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

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

Stress-Strain: Testing Uses Standardized methods developed by ASTM for Tensile Tests it is ASTM E8 • Typical tensile test machine • Typical tensile specimen (ASTM A-bar) gauge length specimen extensometer 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.)

Comparison of Units: SI and Engineering Common (in US) Eng. Common Force Newton (N) Pound-force (lbf) Area mm2 or m2 in2 Stress Pascal (N/m2) or MPa (106 pascals) psi (lbf/in2) or Ksi (1000 lbf/in2) Strain (Unitless!) mm/mm or m/m in/in Conversion Factors SI to Eng. Common Eng. Common to SI N*4.448 = lbf Lbf*0.2248 = N Area I mm2*645.16 = in2 in2 *1.55x10-3 = mm2 Area II m2 *1550 = in2 in2* 6.452x10-4 = m2 Stress I - a Pascal * 1.450x10-4 = psi psi * 6894.76 = Pascal Stress I - b Pascal * 1.450x10-7 = Ksi Ksi * 6.894 x106 = Pascal Stress II - a MPa * 145.03 = psi psi * 6.89x 10-3 = MPa Stress II - b MPa * 1.4503 x 10-1= Ksi Ksi * 6.89 = MPa One other conversion: 1 GPa = 103 MPa

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

we can also see the symbol ‘s’ used for engineering stress • Tensile stress, s: original area before loading Area, A F t s = A o 2 f m N or in lb • Shear stress, t: Area, A F t s = A o  Stress has units: N/m2 (MPa) or lbf/in2 we can also see the symbol ‘s’ used for engineering stress

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

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

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

Figure 6. 2 Load-versus-elongation curve obtained in a tensile test Figure 6.2 Load-versus-elongation curve obtained in a tensile test. The specimen was aluminum 2024-T81.

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

Solving:

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

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

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

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

Other Elastic Properties simple torsion test M t G g • Elastic Shear modulus, G: t = G g • Elastic Bulk modulus, K: pressure test: Init. vol =Vo. Vol chg. = DV P P = - K D V o • Special relations for isotropic materials: 2(1 + n) E G = 3(1 - 2n) K E is Modulus of Elasticity  is Poisson’s Ratio

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.

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

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

vertical).

Lets Try an Example Problem Load (N) len. (mm) len. (m) (l) 50.8 0.0508 12700 50.825 0.050825 2.5E-05 25400 50.851 0.050851 5.1E-05 38100 50.876 0.050876 7.6E-05 50800 50.902 0.050902 0.000102 76200 50.952 0.050952 0.000152 89100 51.003 0.051003 0.000203 92700 51.054 0.051054 0.000254 102500 51.181 0.051181 0.000381 107800 51.308 0.051308 0.000508 119400 51.562 0.051562 0.000762 128300 51.816 0.051816 0.001016 149700 52.832 0.052832 0.002032 159000 53.848 0.053848 0.003048 160400 54.356 0.054356 0.003556 159500 54.864 0.054864 0.004064 151500 55.88 0.05588 0.00508 124700 56.642 0.056642 0.005842 GIVENS (load and length as test progresses):

Leads to the following computed Stress/Strains: e stress (Pa) e str (MPa) e. strain 98694715.7 98.694716 0.000492 197389431 197.38943 0.001004 296084147 296.08415 0.001496 394778863 394.77886 0.002008 592168294 592.16829 0.002992 692417257 692.41726 0.003996 720393712 720.39371 0.005 796551839 796.55184 0.0075 837739398 837.7394 0.01 927885752 927.88575 0.015 997049766 997.04977 0.02 1163354247 1163.3542 0.04 1235626755 1235.6268 0.06 1246506488 1246.5065 0.07 1239512374 1239.5124 0.08 1177342475 1177.3425 0.1 969073311 969.07331 0.115

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

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.)

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.)

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

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.]

Returning to our Example – True Properties T. Stress T. Strain 98.74328592 0.000492 197.5875979 0.001003 296.5271076 0.001495 395.571529 0.002006 593.9401363 0.002988 695.1842003 0.003988 723.9956807 0.004988 802.525978 0.007472 846.1167917 0.00995 941.8040385 0.014889 1016.990761 0.019803 1209.888417 0.039221 1309.764361 0.058269 1333.761942 0.067659 1338.673364 0.076961 1295.076722 0.09531 1080.516741 0.108854 Necking Began

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

Take Logs of both T and T We Compute T = KTn 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

Stress Strain Plot w/ True Values K  2617 MPa

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

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

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

Deformation Mechanisms – from Macro-results to Micro-causes 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 <110> directions (close-packed) => total of 12 slip systems in FCC in BCC & HCP other slip systems occur Adapted from Fig. 7.6, Callister 7e.

Figure 6. 19 Sliding of one plane of atoms past an adjacent one 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.

