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Mechanical Behavior of Materials ME 2105 Dr. R. Lindeke, Ph.D.

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1 Mechanical Behavior of Materials ME 2105 Dr. R. Lindeke, Ph.D.

2 Chapter 6: Behavior Of Material Under Mechanical Loads = Mechanical Properties. Stress and strain: What are they and why are they used instead of load and deformation Elastic behavior: Recoverable Deformation of small magnitude Plastic behavior: Permanent deformation We must consider which materials are most resistant to permanent deformation? Toughness and ductility: Defining how much energy that a material can take before failure. How do we measure them? Hardness: How we measure hardness and its relationship to material strength

3 Stress-Strain : Testing Uses Standardized methods developed by ASTM for Tensile Tests it is ASTM E8 Typical tensile test machine Adapted from Fig. 6.3, Callister 7e. (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.) specimen extensometer Typical tensile specimen (ASTM A-bar) gauge length

4 Comparison of Units: SI and Engineering Common (in US) UnitSIEng. Common ForceNewton (N)Pound-force (lb f ) Areamm 2 or m 2 in 2 StressPascal (N/m 2 ) or MPa (10 6 pascals) psi (lb f /in 2 ) or Ksi (1000 lb f /in 2 ) Strain (Unitless!)mm/mm or m/min/in Conversion FactorsSI to Eng. CommonEng. Common to SI ForceN*4.448 = lbfLbf*0.2248 = N Area Imm 2 *645.16 = in 2 in 2 *1.55x10 -3 = mm 2 Area IIm 2 *1550 = in 2 in 2 * 6.452x10 -4 = m 2 Stress I - aPascal * 1.450x10 -4 = psipsi * 6894.76 = Pascal Stress I - bPascal * 1.450x10 -7 = KsiKsi * 6.894 x10 6 = Pascal Stress II - aMPa * 145.03 = psipsi * 6.89x 10 -3 = MPa Stress II - bMPa * 1.4503 x 10 -1 = KsiKsi * 6.89 = MPa One other conversion: 1 GPa = 10 3 MPa

5 The Engineering Stress - Strain curve Divided into 2 regions ELASTIC PLASTIC

6  Stress has units: N/m 2 (MPa) 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 = we can also see the symbol ‘s’ used for engineering stress

7 Simple tension: cable Note:  = M/A c R here. Where M is the “Moment” A c shaft area & R shaft radius Common States of Stress A o = cross sectional area (when unloaded) FF o   F A o   F s A  M M A o 2R2R F s A c Torsion (a form of shear): drive shaft Ski lift (photo courtesy P.M. Anderson)

8 Linear: Elastic Properties Modulus of Elasticity, E: (also known as Young's modulus) Hooke's Law:  = E   Linear- elastic E  Units: E: [GPa] or [psi] F A o  /2  L LoLo w o Here: The Black Outline is Original, Green is after application of load

9 Typical Example: – Aluminum tensile specimen, square X-Section (16.5 mm on a side) and 150 mm long – Pulled in tension to a load of 66700 N – Experiences elongation of: 0.43 mm Determine Young’s Modulus if all the deformation is recoverable

10 Figure 6.2 Load-versus-elongation curve obtained in a tensile test. The specimen was aluminum 2024-T81.

11 Figure 6.3 Stress-versus-strain curve obtained by normalizing the data of Figure 6.2 for specimen geometry.

12 Solving:

13 (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

14 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  

15 Tensile (parallel to load) strain: Lateral (Normal to load) strain: Shear strain: Strain is always Dimensionless! Engineering Strain (resulting from engineering stress):  90º 90º -  y xx   =  x/y = tan   L o   L  L w o  /2  L L o w o We often see the symbol ‘e’ used for engineering strain Here: The Black Outline is Original, Green is after application of load

16 Figure 6.11 The Poisson’s ratio (ν) characterizes the contraction perpendicular to the extension caused by a tensile stress. metals:  0.33 ceramics:  0.25 polymers:  0.40

17 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 E is Modulus of Elasticity is Poisson’s Ratio


19 Figure 6.4 The Yield Strength is defined relative to the intersection of the stress–strain curve with a “0.2% offset.” Yield strength is a convenient indication of the onset of plastic deformation.

