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Chapter 6: Mechanical Properties. 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?. I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described.
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1. Chapter 6: Mechanical Properties 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

2. 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

3. Elastic Deformation 1. Initial 2. Small load 3. Unload bonds stretch return to initial d F F Linear- elastic Non-Linear- elastic d Elastic means reversible Time independent!

4. Plastic Deformation (Metals) 1. Initial 2. Small load 3. Unload bonds p lanes stretch still & planes sheared shear d plastic d elastic + plastic F F linear linear elastic elastic d d plastic Plastic means permanent!

5. Engineering Stress • Tensile stress, s: • Shear stress, t: F F F t t Area, A F s Area, A F s F t F F t s F t = F lb N t s = A = f or o 2 2 in m A o original area before loading  Stress has units: N/m2 or lbf/in2

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

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

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

9. Engineering Strain d /2 d L e = e = L L w L o w o o o d /2 L g = Dx/y= tan q • Tensile strain: • Lateral strain: d  = L=L-L0 • Shear strain: q x y 90º - q Strain is always dimensionless. 90º

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

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

12. e L n = - e Poisson's ratio, n • Poisson's ratio, n: metals: n ~ 0.33ceramics: n~ 0.25polymers: n ~ 0.40 Units: E: [GPa] or [psi] n: dimensionless

13. 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)

14. T – E diagram Plot of modulus of elasticity versus temperature for tungsten, steel, and aluminum.

15. Other Elastic Properties M t G simple torsion test g M P • Elastic Bulk modulus, K: P P P D V D V P = - K V o V pressure test: Init. vol =Vo. Vol chg. = DV K o • Special relations for isotropic materials: E E = = K G 3(1-2n) 2(1+n) • Elastic Shear modulus, G: t = Gg stress strain

16. Young’s Moduli: Comparison 1200 10 00 Diamond 8 00 6 00 Si carbide 4 00 Tungsten Carbon fibers only Al oxide Molybdenum Si nitride Steel, Ni C FRE(|| fibers)* 2 00 Tantalum <111> Si crystal Platinum A ramid fibers only Cu alloys <100> 10 0 Zinc, Ti 8 0 Silver, Gold A FRE(|| fibers)* Glass - soda Aluminum 6 0 Glass fibers only Magnesium, G FRE(|| fibers)* 4 0 Tin Concrete GFRE* 2 0 CFRE * G FRE( fibers)* G raphite 10 8 C FRE( fibers) * 6 AFRE( fibers) * Polyester 4 PET PS Epoxy only PC 2 PP HDP E 1 0.8 0.6 Wood( grain) PTF E 0.4 LDPE 0.2 Graphite Ceramics Semicond Metals Alloys Composites /fibers Polymers E(GPa) Composite data based on reinforced epoxy with 60 vol% of aligned carbon (CFRE), aramid (AFRE), or glass (GFRE) fibers. 109 Pa

17. Useful Linear Elastic Relationships • Simple torsion: 2 ML FL Fw o d = d = - n a = o o L 4 r p G E A E A o o o F M = moment a = angle of twist d /2 A o d /2 L Lo Lo w o 2ro • Simple tension: • Material, geometric, and loading parameters all contribute to deflection. • Larger elastic moduli minimize elastic deflection.

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

19. Plastic properties of materials • Yield Strength, sy • Tensile Strength, TS • Ductility • Toughness • Resilience, Ur

20. Yield Strength, sy s tensile stress, sy e engineering strain, e = 0.002 p • Stress at which noticeableplastic deformation has occurred. when ep = 0.002 y = yield strength Note: for 2 inch sample  = 0.002 = z/z  z = 0.004 in

