Yield Strength, Tensile Strength, and Ductility

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Yield Strength, Tensile Strength, and Ductility



9.1 Introduction


All solids have an elastic limit beyond which something happens. A totally brittle solid will fracture suddenly (e.g., glass). Most engineering materials do something different; they deform plastically or change their shapes in a permanent way. It is important to know when, and how, they do this—so we can design structures to withstand normal service loads without any permanent deformation; and rolling mills, presses and forging machinery strong enough to deform the materials being rolled, pressed or forged.


To study this, we pull carefully prepared samples in a tensile testing machine, or compress them in a compression machine, and record the stress required to produce a given strain.


9.2 Linear and Nonlinear Elasticity


Figure 9.1 shows the stress–strain curve of a material exhibiting perfectly linear elastic behavior. This is the behavior that is characterized by Hooke’s law (Chapter 3). All solids are linear elastic at small strains—by which we usually mean less than 0.001, or 0.1%. The slope of the stress–strain line, which is the same in compression as in tension, is Young’s modulus, E. The area (shaded) is the elastic energy stored, per unit volume: we can get it all back if we unload the solid, which behaves like a spring.


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Figure 9.1 Stress–strain behavior for a linear elastic solid. The axes are calibrated for a material such as steel.

Figure 9.2 shows a nonlinear elastic solid. Rubbers have a stress–strain curve like this, extending to very large strains (of order 5 or even 8). The material is still elastic: if unloaded much of the energy stored during loading is recovered—that is why catapults can be as lethal as they are.


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Figure 9.2 Stress–strain behavior for a nonlinear elastic solid. The axes are calibrated for a material such as rubber.

9.3 Load–Extension Curves for Nonelastic (Plastic) Behavior


Rubbers are exceptional in behaving elastically to high strains; as we said, almost all materials, when strained by more than about 0.001 (0.1%), do something irreversible: and most engineering materials deform plastically to change their shape permanently. If we load a piece of ductile metal (e.g., copper), in tension, we get the following relationship between the load and the extension (Figure 9.3). This can be demonstrated nicely by pulling a piece of plasticine modelling clay (a ductile non-metallic material). Initially, the plasticine deforms elastically, but at a small strain begins to deform plastically, so if the load is removed, the piece of plasticine is permanently longer than it was at the beginning of the test: it has undergone plastic deformation (Figure 9.4).


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Figure 9.3 Load-extension curve for a bar of ductile metal (e.g., annealed copper) pulled in tension.

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Figure 9.4 Permanent plastic deformation after a sample has yielded and been unloaded.

If you continue to pull, it continues to get longer, at the same time getting thinner because in plastic deformation volume is conserved (matter is just flowing from place to place). Eventually, the plasticine becomes unstable and begins to neck at the maximum load point in the force-extension curve (see Figure 9.3). Necking is an instability which we shall look at in more detail in Chapter 12. The neck grows rapidly, and the load that the specimen can carry through the neck decreases until breakage takes place. The two pieces produced after breakage have a total length that is slightly less than the length just before breakage by the amount of the elastic extension produced by the terminal load.


If we load a material in compression, the force-displacement curve is simply the reverse of that for tension at small strains, but it becomes different at larger strains. As the specimen squashes down, becoming shorter and fatter to conserve volume, the load needed to keep it flowing rises (Figure 9.5). No instability such as necking appears, and the specimen can be squashed almost indefinitely, only being limited by severe cracking in the specimen or plastic flow of the compression plates.


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Figure 9.5 Squashing of the specimen increases the load needed to keep it flowing.

9.4 True Stress–Strain Curves for Plastic Flow


The apparent difference between the curves for tension and compression is due solely to the geometry of testing. If, instead of plotting load, we plot load divided by the actual cross-sectional area of the specimen, A, at any particular elongation or compression, the two curves become much more like one another. In other words, we simply plot true stress (see Chapter 3) as our vertical coordinate (Figure 9.6). This method of plotting allows for the thinning of the material when pulled in tension, or the fattening of the material when compressed.


