Chapter 6: Mechanical Properties
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1 ISSUES TO ADDRESS... Stress and strain Elastic behavior: When loads are small, how much reversible deformation occurs? What material resist reversible deformation better? Plastic behavior: At what point does permanent deformation occur? What materials are most resistant to permanent deformation? Toughness and ductility: What are they and how do we measure them? Hardness Chapter 6: Mechanical Properties Chapter 6-1
2 The Need for Considering Force Same material Intensity or Stress F F D 1 D 2 Common engineering questions: What dimension to use? What material to use? Is the geometric design safe/dangerous? Is the design economical? How much change (elongation) each would experience Chapter 6-2
3 Engineering Stress Normal (tensile/compressive) stress, s F t Shear stress, t Area, A o Area, A o F s s = F n N lb in 2 or f 2 A o m in original area before loading F t t = F s A o F s Stress (tensile or shear) has units of N/m 2 (Pa, or 10 6 Pa, i.e., MPa, or lb f /in 2, i.e., psi) Chapter 6-3
4 Class Exercise on Stress (1) For a cylindrical sample with tensile force of N loaded along its long axis, if the initial cross-section area is 10 cm 2, please calculate the tensile stress σ. s F A 0 s F A N 10 N 7 N cm m m Pa 10 MPa Chapter 6-4
5 Class Exercise on Stress (2) For a straight metal bar with a square cross-section, if knowing the tensile stress is along its long axis and it is 10 MPa and the square crosssection has initial edge length of 1 cm, please calculate the tensile force applied. 1 cm 1 cm F s A F s A s b Pa (1cm 1cm ) 10 Pa 10 m 1000 N Chapter 6-5
6 Common States of Stress (1) Simple tension: cable uniaxial F F A o = cross sectional area (when unloaded) s F A o s s Ski lift (photo courtesy P.M. Anderson) Chapter 6-6
7 Common States of Stress (2) Simple compression: uniaxial A o Canyon Bridge, Los Alamos, NM (photo courtesy P.M. Anderson) Balanced Rock, Arches National Park (photo courtesy P.M. Anderson) s F A o Note: s < 0 here by convention Chapter 6-7
8 OTHER COMMON STRESS STATES (ii) Bi-axial tension: Hydrostatic compression: Pressurized tank (photo courtesy P.M. Anderson) s q > 0 Fish under water (photo courtesy P.M. Anderson) s z > 0 s < 0 h Chapter 6-8
9 The Need for Considering Relative Dimension Change or Strain Same material Different material F F F F F D 2 D 2 D 1 D 2 D 2 Common engineering questions: What is the absolute change in length? What is the relative change in length? In what way different materials behave differently when faced with same loading configuration? Chapter 6-9
10 Engineering Strain Tensile strain: e d L o w o d /2 L o For typical materials, when pulled along one direction, it contracts laterally at the same time Shear strain: q x d L /2 g = x/y = tan q y 90º 90º - q Strain is always dimensionless. Adapted from Fig. 6.1(a) and (c), Callister & Rethwisch 8e. Chapter 6-10
11 Class Exercise on Strain A cylindrical sample is subject to tensile force along its long axis. Its original length is 50 mm. After the tensile force is applied, its length become mm. Please calculate the tensile strain along its long axis e L L 0 e L L 0 L L 0 L mm mm.00 mm 0.05 mm 50 mm 0.1 % Chapter 6-11
12 Stress-Strain Testing Typical tensile test machine Typical tensile specimen extensometer specimen Adapted from Fig. 6.2, Callister & Rethwisch 8e. gauge length Adapted from Fig. 6.3, Callister & Rethwisch 8e. (Fig. 6.3 is taken from H.W. Hayden, W.G. Moffatt, and J. Wulff, The Structure and Properties of Materials, Vol. III, Mechanical Behavior, p. 2, John Wiley and Sons, New York, 1965.) Chapter 6-12
13 Materials Deformation under Stress 1 st Stage Elastic Deformation 1. Initial 2. Small load 3. Unload bonds stretch d return to initial Elastic means reversible! F σ Linearelastic ε Chapter 6-13