20 Figure 6.10 For a low-carbon steel, the stress-versus- strain curve includes both an upper and lower yield point.

21 Figure 6.5 Elastic recovery occurs when stress is removed from a specimen that has already undergone plastic deformation.

22 vertical).

23 Lets Try an Example Problem Load (N)len. (mm)len. (m)  (l) 050.80.05080 1270050.8250.0508252.5E-05 2540050.8510.0508515.1E-05 3810050.8760.0508767.6E-05 5080050.9020.0509020.000102 7620050.9520.0509520.000152 8910051.0030.0510030.000203 9270051.0540.0510540.000254 10250051.1810.0511810.000381 10780051.3080.0513080.000508 11940051.5620.0515620.000762 12830051.8160.0518160.001016 14970052.8320.0528320.002032 15900053.8480.0538480.003048 16040054.3560.0543560.003556 15950054.8640.0548640.004064 15150055.880.055880.00508 12470056.6420.0566420.005842 GIVENS (load and length as test progresses):

24 Leads to the following computed Stress/Strains: e stress (Pa)e str (MPa)e. strain 000 98694715.798.6947160.000492 197389431197.389430.001004 296084147296.084150.001496 394778863394.778860.002008 592168294592.168290.002992 692417257692.417260.003996 720393712720.393710.005 796551839796.551840.0075 837739398837.73940.01 927885752927.885750.015 997049766997.049770.02 11633542471163.35420.04 12356267551235.62680.06 12465064881246.50650.07 12395123741239.51240.08 11773424751177.34250.1 969073311969.073310.115

25 Leads to the Eng. Stress/Strain Curve: T. Str.  1245 MPa Y. Str.  742 MPa %el  11.5% F. Str  970 MPa E  195 GPa (by regression)


27 Figure 6.7 Neck down of a tensile test specimen within its gage length after extension beyond the tensile strength. (Courtesy of R. S. Wortman.)

28 Figure 6.8 True stress (load divided by actual area in the necked-down region) continues to rise to the point of fracture, in contrast to the behavior of engineering stress. (From R. A. Flinn and P. K. Trojan, Engineering Materials and Their Applications, 2nd ed., Houghton Mifflin Company, 1981, used by permission.)

29 True Stress & Strain Note: Stressed Area changes when sample is deformed (stretched) True stress True Strain Adapted from Fig. 6.16, Callister 7e.

30 Figure 6.25 Forest of dislocations in a stainless steel as seen by a transmission electron microscope [Courtesy of Chuck Echer, Lawrence Berkeley National Laboratory, National Center for Electron Microscopy.]

31 Returning to our Example – True Properties T. StressT. Strain 00 98.743285920.000492 197.58759790.001003 296.52710760.001495 395.5715290.002006 593.94013630.002988 695.18420030.003988 723.99568070.004988 802.5259780.007472 846.11679170.00995 941.80403850.014889 1016.9907610.019803 1209.8884170.039221 1309.7643610.058269 1333.7619420.067659 1338.6733640.076961 1295.0767220.09531 1080.5167410.108854 Necking Began

32 Strain Hardening Curve fit to the stress-strain response:  T  K  T  n “true” stress (F/A i ) “true” strain: ln(L i /L o ) hardening exponent: n =0.15 (some steels) n = 0.5 (some coppers) An increase in  y due to continuing plastic deformation.   large Strain hardening small Strain hardening  y 0  y 1

33 We Compute  T = K  T n to complete our True Stress True Strain plot (plastic data to necking) Take Logs of both  T and  T Regress the values from Yielding to Necking Gives a value for n (slope of line) and K (its  T when  T =1) Plot as a line beyond necking start

34 Stress Strain Plot w/ True Values n  0.242K  2617 MPa


36 TOUGHNESS High toughness =High yield strength and ductility Dynamic (high strain rate) loading condition (Impact test) 1. Specimen with notch- Notch toughness 2. Specimen with crack- Fracture toughness Is a measure of the ability of a material to absorb energy up to fracture Important Factors in determining Toughness: 1. Specimen Geometry & 2. Method of load application Static (low strain rate) loading condition (tensile stress-strain test) 1. Area under stress vs strain curve up to the point of fracture.