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Figure 9.6


But the two curves still do not exactly match, as Figure 9.6 shows. The reason is that a displacement of (for example) u = L0/2 in tension and in compression gives different strains; it represents a drawing out of the tensile specimen from L0 to 1.5L0, but a squashing down of the compressive specimen from L0 to 0.5L0. The material of the compressive specimen has thus undergone much more plastic deformation than the material in the tensile specimen, and can hardly be expected to be in the same state, or to show the same resistance to plastic deformation. The two conditions can be compared properly by taking small strain increments


δɛ=δuL=δLL



si1_e  (9.1)


about which the state of the material is the same for either tension or compression (Figure 9.7).


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Figure 9.7


This is the same as saying that a decrease in length from 100 mm (L0) to 99 mm (L), or an increase in length from 100 mm (L0) to 101 mm (L) both represent a 1% change in the state of the material. Actually, they do not give exactly the same state in both cases, but they do in the limit


dɛ=dLL



si2_e  (9.2)


Then, if the stresses in compression and tension are plotted against


ɛ=L0LdLL=lnLL0



the two curves exactly mirror one another (Figure 9.8). The quantity ɛ is called the true strain (to be contrasted with the nominal strain u/L0 defined in Chapter 3) and the matching curves are true stress/true strain (σ/ɛ) curves.


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Figure 9.8


Now, a final catch. From our original load-extension or load-compression curves we can easily calculate ɛ, simply by knowing L0 and taking natural logs. But how do we calculate σ? Because volume is conserved during plastic deformation we can write, at any strain,


A0L0=AL



si4_e


provided the extent of plastic deformation is much greater than the extent of elastic deformation (volume is only conserved during elastic deformation if Poisson’s ratio v = 0.5; and it is near 0.33 for most materials). Thus


A=A0L0L



and


σ=FA=FLA0L0



all of which we know or can measure easily.


9.5 Plastic Work


When metals are rolled or forged, or drawn to wire, or when polymers are injection molded or pressed or drawn, energy is absorbed. The work done on a material to change its shape permanently is called the plastic work; its value, per unit volume, is the area of the crosshatched region shown in Figure 9.8; it may easily be found (if the stress–strain curve is known) for any amount of permanent plastic deformation, ɛ′. Plastic work is important in metal and polymer forming operations because it determines the forces that the rolls, or press or molding machine must exert on the material.


9.6 Tensile Testing


The plastic behavior of a material is usually measured by conducting a tensile test. Tensile testing equipment is standard in engineering laboratories. Such equipment produces a load/displacement (F/u) curve for the material, which is then converted to a nominal stress/nominal strain, or σn/ɛn, curve (see Figure 9.9), where


σn=FA0



si7_e  (9.6)


and


ɛn=uL0



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Figure 9.9


Naturally, because A0 and L0 are constant, the shape of the σn/ɛn curve is identical to that of the load-extension curve. But the σn/ɛn plotting method allows us to compare data for specimens having different (though now standardized) A0 and L0, and thus to examine the properties of material, unaffected by specimen size. The advantage of keeping the stress in nominal units and not converting to true stress (as shown before) is that the onset of necking can clearly be seen on the σn/ɛn curve.


Now, we define the quantities usually listed as the results of a tensile test. The easiest way to do this is to show them on the σn/ɛn curve itself (Figure 9.10). They are:



  •  σy = yield strength (F/A0 at onset of plastic flow)
  •  σ0.1% = 0.1% proof stress (F/A0 at a permanent strain of 0.1%) (0.2% proof stress is often quoted instead; proof stress is useful for characterizing yield of a material that yields gradually, and does not show a distinct yield point)
  •  σTS = tensile strength (F/A0 at onset of necking)
  •  ɛf = (plastic) strain after fracture, or tensile ductility; the broken pieces are put together and measured, and ɛf calculated from (LL0)/L0, where L is the length of the assembled pieces

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Figure 9.10


9.7 Data


Data for yield strength, tensile strength, and tensile ductility are given in Table 9.1 and shown on the bar chart (Figure 9.11). Like moduli, they span a range of about 106: from about 0.1 MN m-2 (for polystyrene foams) to nearly 105 MN m-2 (for diamond).