14 Hooke s Law & Elastic Modulus Hooke's Law: s = E e E : Modulus of Elasticity also known as Young's modulus or elastic modulus Hooke s law only applicable to initial stage, elastic (or reversible) deformation!! Units: E: [GPa] or [psi] s Linear Slope = E e F F simple tension test Chapter 6-14
15 Class Exercise A steel bar 100 mm long and having a square cross section 20 mm on an edge is pulled in tension with a load of 89,000 N, and experiences an elongation of 0.10 mm. Assuming that the deformation is entirely elastic, calculate the elastic modulus of the steel E. Knowing Hooke s law is s = E e F E = s e = A 0 l = Fl 0 b 0 2 l l 0 = (89, 000 N ) ( m) ( m) 2 ( m) = N/m2 = 223 GPa Chapter 6-15
16 Examples Examples when modulus E needs to be low: (Rubber) band for hair O-ring for sealing Examples when elastic modulus needs to be high: Wheels for trains Building/bridges Chapter 6-16
17 E(GPa) 10 9 Pa Range of Young s Moduli for Common Engineering Materials Metals Alloys Tungsten Molybdenum Steel, Ni Tantalum Platinum Cu alloys Zinc, Ti Silver, Gold Aluminum Magnesium, Tin Ceramics & other non-metal inorganic materials Diamond Si carbide Al oxide Si nitride <111> Si crystal <100> Glass - soda Concrete G raphite Polymers Composites /fibers Polyester PET PS PC PP HDP E PTF E LDPE Carbon fibers only C FRE( fibers)* A ramid fibers only A FRE( fibers)* Glass fibers only G FRE( fibers)* GFRE* CFRE * G FRE( fibers)* C FRE( fibers) * AFRE( fibers) * Epoxy only Wood( grain) Based on data in Table B.2, Callister & Rethwisch 8e. Composite data based on reinforced epoxy with 60 vol% of aligned carbon (CFRE), aramid (AFRE), or glass (GFRE) fibers. Chapter 6-17
18 e n L e If material is isotropic and pulling along z direction, e e x y n e e metals: n ~ 0.33 ceramics: n ~ 0.25 polymers: n ~ 0.40 z Units: n: dimensionless z Poisson's ratio, n For typical materials, when pulled along one direction, it contracts laterally at the same time a material property e Poisson's ratio, n: L describing such a behavior -n Latitude strain n < 0.50 density decreases (often for tension) n = 0.50 if no density change e Chapter 6-18 Strain along tensile direction
19 Shear Modulus Elastic Shear modulus, G: t = G g t G g τ τ Special relations for isotropic materials: G E 2(1 + n) Chapter 6-19
20 Class Exercise Explain all the terms in Hooke s Law including their physical meaning and units s = E e If you know tensile load force F and sample initial cross-section area A 0, how to calculate tensile stress? If you know sample initial length l 0 and change in sample length Δl due to tensile force, how to calculate tensile strain? Chapter 6-20
21 Class Exercise Under tensile load of 1000N, for a barshaped sample with square crosssection and edge length of 1cm, if the bar sample initial length is 50mm, please calculate the change in sample length if the sample material is a) Steel with elastic modulus of 200 GPa 1 cm 1 cm b) Polyethylene with elastic modulus of 1 GPa. Assuming for both sample, the deformation is well within elastic deformation limit Chapter 6-21
22 Stress: s s E e Solution for class exercise F 1000 N 1000 N N / m A 1cm 10 m For steel: Pa e steel s E steel 7 10 Pa Pa 5 e d L 0 d e L mm mm 2. 5 m steel steel Chapter 6-22
23 Solution for class exercise (continued) For polyethylene bar, stress is the same s F 1000 N 1000 N N / m A 1cm 10 m Pa s E e e PE s E PE Pa Pa 0.01 e d L 0 d e L mm 0.5 mm 500 PE PE 0 m Chapter 6-23
24 Materials Deformation under Stress 2 nd Stage Plastic (Permanent) Deformation In tension test, for certain materials (especially metals): When stress is low elastic, reversible deformation When stress increases further plastic, irreversible, or permanent deformation starts to occur, also non-linear engineering stress, s Elastic+Plastic at larger stress Elastic initially permanent (plastic) irreversible after load is removed e p engineering strain, e plastic strain Adapted from Fig. 6.10(a), Callister & Rethwisch 8e. Chapter 6-24
