plastic properties of materials Yield Strength, y Tensile Strength, TS Ductility Toughness Resilience, U r 22

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Chapter Stress at which noticeable plastic deformation has occurred. when p = Yield Strength, y y = yield strength Note: for 2 inch sample = = z/z z = in tensile stress, engineering strain, yy p = 0.002

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Chapter Room T values a = annealed hr = hot rolled ag = aged cd = cold drawn cw = cold worked qt = quenched & tempered Yield Strength : Comparison

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Chapter Tensile Strength, TS Metals: occurs when noticeable necking starts. Polymers: occurs when polymer backbone chains are aligned and about to break. yy strain Typical response of a metal F = fracture or ultimate strength Neck – acts as stress concentrator engineering TS stress engineering strain Maximum stress on engineering stress-strain curve.

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Chapter Tensile Strength : Comparison Room Temp. values a = annealed hr = hot rolled ag = aged cd = cold drawn cw = cold worked qt = quenched & tempered AFRE, GFRE, & CFRE = aramid, glass, & carbon fiber-reinforced epoxy composites, with 60 vol% fibers.

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Chapter Plastic tensile strain at failure: Ductility Another ductility measure: 100x A AA RA% o fo - = x 100 L LL EL% o of Engineering tensile strain, Engineering tensile stress, smaller %EL larger %EL LfLf AoAo AfAf LoLo Percent reduction of area Percent elongation Ductility can be expressed quantita brittle ductile

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Chapter 6 - Ductility Brittle materials are approximately considered to be those having a fracture strain of less than about 5%. Brittle materials have little or no plastic deformation upon fracture 28

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Chapter Energy to break a unit volume of material or it is a measure of the ability of a material to absorb energy up to fracture( or impact resistance) Approximate by the area under the stress-strain curve. Toughness Brittle fracture: elastic energy( no apparent plastic deformation takes place fracture) Ductile fracture: elastic + plastic energy(extensive plastic deformation takes place before fracture) very small toughness (unreinforced polymers) Engineering tensile strain, Engineering tensile stress, small toughness (ceramics) large toughness (metals)

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Chapter Fracture toughness is a property indicative of a material’s resistance to fracture when a crack is present For a material to be tough, it must display both strength and ductility; and often, ductile materials are tougher than brittle ones

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Chapter Resilience, U r Ability of a material to store energy when it is deformed elastically, and then, upon unloading, to have this energy recovered –Energy stored best in elastic region If we assume a linear stress-strain curve this simplifies to yyr 2 1 U Area under - curve taken to yielding(the shaded area)

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Chapter Thus, resilient materials are those having high yield strengths and low module of elasticity; such alloys would be used in spring applications. Modulus of resilience ( is the area under curve)

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Chapter Elastic Strain Recovery

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Chapter Hardness Resistance to permanently indenting the surface, or is a measure of a materials resistance to deformation by surface indentation or by abrasion. Large hardness means: --resistance to plastic deformation or cracking in compression. --better wear properties. e.g., 10 mm sphere apply known force measure size of indent after removing load d D Smaller indents mean larger hardness. increasing hardness most plastics brasses Al alloys easy to machine steelsfile hard cutting tools nitrided steelsdiamond

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Chapter Hardness: Measurement Rockwell –No major sample damage –Each scale runs to 130 but only useful in range –Minor load 10 kg –Major load 60 (A), 100 (B) & 150 (C) kg A = diamond, B = mm ball of steel, C = diamond the hardness can be taken directly from the machine, so it is a quick test HB = Brinell Hardness - P= 500, 1500, 3000 kg –TS (psia) = 500 x HB –TS (MPa) = 3.45 x HB (The HB and tensile strength relationship)

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Chapter Hardness: Measurement Table 6.5

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Chapter 6 - problem MaterialYielding strength ( MPa) Tensile strength ( MPa) Strain at fracture Fracture strength (MPa) Elastic modulus (GPa) A B C D EFracture before yielding a) Which experience the greatest percent reduction in area? Why? b) Which is the strongest? Why? c) Which is the stiffest? Why? d) Which is the hardest? Why?