Table 9.1














































































































































































































































































































































































Yield Strength, σy, Tensile Strength, σTS, and Tensile Ductility, ɛf
Material σ (MN m-2) σTS (MN m-2) ɛf
Diamond 50,000 0
Boron carbide, B4C 14,000 (330) 0
Silicon carbide, SiC 10,000 (200–800) 0
Silicon nitride, Si3N4 9600 (200–900) 0
Silica glass, SiO2 7200 (110) 0
Tungsten carbide, WC 5600–8000 (80–710) 0
Niobium carbide, NbC 6000 0
Alumina, Al2O3 5000 (250–550) 0
Beryllia, BeO 4000 (130–280) 0
Syalon (Si-Al-O-N ceramic) 4000 (945) 0
Mullite 4000 (128–140) 0
Titanium carbide, TiC 4000 0
Zirconium carbide, ZrC 4000 0
Tantalum carbide, TaC 4000 0
Zirconia, ZrO2 4000 (100–700) 0
Soda glass (standard) 3600 (50–70) 0
Magnesia, MgO 3000 (100) 0
Cobalt and alloys 180–2000 500–2500 0.01–6
Low-alloy steels (water-quenched and tempered) 500–1900 680–2400 0.02–0.3
Pressure-vessel steels 1500–1900 1500–2000 0.3–0.6
Stainless steels, austenitic 286–500 760–1280 0.45–0.65
Boron/epoxy composites 725–1730
Nickel alloys 200–1600 400–2000 0.01–0.6
Nickel 70 400 0.65
Tungsten 1000 1510 0.01–0.6
Molybdenum and alloys 560–1450 665–1650 0.01–0.36
Titanium and alloys 180–1320 300–1400 0.06–0.3
Carbon steels (water-quenched and tempered) 260–1300 500–1880 0.2–0.3
Tantalum and alloys 330–1090 400–1100 0.01–0.4
Cast irons 220–1030 400–1200 0–0.18
Copper alloys 60–960 250–1000 0.01–0.55
Copper 60 400 0.55
Cobalt/tungsten carbide cermets 400–900 900 0.02
CFRPs 670–640
Brasses and bronzes 70–640 230–890 0.01–0.7
Aluminum alloys 100–627 300–700 0.05–0.3
Aluminum 40 200 0.5
Stainless steels, ferritic 240–400 500–800 0.15–0.25
Zinc alloys 160–421 200–500 0.1–1.0
Concrete, steel reinforced 50–200 0.02
Alkali halides 200–350 0
Zirconium and alloys 100–365 240–440 0.24–0.37
Mild steel 220 430 0.18–0.25
Iron 50 200 0.3
Magnesium alloys 80–300 125–380 0.06–0.20
GFRPs 100–300
Beryllium and alloys 34–276 380–620 0.02–0.10
Gold 40 220 0.5
PMMA 60–110 110 0.03–0.05
Epoxies 30–100 30–120
Polyimides 52–90
Nylons 49–87 100
Ice 85 (6) 0
Pure ductile metals 20–80 200–400 0.5–1.5
Polystyrene 34–70 40–70
Silver 55 300 0.6
ABS/polycarbonate 55 60
Common woods (|| to grain) 35–55
Lead and alloys 11–55 14–70 0.2–0.8
Acrylic/PVC 45–48
Tin and alloys 7–45 14–60 0.3–0.7
Polypropylene 19–36 33–36
Polyurethane 26–31 58
Polyethylene, high density 20–30 37
Concrete, non-reinforced 20–30 (1–5) 0
Natural rubber 30 8.0
Polyethylene, low density 6–20 20
Common woods (⊥ to grain) 4–10
Ultrapure f.c.c. metals 1–10 200–400 1–2
Foamed polymers, rigid 0.2–10 0.2–10 0.1–1
Polyurethane foam 1 1 0.1–1


t0010_at0010_b


Note: Bracketed σTS data for brittle materials refer to the modulus of rupture σr (see Chapter 16).