25 Stress at which plastic (irreversible or permanent) deformation starts to occur In practice, often use offset yield point for a given strain, conventionally or 0.2% s y tensile stress, s Yield Strength, s y s y = yield strength e p = engineering strain, e Adapted from Fig. 6.10(a), Callister & Rethwisch 8e. Chapter 6-25
26 Examples When you want high yield strength Auto-body iphone frame When you want low yield strength Aluminum food wrap Copper wires Chapter 6-26
27 Yield Strength for Common Engineering Materials since in tension, fracture usually occurs before yield. Hard to measure, in ceramic matrix and epoxy matrix composites, since in tension, fracture usually occurs before yield Metals/ Alloys Steel (4140) qt Ceramics & other non-metal inorganic materials Polymers Composites/ fibers Yield strength, s y (MPa) Ti (5Al-2.5Sn) W (pure) a Cu (71500) Mo (pure) cw Steel (4140) a Steel (1020) cd Al (6061) ag Steel (1020) hr Ti (pure) Ta (pure) a Cu (71500) hr Al (6061) a Hard to measure, PC Nylon 6,6 PET humid PVC PP H DPE dry Room temperature values Based on data in Table B.4, Callister & Rethwisch 8e. a = annealed hr = hot rolled ag = aged cd = cold drawn cw = cold worked qt = quenched & tempered 20 Tin (pure) LDPE Chapter 6-27
28 VMSE: Virtual Tensile Testing Chapter 6-28
29 engineering stress (Ultimate) Tensile Strength, TS Maximum stress on engineering stress-strain curve. Highest stress level beyond which a material will fracture Tensile strength TS s y s y Typical response of a metal strain engineering strain Metals: TS often corresponds to stress when noticeable necking starts. Polymers: TS corresponds to stress when polymer backbone chains are aligned and about to break. Adapted from Fig. 6.11, Callister & Rethwisch 8e. F = fracture Neck acts as stress concentrator, i.e., localized deformation Chapter 6-29
30 Tensile strength, TS (MPa) Tensile Strength for Common Engineering Materials Metals/ Alloys Steel (4140) qt W (pure) Ti (5Al-2.5Sn) Steel (4140) a a Cu (71500) Cu (71500) hr cw Steel (1020) Al (6061) Ti (pure) ag a Ta (pure) Al (6061) a Ceramics & other non-metal inorganic materials Diamond Si nitride Al oxide Si crystal <100> Glass-soda Concrete Graphite Polymers Nylon 6,6 PC PET PVC PP L DPE H DPE Composites/ fibers C fibers Aramid fib E-glass fib A FRE ( fiber) GFRE ( fiber) C FRE ( fiber) wood( fiber) GFRE ( fiber) C FRE ( fiber) A FRE( fiber) wood ( fiber) Room temperature values Based on data in Table B.4, Callister & Rethwisch 8e. 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. Chapter 6-30
31 Examples When you want high tensile strength TS Cables to draw bridges and others Axis for bicycle pedals When you want low tensile strength TS Plastic bag-roll Plastic connector for product label for clothes Chapter 6-31
32 Plastic tensile strain at failure (fracture or breakage). Often defined as fracture strain, or percentage elongation (EL) at fracture E ngineering tensile stress, s Ductility smaller %EL larger %EL % EL L o L f L o A o L o A f x 100 L f Adapted from Fig. 6.13, Callister & Rethwisch 8e. Engineering tensile strain, e Another ductility measurement by (relative) reduction in crosssection area % RA = A o - A o A f x 100 Chapter 6-32
33 Example Tensile Test Results for Different Metals Chapter 6-33
34 Tensile Test for Different Grades of Al (left) and Ti (right) Alloys Al alloys Ti alloys default.asp?catid=217&pageid= Different grades of Al alloys have same elastic modulus E, but different yield strength σ y and tensile strength TS Different grades of Ti alloys have same elastic modulus E, but different yield strength σ y and tensile strength TS Chapter 6-34