37 Figure 6.9 The toughness of an alloy depends on a combination of strength and ductility.

38 Energy to break a unit volume of material Approximate by the area under the stress-strain curve. Toughness – Generally considering various materials Brittle fracture: elastic energy Ductile fracture: elastic + plastic energy very small toughness (unreinforced polymers) Engineering tensile strain,  Engineering tensile stress,  small toughness (ceramics) large toughness (metals) Adapted from Fig. 6.13, Callister 7e.

39 Slip Systems – Slip plane - plane allowing easiest slippage Wide interplanar spacings - highest planar densities – Slip direction - direction of movement - Highest linear densities – FCC Slip occurs on {111} planes (close-packed) in directions (close-packed) => total of 12 slip systems in FCC – in BCC & HCP other slip systems occur Deformation Mechanisms – from Macro-results to Micro-causes Adapted from Fig. 7.6, Callister 7e.

40 Figure 6.19 Sliding of one plane of atoms past an adjacent one. This high-stress process is necessary to plastically (permanently) deform a perfect crystal.


42 Note: By definition is the angle between the stress direction and Slip direction;  is the angle between the normal to slip plane and stress direction Stress and Dislocation Motion Crystals slip due to a resolved shear stress,  R. Applied tension can produce such a stress. slip plane normal, n s Resolved shear stress: RR =F s /A/A s slip direction ASAS RR RR FSFS slip direction Relation between  and RR RR = FSFS /AS/AS Fcos A/cos  F FSFS  nSnS ASAS A Applied tensile stress: = F/A  slip direction F A F

43 Condition for dislocation motion: Crystal orientation can make it easy or hard to move dislocation 10 -4 GPa to 10 -2 GPa typically Critical Resolved Shear Stress RR = 0 =90°  RR =  /2 =45°   RR = 0  =90° 

44 Generally: Resolved  (shear stress) is maximum at =  = 45  And  CRSS =  y /2 for dislocations to move (in single crystals)

45 Determining  and angles for Slip in Crystals (single X-tals this is easy!)  and angles are respectively angle between tensile direction and Normal to Slip plane and angle between tensile direction and slip direction (these slip directions are material dependent) In General for cubic xtals, angles between directions are given by: Thus for metals we compare Slip System (normal to slip plane is a direction with exact indices as plane) to applied tensile direction using this equation to determine the values of  and to plug into the  R equation to determine if slip is expected As we saw earlier!

46 Stronger since grain boundaries pin deformations Slip planes & directions (,  ) change from one crystal to another.  R will vary from one crystal to another. The crystal with the largest  R yields first. Other (less favorably oriented) crystals yield (slip) later. (courtesy of C. Brady, National Bureau of Standards [now the National Institute of Standards and Technology, Gaithersburg, MD].) Slip Motion in Polycrystals  300  m

47 After seeing the effect of poly crystalline materials we can say (as related to strength): Ordinarily ductility is sacrificed when an alloy is strengthened. The relationship between dislocation motion and mechanical behavior of metals is significance to the understanding of strengthening mechanisms. The ability of a metal to plastically deform depends on the ability of dislocations to move. Virtually all strengthening techniques rely on this simple principle: Restricting or Hindering dislocation motion renders a material harder and stronger. We will consider strengthening single phase metals by: grain size reduction, solid-solution alloying, and strain hardening

48 Hardness Resistance to permanently (plastically) indenting the surface of a product. Large hardness means: --resistance to plastic deformation or cracking in compression. --better wear properties. e.g., Hardened 10 mm sphere apply known force measure size of indentation 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 Hardness tests cause deformation that is the same as tensile testing – but are non-destructive and cheap to perform

49 Hardness: Common Measurement Systems

50 Comparing Hardness Scales:

51 Correlation Between Hardness and Tensile Strength Both measures the resistance to plastic deformation of a material.

52 Figure 6.29 (a) Plot of data from Table 6.10. A general trend of BHN with T.S. is shown. (b) A more precise correlation of BHN with T.S. (or Y.S.) is obtained for given families of alloys. [Part (b) from Metals Handbook, 9th ed., Vol. 1, American Society for Metals, Metals Park, OH, 1978.]