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Figure 9.11 Bar chart of data for yield strength, σy.

Most ceramics have enormous yield stresses. In a tensile test, at room temperature, ceramics almost all fracture long before they yield: this is because their fracture toughness, which we will discuss later, is very low. Because of this, you cannot measure the yield strength of a ceramic by using a tensile test. Instead, you have to use a test that somehow suppresses fracture: a compression test, for instance. The best and easiest is the hardness test.


Pure metals are very soft indeed, and have a high ductility. This is what, for centuries, has made them so attractive at first for jewellery and weapons, and then for other implements and structures: they can be worked to the shape that you want them in; furthermore, their ability to work-harden means that, after you have finished, the metal is much stronger than when you started. By alloying, the strength of metals can be further increased, though—in yield strength—the strongest metals still fall short of most ceramics.


Polymers, in general, have lower yield strengths than metals. The very strongest barely reach the strength of aluminum alloys. They can be strengthened, however, by making composites out of them: GFRP has a strength only slightly inferior to aluminum, and CFRP is substantially stronger.


Worked Example


The hardness test is used for estimating the yield strengths of hard brittle materials. It is also widely used as a simple non-destructive test on material or finished components to check whether they meet the specified mechanical properties. Figure 9.12 shows how one type of test works (the diamond pyramid test). A very hard (diamond) pyramid “indenter” (with a standardized angle of 136° between opposite faces) is pressed into the surface of the material with a force F. Material flows away from underneath the indenter, so when the indenter is removed again, a permanent “indent” is left. The diagonals of the indent are measured with a microscope, and the total surface (or contact) area A of the indent is found from look-up tables. The hardness H is then given by H = F/A. Obviously, the softer the material, the larger the indents.


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Figure 9.12 The hardness test.

The yield strength can be estimated from the relation (derived in Chapter 12)


H=3σy



si9_e  (9.8)


When flowing away from the indenter, the material experiences an average nominal strain of 0.08. This means that Equation (9.8) is only accurate for materials which do not work harden. For materials that do, the hardness test gives the yield strength of material that has been work hardened by 8%. Good correlations can also be obtained between hardness and tensile strength. For example,


σTS=0.33H



si10_e  (9.9)


holds well (to ± 10%) for all types of steel.


The indentation hardness is defined slightly differently—still as H = F/A, but A is now the area of the indent measured at the surface of the material (which is slightly less than that measured on the faces of the indentation).


Examples




  1. 9.1 Define nominal stress σn and true stress σ.
  2. 9.2 Define nominal strain ɛn and true strain ɛ.
  3. 9.3 Why is volume conserved during plastic deformation?
  4. 9.4 Show that σ = σn(1 + ɛn), where σ is true stress, σn is nominal stress, and ɛn is nominal strain.
  5. 9.5 Show that ɛ =  ln (1 + ɛn), where ɛ is true strain and ɛn is nominal strain.
  6. 9.6 For small strains, show that ɛɛn and σσn.
  7. 9.7 Show that the work per unit volume required to extend a specimen to a nominal strain ɛn is given by

    U=0ɛnσndɛn


    Explain why this work must be the same as

    U=0ɛσdɛ



  8. 9.8 Show that the work per unit volume required to extend a specimen to a nominal strain ɛn in the linear elastic region is given by