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Chapter a)Material B will experience the greatest percent area reduction since it has the highest strain at fracture, and, therefore is most ductile. b)Material D is the strongest because it has the highest yield and tensile strengths. c)Material E is the stiffest because it has the highest elastic modulus. stiffness=E=σ/ε, the higher E the material more stiffest d)Material D is the hardest because it has the highest tensile strength.

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Chapter Problem:

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Chapter True Stress & Strain Note: Surf.Are. changes when sample stretched True stress True Strain

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Chapter Hardening Curve fit to the stress-strain response: T K T n “true” stress (F/A i ) “true” strain: ln(L/L o ) hardening exponent: n =0.15 (some steels) to n =0.5 (some coppers) An increase in y due to plastic deformation. large hardening small hardening y 0 y 1 The region of the true stress-strain curve from the onset of plastic deformation to the point at which necking begins may be approximated by (n and K constants)

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Chapter Design uncertainties mean we do not push the limit. Factor of safety, N Often N is between 1.2 and 4 Design or Safety Factors Safe stress, or The choice of an appropriate value of N is necessary. If N is too large, then component overdesign will result, that is, either too much material or a material having a higher- than-necessary strength will be used. Values normally range between1.2 and 4.0. Selection of N will depend on a number of factors, including economics, previous experience, the accuracy with which mechanical forces and material properties may be determined, and, most important, the consequences of failure in terms of loss of life and/or property damage.

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Chapter Example: Calculate a diameter, d, to ensure that yield does not occur in the 1045 carbon steel rod below. Use a factor of safety of plain carbon steel: y = 310 MPa TS = 565 MPa F = 220,000N d L o d = m = 6.7 cm

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Chapter problem

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Chapter

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Chapter Stress and strain: These are size-independent measures of load and displacement, respectively. Elastic behavior: This reversible behavior often shows a linear relation between stress and strain. To minimize deformation, select a material with a large elastic modulus (E or G). Toughness: The energy needed to break a unit volume of material. Ductility: The amount of plastic strain that has occurred at fracture. Summary Plastic behavior: This permanent deformation behavior occurs when the tensile (or compressive) uniaxial stress reaches y.

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Chapter 6 - problem A brass alloy is known to have a yield strength of 240 MPa (35,000 psi), a tensile strength of 310 MPa (45,000 psi), and an elastic modulus of 110 GPa ( psi). A cylindrical specimen of this alloy 15.2 mm (0.60 in.) in diameter and 380 mm (15.0 in.) long is stressed in tension and found to elongate 1.9 mm (0.075 in.). On the basis of the information given, is it possible to compute the magnitude of the load that is necessary to produce this change in length? If so, calculate the load. If not, explain why. 47

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Chapter We are asked to ascertain whether or not it is possible to compute, for brass, the magnitude of the load necessary to produce an elongation of 1.9 mm (0.075 in.). It is first necessary to compute the strain at yielding from the yield strength and the elastic modulus, and then the strain experienced by the test specimen. Then, if ε(test) < ε(yield) deformation is elastic, and the load may be computed using Equations 6.1 and 6.5. However, if ε(test) > ε(yield) computation of the load is not possible inasmuch as deformation is plastic and we have neither a stress-strain plot nor a mathematical expression relating plastic stress and strain. We compute these two strain values as

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Chapter Therefore, computation of the load is not possible since ε(test) > ε(yield).

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Chapter 6 - problem A cylindrical metal specimen having an original diameter of 12.8 mm(0.505 in.) and gauge length of mm (2.000 in.) is pulled in tension until fracture occurs. The diameter at the point of fracture is 8.13 mm (0.320 in.), and the fractured gauge length is mm (2.920 in.). Calculate the ductility in terms of percent reduction in area and percent elongation. 50

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Chapter This problem calls for the computation of ductility in both percent reduction in area and percent elongation. Percent reduction in area is computed using Equation 6.12 as in which d0 and df are, respectively, the original and fracture cross-sectional areas. Thus, While, for percent elongation, we use Equation 6.11 as

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