Stress and Dislocation Motion • Crystals slip due to a resolved shear stress, tR. • Applied tension can produce such a stress. Applied tensile stress: = F/A s direction slip F A slip plane normal, ns Resolved shear stress: tR = F s /A direction slip AS FS direction slip Relation between s and tR = FS /AS F cos l A / f nS AS 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

Critical Resolved Shear Stress • Condition for dislocation motion: 10-4 GPa to 10-2 GPa typically • Crystal orientation can make it easy or hard to move dislocation tR = 0 l =90° s tR = s /2 l =45° f tR = 0 f =90° s

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

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!

Slip Motion in Polycrystals 300 mm • Stronger since grain boundaries pin deformations • Slip planes & directions (l, f) change from one crystal to another. • tR will vary from one • The crystal with the largest tR 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].)

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

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 steels file hard cutting tools nitrided diamond Hardness tests cause deformation that is the same as tensile testing – but are non-destructive and cheap to perform

Hardness: Common Measurement Systems

Comparing Hardness Scales:

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

Figure 6. 29 (a) Plot of data from Table 6. 10 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.]

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

USE CARE – PROBLEMS CAN HAPPEN! 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.

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)

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

Measuring Strength s = 1.5Ff L bd 2 Ff L pR3 x F F • 3-point bend test to measure room T strength. F L/2 d = midpoint deflection cross section R b d rect. circ. location of max tension Adapted from Fig. 12.32, Callister 7e. • Flexural strength: • Typ. values: Data from Table 12.5, Callister 7e. rect. s fs = 1.5Ff L bd 2 Ff L pR3 Si nitride Si carbide Al oxide glass (soda) 250-1000 100-820 275-700 69 304 345 393 Material (MPa) E(GPa) x F Ff d

Mechanical Issues: Properties are significantly dependent on processing – and as it relates to the level of Porosity: E = E0(1-1.9P+0.9P2) – P is fraction porosity fs = 0e-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!)

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

Tensile Response: Brittle & Plastic (MPa) fibrillar structure near failure Initial Near Failure x brittle failure onset of crystalline regions align necking plastic failure x crystalline regions slide amorphous regions elongate aligned, cross- linked case networked unload/reload e semi- crystalline case 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.)

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 -- 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.)

Tensile Response: Elastomer Case (MPa) final: chains are straight, still cross-linked x brittle failure 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.) plastic failure x x elastomer initial: amorphous chains are kinked, cross-linked. Deformation is reversible! e • Compare to responses of other polymers: -- brittle response (aligned, crosslinked & networked polymer) -- plastic response (semi-crystalline polymers)

TABLE 6.7 (continued)

T and Strain Rate: Thermoplastics (MPa) • Decreasing T... -- increases E -- increases TS -- decreases %EL • Increasing strain rate... -- same effects as decreasing T. 20 4 6 8 Data for the 4°C semicrystalline polymer: PMMA 20°C (Plexiglas) 40°C to 1.3 60°C e 0.1 0.2 0.3 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.)

Time Dependent Deformation • Stress relaxation test: • Data: Large drop in Er for T > Tg. (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 -1 -3 5 60 100 140 180 rigid solid (small relax) transition region T(°C) Tg Er (10s) in MPa viscous liquid (large relax) -- strain to eo and hold. -- observe decrease in stress with time. time strain tensile test eo s(t) • Relaxation modulus: • Sample Tg(C) values: PE (low density) PE (high density) PVC PS PC - 110 - 90 + 87 +100 +150

Creep Sample deformation at a constant stress (s) vs. time s s,e t 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. Adapted from Fig. 8.28, Callister 7e.

Figure 6. 33 Mechanism of dislocation climb Figure 6.33 Mechanism of dislocation climb. Obviously, many adjacent atom movements are required to produce climb of an entire dislocation line.

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

Secondary Creep • Strain rate is constant at a given T, s -- strain hardening is balanced by recovery stress exponent (material parameter) activation energy for creep (a material parameter) strain rate material const. applied stress • Strain rate increases for higher T, s 10 2 4 -2 -1 1 Steady state creep rate (%/1000hr) e s Stress (MPa) 427°C 538 °C 649 Adapted from Fig. 8.31, Callister 7e. (Fig. 8.31 is from Metals Handbook: Properties and Selection: Stainless Steels, Tool Materials, and Special Purpose Metals, Vol. 3, 9th ed., D. Benjamin (Senior Ed.), American Society for Metals, 1980, p. 131.)

Figure 6.31 Typical creep test.

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.)

Creep Failure • Failure: along grain boundaries. • Time to rupture, tr time to failure (rupture) function of applied stress temperature applied stress g.b. cavities • Time to rupture, tr • Estimate rupture time S-590 Iron, T = 800°C, s = 20 ksi 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 12 20 24 28 16 data for S-590 Iron 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: tr = 233 hr 24x103 K-log hr “L” is the Larson- Miller parameter

Considering a Problem S-590 steel subject to stress of 55 MPa Using Data in Fig 8.32 (handout), L  26.2x103 Determine the temperature for creep at which the component fails at 200 hours