21. Yield Strength : Comparison Graphite/ Metals/ Composites/ Ceramics/ Polymers Alloys fibers Semicond 2 0 00 qt Steel (4140) 10 00 a Ti (5Al-2.5Sn) W (pure) (MPa) 7 00 6 00 cw Cu (71500) 5 00 Mo (pure) a Steel (4140) 4 00 cd y Steel (1020) s , 3 00 ag Al (6061) hr Steel (1020) 2 00 ¨ a Ti (pure) Ta (pure) Hard to measure, hr Cu (71500) Hard to measure in tension, fracture usually occurs before yield. in ceramic matrix and epoxy matrix composites, since since in tension, fracture usually occurs before yield. 100 dry Yield strength, 70 PC 60 Nylon 6,6 a Al (6061) PET 50 humid PVC 40 PP 30 H DPE 20 LDPE Tin (pure) 10 Room T values a = annealed hr = hot rolled ag = aged cd = cold drawn cw = cold worked qt = quenched & tempered

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

23. Graphite/ Metals/ Composites/ Ceramics/ Polymers Alloys fibers Semicond 5000 C fibers Aramid fib 3000 E-glass fib 2000 qt Steel (4140) (MPa) A FRE (|| fiber) 1000 Diamond W (pure) GFRE (|| fiber) a Ti (5Al-2.5Sn) C FRE (|| fiber) a Steel (4140) cw Si nitride Cu (71500) hr Cu (71500) Al oxide Steel (1020) 300 ag Al (6061) a Ti (pure) 200 Ta (pure) a Al (6061) Si crystal wood(|| fiber) 100 <100> strength, TS Nylon 6,6 Glass-soda PET PC PVC GFRE ( fiber) 40 Concrete PP C FRE ( fiber) 30 A FRE( fiber) H DPE Graphite 20 L DPE 10 Tensile wood ( fiber) 1 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.

24. Ductility - L L = x 100 f o % EL L o smaller %EL E ngineering tensile Ao s stress, Lo larger %EL Af Lf e Engineering tensile strain, - A A • Another ductility measure: = o f % RA x 100 A o Ductility can be expressed quantita • Plastic tensile strain at failure: Percent elongation brittle ductile Percent reduction of area

25. 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

26. Toughness small toughness (ceramics) E ngineering tensile large toughness (metals) s stress, very small toughness (unreinforced polymers) e Engineering tensile strain, • 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. Brittle fracture: elastic energy( no apparent plastic deformation takes place fracture) Ductile fracture: elastic + plastic energy(extensive plastic deformation takes place before fracture)

27. 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

28. 1 @ s e U r y y 2 Resilience, Ur • 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 Area under -  curve taken to yielding(the shaded area)

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

30. Elastic Strain Recovery

31. measure size apply known force e.g., of indent after 10 mm sphere removing load Smaller indents d D mean larger hardness. most brasses easy to machine cutting nitrided plastics Al alloys steels file hard tools steels diamond increasing hardness 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.

32. Hardness: Measurement • Rockwell • No major sample damage • Each scale runs to 130 but only useful in range 20-100. • Minor load 10 kg • Major load 60 (A), 100 (B) & 150 (C) kg • A = diamond, B = 1.588 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)

33. Hardness: Measurement Table 6.5

34. problem 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?

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

36. Problem:

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

38. s s large hardening y 1 s y small hardening 0 e hardening exponent: ( ) n n = 0.15 (some steels) s = e K T T to n = 0.5 (some coppers) “true” strain: ln(L/Lo) “true” stress (F/Ai ) Hardening • An increase in sy due to plastic deformation. • Curve fit to the stress-strain response: 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)

39. Design or Safety Factors • Design uncertainties mean we do not push the limit. • Factor of safety, N Often N is between 1.2 and 4 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.

40. d 1045 plain carbon steel: L o s = 310 MPa y 5 TS = 565 MPa F = 220,000N 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 5. d = 0.067 m = 6.7 cm

41. problem

42. 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 sy. • Toughness: The energy needed to break a unit volume of material. • Ductility: The amount of plastic strain that has occurred at fracture.

43. 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 (16.0 10 6 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.

44. 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

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

46. problem A cylindrical metal specimen having an original diameter of 12.8 mm(0.505 in.) and gauge length of 50.80 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 74.17 mm (2.920 in.). Calculate the ductility in terms of percent reduction in area and percent elongation.