    Uel=σn2/2E



  9. 9.9 How would you determine the 0.1% proof stress for a metal?
  10. 9.10 How would you determine the tensile strength σTS for a metal? Is it a true stress or a nominal stress?
  11. 9.11 Define (a) ɛf, the strain after fracture (tensile ductility), and (b) the reduction in area at break. Why, in a tensile test on a rod of metal, are (a) and (b) generally not equal to one another?
  12. 9.12 Define (a) diamond-pyramid hardness, and (b) indentation hardness.
  13. 9.13 Diamond-pyramid hardness values are usually quoted in units of kg mm–2. A sample of steel has a diamond-pyramid hardness of 200. Estimate the tensile strength of the steel.
  14. 9.14 Referring to Example 9.13, estimate the yield strength of the steel.
  15. 9.15 Comparing the answers to Examples 9.13 and 9.14, we see that the estimate of the yield strength is greater than the estimate of the tensile strength. There are two reasons for this: what are they?
  16. 9.16 Why is the hardness test used for estimating the yield stress of ceramics?
  17. 9.17 Nine strips of pure, fully annealed copper were deformed plastically by being passed between a pair of rotating rollers so that the strips were made thinner and longer. The increases in length produced were 1, 10, 20, 30, 40, 50, 60, 70, and 100%, respectively. The diamond–pyramid hardness of each piece was measured after rolling. The results are given in the following table:

    Assuming that the hardness test creates a further nominal strain of 0.08, and the hardness value is 3.0 times the true stress, construct the curve of nominal stress against nominal strain.
    [Hint: Add 0.08 to each value of nominal strain in the table.]
  18. 9.18 Using the plot of nominal stress against nominal strain from Example 9.1, find:

    1. a. The tensile strength of copper
    2. b. The strain at which tensile failure commences
    3. c. The percentage reduction in cross-sectional area at this strain
    4. d. The work required to initiate tensile failure in a cubic metre of annealed copper

  19. 9.19 Why can copper survive a much higher extension during rolling than during a tensile test?
  20. 9.20 The following data were obtained in a tensile test on a specimen with 50 mm gauge length and a cross-sectional area of 160 mm2.

    Find the maximum allowable working stress if this is to equal

    1. a. 0.25 × tensile strength
    2. b. 0.6 × 0.1% proof stress

  21. 9.21 One type of hardness test involves pressing a hard sphere (radius r) into the test material under a fixed load F, and measuring the depth, h, to which the sphere sinks into the material, plastically deforming it. Show that the indentation hardness, H, of the material is given by

    H=F2πrh


    where h ≪ r.
  22. 9.22 The following diagram shows the force-extension characteristics of a bungee rope. The length of the rope under zero load is 15 m. One end of the rope is attached to a bridge deck, and the other end is attached to a person standing on the deck. The person then jumps off the deck, and descends vertically until arrested by the rope.
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    1. a. Find the maximum mass of the person for a successful arrest, mmax.
    2. b. The diagram shows that the unloading line is 0.5 kN below the loading line. Comment on the practical significance of this to bungee jumping.