53 HB = Brinell Hardness TS (psia) = 500 x HB TS (MPa) = 3.45 x HB


55 Inaccuracies in Rockwell (Brinell) hardness measurements may occur due to:  An indentation is made too near a specimen edge.  Two indentations are made too close to one another.  Specimen thickness should be at least ten times the indentation depth.  Allowance of at least three indentation diameters between the center on one indentation and the specimen edge, or to the center of a second indentation.  Testing of specimens stacked one on top of another is not recommended.  Indentation should be made into a smooth flat surface. USE CARE – PROBLEMS CAN HAPPEN!

56 Mechanical Properties - Ceramics We know that ceramics are more brittle than metals. Why? Consider method of deformation – slippage along slip planes in ionic solids this slippage is very difficult too much energy needed to move one anion past another anion (like charges repel)

57 Room T behavior is usually elastic, with brittle failure. 3-Point Bend Testing often used. --tensile tests are difficult for brittle materials! Adapted from Fig. 12.32, Callister 7e. Measuring Ceramic Strength: Elastic Modulus F L/2 d = midpoint deflection cross section R b d rect.circ. Determine elastic modulus according to: F x linear-elastic behavior  F  slope = E = F  L 3 4bd 3 = F  L 3 12  R 4 rect. cross section circ. cross section

58 3-point bend test to measure room T strength. Adapted from Fig. 12.32, Callister 7e. Measuring Strength F L/2 d = midpoint deflection cross section R b d rect.circ. location of max tension Flexural strength: Typ. values: Data from Table 12.5, Callister 7e. rect.  fs  1.5F f L bd 2  F f L R3R3 Si nitride Si carbide Al oxide glass (soda) 250-1000 100-820 275-700 69 304 345 393 69 Material  fs (MPa) E(GPa) x F FfFf  fs 

59 Mechanical Issues: Properties are significantly dependent on processing – and as it relates to the level of Porosity: E = E 0 (1-1.9P+0.9P 2 ) – P is fraction porosity  fs =  0 e -nP --  0 & n are empirical values Because the very unpredictable nature of ceramic defects, we do not simply add a factor of safety for tensile loading We may add compressive surface loads We often choose to avoid tensile loading at all – most ceramic loading of any significance is compressive (consider buildings, dams, brigdes and roads!)



62 Mechanical Properties – of Polymers i.e. stress-strain behavior of polymers brittle polymer plastic elastomer  FS of polymer ca. 10% that of metals Strains – deformations > 1000% possible (for metals, maximum strain ca. 100% or less) elastic modulus – less than metal Adapted from Fig. 15.1, Callister 7e.

63 Tensile Response: Brittle & Plastic brittle failure plastic failure  (MPa)  x x crystalline regions slide fibrillar structure near failure crystalline regions align onset of necking Initial Near Failure semi- crystalline case aligned, cross- linked case networked case amorphous regions elongate unload/reload Stress-strain curves adapted from Fig. 15.1, Callister 7e. Inset figures along plastic response curve adapted from Figs. 15.12 & 15.13, Callister 7e. (Figs. 15.12 & 15.13 are from J.M. Schultz, Polymer Materials Science, Prentice-Hall, Inc., 1974, pp. 500- 501.)

64 Predeformation by Drawing Drawing…(ex: monofilament fishline) -- stretches the polymer prior to use -- aligns chains in the stretching direction Results of drawing: -- increases the elastic modulus (E) in the stretching direction -- increases the tensile strength (TS) in the stretching direction -- decreases ductility (%EL) Annealing after drawing... -- decreases alignment -- reverses effects of drawing. Comparable to cold working in metals! Adapted from Fig. 15.13, Callister 7e. (Fig. 15.13 is from J.M. Schultz, Polymer Materials Science, Prentice- Hall, Inc., 1974, pp. 500-501.)

