mechanical-properties-and-working-of-metals-and-alloys_compress

1.13 Notch Tensile Test 81 Notch strength, Snet ¼ original maximum load root of the notch ¼ ðPmaxÞnet ð1:164Þ cross-sectional area at the ðA0Þnet ð1:165Þ Notch strength ratio, NSR ¼ strength at maximum load for notched specimen ¼ Snet ultimate tensile strength for smooth specimen Su The value of notch strength, Snet; for a notched material strength obtained from the notch tensile test of embrittled possessing some amount of ductility will be more than the steel shows a trough revealing the metallurgical embrittle- ultimate tensile strength ðSuÞ of a smooth specimen of the ment. For further detailed study on notch tensile testing, the same material, due to the development of the plastic constraint reader may see the reference (Lubahn 1957). in the notched region similar to that occurred at the neck, as discussed in Sect. 1.6.2.7. If the material is brittle, then 1.14 Tensile Fracture Snet\\Su: When the high strength or the hardness of a material or the presence of some metallurgical variables or any other Two general classifications of fracture are brittle fracture factors limits the plastic deformation at the notch-root, the and ductile fracture. Different types of loading can produce material becomes susceptible to brittle fracture. However, the various kinds of fracture. In this section, it will be assumed notch sensitivity can easily be detected by the evaluation of that fracture occurs only due to application of a uniaxial NSR. If NSR\\1; then the material is notch-brittle, otherwise static tensile stress at ambient temperature. Brittle tensile a notch-tough material will have NSR [ 1: Another tensile fracture occurs by the stress-assisted separation of atomic property that is determined from the notch tensile test is the bonds across atomic plane, which is called the fracture plane. percentage notch ductility usually measured by percentage Single- and poly-crystalline brittle metals exhibits a flat and reduction of area at the root of the notch. bright fracture surface lying normal to the applied uniaxial tensile stress, as shown in Fig. 1.61a. Characteristics of Figure 1.59 compares the variations of notch strengths brittle fractures of single crystals and polycrystals are as and percentage notch ductility of two types of steel as a follows: function of their tensile strengths, which clearly reveals that steel A is more notch brittle than steel B, i.e., the notch • No gross plastic deformation is evident, although there is sensitivity of steel A is higher than that of steel B, although a very little plastic deformation in the microscopic level they have the same notch strength up to a certain level of in the metals and a plastically deformed metallic thin their tensile strengths. The above steels become sensitive to layer has been detected with X-ray diffraction analysis. notch at the strength levels where the notch strengths start to decrease, or more appropriately where NSR < 1. • The surface of brittle cleavage fracture produced is flat and has a granular appearance, as observed in cast iron. The metallurgical embrittlement developed on tempering a hardened alloy steel in the temperature range of 330 to • The fracture surface appears as bright due to light 480C can be detected by the sesitivity of the notch strength, reflection from flat surface. as illustrated in Fig. 1.60 (Espey et al. 1959). Note that the determination of UTS (ultimate tensile strength) or yield • Brittle fracture is initiated by some flaws or imperfections strength or the measurement of conventional elongation in the material, such as microscopic cracks or holes. This from the tensile test on the smooth embrittled alloy steel flaw causes stress concentration, i.e., a rise of stress to a specimen cannot detect the embrittlement, while the notch high value at its tip, just like a notch. If this local stress at

82 1 Tension Tensile B 0.2% offset yield Notch strength Strength A Notch 0 0 Embrittlement 0 Tensile strength 200 zone 330 480 Tempering temperature, °C Notch ductility, % Elongation (50 mm gage length)Elongation, % B A 0 200 330 480 Tempering temperature, °C 0 Fig. 1.60 Schematically showing the effect of tempering temperature 0 Tensile strength on notched and unnotched tensile properties of hardened alloy steel during tempering. The notch strength shows a trough revealing Fig. 1.59 Schematic variations of notch strengths and percentage metallurgical embrittlement developed in alloy steel during tempering notch ductility of two steels A and B as a function of their tensile in the temperature range of 330–480 °C, while unnotched tensile strengths, showing that the notch sensitivity of steel A is higher than properties cannot detect the embrittlement (Espey et al. 1959) that of steel B polycrystals. Various forms of ductile fractures usually the tip of the flaw becomes high enough so that it equals observed in metals are shown in Figs. 1.61b–d and 1.62, the theoretical cohesive strength of the material, there which are as follows: will be breaking of the interatomic bonds in that area. Hence the crack, thus formed, will propagate at a very (a) Single crystals: rapid rate across the member and cause complete sepa- (1) Slant or shearing fracture: Shear stress causes ration of the member leading to fracture almost extensive slip in a single crystal on its active slip immediately. planes and finally separation occurs by shear along • Brittle fracture occurs all on a sudden and must be a plane usually making an angle of 45° with the avoided at any cost. axis of applied load. This results in a shear frac- ture or slant fracture, as shown in Fig. 1.61b, The process leading to ductile fracture involves plastic example of which is the single crystal of HCP deformation by atomic shear. Ductile fractures can have metal where slip may occur on successive basal many varieties that depend on whether the metals are single planes before the final separation by shear. This crystals or polycrystals and the extent of ductility of the type of fracture can occur when the single crystal is so oriented that only one slip system is active.