Answers




  1. 9.1 See relations at the end of this chapter.
  2. 9.2 See relations at the end of this chapter.
  3. 9.3 See Section 9.3, paragraph 2.
  4. 9.4 See relations at the end of this chapter.
  5. 9.5 See relations at the end of this chapter.
  6. 9.6 See relations at the end of this chapter.
  7. 9.7 See relations at the end of this chapter. The two forms of the integral must be the same, because the same amount of work is put into straining the specimen, no matter whether stress and strain are expressed as nominal or true.
  8. 9.8 See relations at the end of this chapter.
  9. 9.9 See Figure 9.10.
  10. 9.10 See Figure 9.10, and relations at the end of this chapter. It is a nominal stress.
  11. 9.11 See relations at the end of this chapter. Regarding (b), the necked region has the same dimensions whatever the length of the specimen, provided the specimen is long enough so the ends of the specimen do not interact with the necked region. Regarding (a), the longer the specimen, the smaller the value of (a). This is why test specimens have a standard length and diameter, to give comparable values of ɛf (as in Table 9.1).
  12. 9.12 See Worked Example.
  13. 9.13 648 MN m–2. See Equation (9.9), and remember to convert from kgf to N.
  14. 9.14 654 MN m–2.
  15. 9.15 (i) Work hardening in the zone beneath the diamond pyramid (see Worked Example). (ii) The tensile strength is the nominal stress, and the true stress is larger because the specimen has thinned down.
  16. 9.16 See Section 9.7, paragraph 2.
  17. 9.17 From the relations at the end of this chapter, σn=σ1+ɛnsi15_e. Values of σ are obtained by dividing the hardness values by 3. ɛn is the nominal strain given in the table plus 0.08.
  18. 9.18 (a) When the curve is drawn, the tensile strength (stress maximum) is 217 MN m–2. (b) The corresponding strain is 0.6. (c) From the relations at the end of this chapter, A0 = A(1 + ɛn), so (A0A)/A0 = 0.38, or 38%. (d) Work out the area beneath the curve up to a strain of 0.6 (count squares, or integrate numerically) to get 109 MJ.
  19. 9.19 Because the copper is being deformed in compression, it cannot fail by necking (see Figure 9.3).
  20. 9.20 (a) Plot the load/extension curve, and determine the maximum load. Hence, find the tensile strength. 0.25 of this = 160 MN m–2. (b) Plot the first six data points of the load/extension curve with an expanded extension scale. Find the 0.1% proof load as shown in Figure 9.10. Hence find the proof stress. 0.6 of this = 131 MN m–2.
  21. 9.22 (a) Equate potential energy lost to elastic energy stored in rope. Note that distance fallen = 30 m, and use 9.81 for g. Answer = 89 kg. (b) 29% of the energy input is lost when the rope returns to zero load. If energy were not dissipated in this way, the jumper would never come to rest, and could also collide with the bridge at the top of the rebound.

Revision of Terms and Useful Relations


σn, nominal stress σn =  F/A0


σ, true stress σ = F/A


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ɛn, nominal strain


ɛn=uL0,orLL0L0orLL01



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Relations between σn, σ and ɛn


Assuming constant volume (valid if υ = 0.5 or, if not, plastic deformation ≫ elastic deformation):


A0L0=AL;A0=ALL0=A(1+ɛn)



Thus


σ=FA=FA0(1+ɛn)=σn(1+ɛn)



ɛ, true strain and the relation between ɛ and ɛn


ɛ=L0LdLL=lnLL0



Thus


ɛ=ln(1+ɛn)




Small strain condition


For small ɛn


ɛɛn,fromɛ=ln(1+ɛn)σσn,fromσ=σn(1+ɛn)



Thus, when dealing with most elastic strains (but not in rubbers), it is immaterial whether ɛ or ɛn, or σ or σn, are chosen.


Energy


The energy expended in deforming a material per unit volume is given by the area under the stress–strain curve.


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For linear elastic strains, and only linear elastic strains


σnɛn=E,andUel=σndɛn=σndσnE=σn22E



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Elastic limit


In a tensile test, as the load increases, the specimen at first is strained elastically, that is reversibly. Above a limiting stress—the elastic limit—some of the strain is permanent; this is plastic deformation.



  • Yielding The change from elastic to measurable plastic deformation.
  • Yield strength The nominal stress at yielding. In many materials this is difficult to spot on the stress–strain curve and in such cases it is better to use a proof stress.
  • Proof stress The stress that produces a permanent strain equal to a specified percentage of the specimen length. A common proof stress is one that corresponds to 0.2% permanent strain.

Strain hardening (work-hardening)


The increase in stress needed to produce further strain in the plastic region. Each strain increment strengthens or hardens the material so that a larger stress is needed for further strain.


σTS, tensile strength (ultimate tensile strength, or UTS)


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ɛf, strain after fracture, or tensile ductility


The permanent extension in length (measured by fitting the broken pieces together) expressed as a percentage of the original gauge length.


LbreakL0L0×100



Reduction in area at break


The maximum decrease in cross-sectional area at the fracture expressed as a percentage of the original cross-sectional area.


Strain after fracture and percentage reduction in area are used a measures of ductility, that is the ability of a material to undergo large plastic strain under stress before it fractures.


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Aug 9, 2021 | Posted by in General Engineer | Comments Off on Yield Strength, Tensile Strength, and Ductility
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