65 Compare to responses of other polymers: -- brittle response (aligned, crosslinked & networked polymer) -- plastic response (semi-crystalline polymers) Stress-strain curves adapted from Fig. 15.1, Callister 7e. Inset figures along elastomer curve (green) adapted from Fig. 15.15, Callister 7e. (Fig. 15.15 is from Z.D. Jastrzebski, The Nature and Properties of Engineering Materials, 3rd ed., John Wiley and Sons, 1987.) Tensile Response: Elastomer Case  (MPa)  initial: amorphous chains are kinked, cross-linked. x final: chains are straight, still cross-linked elastomer Deformation is reversible! brittle failure plastic failure x x



68 TABLE 6.7 (continued)

69 Decreasing T... -- increases E -- increases TS -- decreases %EL Increasing strain rate... -- same effects as decreasing T. Adapted from Fig. 15.3, Callister 7e. (Fig. 15.3 is from T.S. Carswell and J.K. Nason, 'Effect of Environmental Conditions on the Mechanical Properties of Organic Plastics", Symposium on Plastics, American Society for Testing and Materials, Philadelphia, PA, 1944.) T and Strain Rate: Thermoplastics 20 40 60 80 0 4°C 20°C 40°C 60°C to 1.3  (MPa)  Data for the semicrystalline polymer: PMMA (Plexiglas)

70 Stress relaxation test: -- strain to   and hold. -- observe decrease in stress with time. Relaxation modulus: Sample T g (  C) values: PE (low density) PE (high density) PVC PS PC - 110 - 90 + 87 +100 +150 Time Dependent Deformation time strain tensile test oo (t)(t) Data: Large drop in E r for T > T g. (amorphous polystyrene) Adapted from Fig. 15.7, Callister 7e. (Fig. 15.7 is from A.V. Tobolsky, Properties and Structures of Polymers, John Wiley and Sons, Inc., 1960.) 10 3 1 10 -3 10 5 60100140180 rigid solid (small relax) transition region T(°C) TgTg E r (10s) in MPa viscous liquid (large relax)

71 Creep Sample deformation at a constant stress (  ) vs. time Adapted from Fig. 8.28, Callister 7e. Primary Creep: slope (creep rate) decreases with time. Secondary Creep: steady-state i.e., constant slope. Tertiary Creep: slope (creep rate) increases with time, i.e. acceleration of rate.   0 t

72 Figure 6.33 Mechanism of dislocation climb. Obviously, many adjacent atom movements are required to produce climb of an entire dislocation line.

73 Controls failure at elevated temperature, (T > 0.4 T m ) Adapted from Figs. 8.29, Callister 7e. Creep elastic primary secondary tertiary

74 Strain rate is constant at a given T,  -- strain hardening is balanced by recovery stress exponent (material parameter) strain rate activation energy for creep (a material parameter) applied stress material const. Secondary Creep


76 Figure 6.31 Typical creep test.

77 Figure 6.37 Creep rupture data for the nickel-based superalloy Inconel 718. (From Metals Handbook, 9th ed., Vol. 3, American Society for Metals, Metals Park, OH, 1980.)

78 Creep Failure Estimate rupture time S-590 Iron, T = 800°C,  = 20 ksi Failure: along grain boundaries. time to failure (rupture) function of applied stress temperature applied stress g.b. cavities Time to rupture, t r From V.J. Colangelo and F.A. Heiser, Analysis of Metallurgical Failures (2nd ed.), Fig. 4.32, p. 87, John Wiley and Sons, Inc., 1987. (Orig. source: Pergamon Press, Inc.) 1073K Ans: t r = 233 hr 24x10 3 K-log hr Adapted from Fig. 8.32, Callister 7e. (Fig. 8.32 is from F.R. Larson and J. Miller, Trans. ASME, 74, 765 (1952).) L(10 3 K-log hr) Stress, ksi 100 10 1 1220242816 data for S-590 Iron 20 “L” is the Larson- Miller parameter

79 Considering a Problem S-590 steel subject to stress of 55 MPa Using Data in Fig 8.32 (handout), L  26.2x10 3 Determine the temperature for creep at which the component fails at 200 hours

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