35 Example Tensile Test Results for Some Polymers icalchemical/chem htm PMMA Polymethylmethacrylate, PA6 Polyamide/Nylon, ABS Acrylonitrile Butadiene Styrene, PP Polypropylene, HDPE High Density Polyethylene, LDPE Low Density Polyethylene). Chapter 6-35
36 Class Exercise For tensile engineering stress strain curve for brass, determine (a) Elastic modulus; (b) (Offset) yield strength and tensile strength; (c) Maximum load by a cylinder specimen with original diameter of 12.8 mm; (d) Change in length for a 250 mm sample subject to 345 MPa Chapter 6-36
37 Class Exercise (a) Elastic modulus: Slope for initial linear part: E 80MPa/0.001 = 80GPa (b) (Offset) Yield strength: 250 MPa; Tensile strength: 450 MPa (c) F max =σ TS A 0 = Pa (12.8/ m) 2 = 57906N (d) δ = L 0 ε 250 mm 0.07 = 17.5 mm Chapter 6-37
38 Toughness Energy needed to break or fracture an engineering material Approximate by the area under the stress-strain curve. E ngineering tensile stress, s small toughness (ceramics) large toughness (metals) Adapted from Fig. 6.13, Callister & Rethwisch 8e. small toughness (unreinforced polymers) Engineering tensile strain, e Brittle fracture: elastic energy only Ductile fracture: elastic energy + plastic energy Chapter 6-38
39 Hardness Resistance to local plastic deformation 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. most plastics brasses Al alloys easy to machine steels file hard cutting tools nitrided steels diamond increasing hardness Chapter 6-39
40 True Stress & Engineering Stress Note: Sample cross-section area actually reduces when sample is under tensile load True stress s F T A t s s 1 + T True stress is usually higher than engineering stress e Adapted from Fig. 6.16, Callister & Rethwisch 8e. Chapter 6-40
41 Summary 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). Plastic behavior: This permanent deformation behavior occurs when the tensile (or compressive) uniaxial stress reaches s y. Toughness: The energy needed to break a unit volume of material. Ductility: The plastic strain at failure. Hardness: Resistance to local plastic deformation Chapter 6-41
42 HW for Chapter 6 Read chapter 6 and give a statement confirming the reading Calister 8ed: 6.4; 6.24; 6.29(a-e); Chapter 6-42
43 Calister A cylindrical specimen of a titanium alloy having an elastic modulus of 107 GPa and an original diameter of 3.8 mm will experience only elastic deformation when a tensile load of 2000 N is applied. Compute the maximum length of the specimen before deformation if the maximum allowable elongation is 0.42 mm. Chapter 6-43
44 Calister A cylindrical rod 380 mm long, having a diameter of 10.0 mm, is to be subjected to a tensile load. If the rod is to experience neither plastic deformation nor an elongation of more than 0.9 mm when the applied load is 24,500 N, which of the four metals or alloys listed below are possible candidates? Justify your choice(s). Material Modulus of Elasticity (GPa) Yield Strength (MPa) Tensile Strength (MPa) Aluminum alloy Brass alloy Copper Steel alloy Chapter 6-44
45 6.29 A cylindrical specimen of aluminum having a diameter of 12.8 mm and a gauge length of mm is pulled in tension. Use the load elongation characteristics tabulated below to complete parts (a) through (f). (a) Plot the data as engineering stress versus engineering strain. (b) Compute the modulus of elasticity. (c) Determine the yield strength at a strain offset of (d) Determine the tensile strength of this alloy. (e) What is the approximate ductility, in percent elongation? Load Length N lb f mm in ,330 1, ,100 3, ,100 5, ,400 6, ,400 7, ,400 8, ,300 9, ,800 10, ,200 10, ,300 10, ,500 10, ,100 10, ,800 10, ,600 9, ,400 8, Fracture Chapter 6-45
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