1.14 Tensile Fracture (a) (b) 83 σ Fig. 1.61 Various usual types of σ σ fractures under uniaxial tensile loading condition at ambient (b)(i) (b)(ii) temperature in a single- and poly-crystalline brittle metals, σ σ exhibiting a flat fracture surface σ lying normal to the tensile axis, (c) b ductile single crystals oriented for single slip, exhibiting slant σ fracture along shear plane making an angle of 45° with the tensile axis, c ductile single crystals due to multiple slip, exhibiting rupture (100% R. A.), d highly ductile polycrystals, exhibiting rupture (100% R. A.) followed by necking. The successive stages in shape change leading to fracture are shown in the series of figures b(i)–(ii), c(i)–(iii), and d(i)–(iii) σ σ (c)(i) (c)(ii) (c)(iii) σ σ σ (d) σ σ σ (d)(i) (d)(ii) (d)(iii) σ σ σ

84 (a) (b) 1 Tension Fig. 1.62 Schematic illustration (c) of successive stages in the formation of a cup–and–cone fracture in moderately ductile polycrystals (d) (e) Cup Slant shear lip Fibrous fracture Cone (2) Rupture: Rupture corresponds to a reduction in 100%, and is usually termed as rupture as shown area at fracture approaching 100%. When multiple in Fig. 1.61d, example of which is gold or lead. slip processes are operative in a single crystal (2) Cup-and-cone fracture: Moderately ductile poly- rupture can occur, as shown in Fig. 1.61c. crystalline metals like mild steel produce “cup-and-cone” fracture, the development of (b) Necking followed by fracture in polycrystals: A local- which consists of several stages as illustrated in ized reduction in the cross-sectional area of tensile Fig. 1.62. At certain point, a triaxial state of tensile specimen, called necking, occurs in a polycrystalline stress developed in the necked section causes small metal before it fractures. Types of fracture depend on micro-structural particles to either break or detach the extent of ductility of the polycrystals, as described from the matrix. This results in the nucleation of below: fine cavities or micro-voids most commonly at (1) Rupture or completely ductile fracture: The necked second-phase particles in the necked region. The sections of highly ductile polycrystalline metals or process of crack initiation can be suppressed by materials of ultra high purity having no inclusion imposing hydrostatic pressure. In fact, Bridgman or pores may actually be thinned down to a line (in established that under application of a sufficiently three dimension) or a point (in two dimension) high hydrostatic pressure, necking can continue prior to their fractures. This type of fracture occurs steadily to cause thinning down of the specimen to in a completely ductile manner, which corresponds a line (or a point) before fracture. However with to a reduction in area at fracture approaching

1.14 Tensile Fracture 85 continuation of strain, the micro-voids undergo a (a) 0.2% offset yield strength. slow, stable growth and ultimately coalescence to (b) UTS. produce a disk-shaped central crack oriented per- (c) (i) Fracture strength and (ii) applied true fracture pendicular to the applied tensile axis. This crack propagates outward in a direction normal to the axis strength. of applied stress, causing to diminish the radial and (d) (i) Elongation and (ii) true strain at fracture. circumferential tensile stresses. When the crack has (e) (i) Reduction of area and (ii) true reduction of area at approached the periphery of the specimen there is only a thin section of material left at the surface to fracture. support the applied load, which is in a state very (f) Zero-gage-length elongation. close to the single crystal subjected to uniaxial tension. Finally, the propagation of crack along a Solution conical shear plane oriented at about 45° to the applied stress axis produces the classical cup-cone Given that L0 ¼ 0:1m; Lf ¼ 0:13 m; Df ¼ 0:021 m; Py ¼ form of ductile fracture as shown in Fig. 1.62e. The 160 Â 103 N, Pmax ¼ 200 Â 103 N, Pf ¼ 170 Â 103 N: As central region of this fracture has a zigzag contour per ASTM specification, L0=D0 ¼ 4; so the initial diameter with a fibrous appearance and so it is called the of specimen is: D0 ¼ 0:1=4 ¼ 0:025 m: fibrous fracture or fibrous zone. The fibrous zone usually has a set of peripheral ridges that indicate Initial cross-sectional area of specimen, slight rise and fall in the direction of stable crack growth. The outer zone of this fracture formed by A0 ¼ p ðD0Þ2¼ p ð0:025Þ2 m2 ¼ 490:874 Â 10À6 m2: shearing action produces a smooth slant shear lip, 4 4 known as shear fracture or slant fracture, which spreads faster than the fibrous fracture. Cross-sectional area of specimen at fracture, Characteristics of a ductile fracture are as follows: Af ¼ p À Á2 4 Df • During as well as prior to the crack propagation, there is a substantial amount of gross plastic deformation that is ¼ p ð0:021Þ2 m2 usually seen at the fracture surfaces. 4 • Tearing of the metal occurs slowly with a slow rate of ¼ 346:36 Â 10À6 m2: crack propagation and usually an appreciable amount of energy is expended in the process of fracture. (a) 0.2% offset yield strength is: • The appearance of a fracture surface at low magnification S0 ¼ Py ¼ 160 Â 103 N mÀ2 is gray and fibrous. A0 490:874 Â 10À6 Fracture surfaces often contain a mixture of fibrous, shear ¼ 325:95 Â 106 Pa ¼ 325 :95 MPa: lip and granular fracture zones. The relative amount of these zones usually depends on the strength of the material and the (b) UTS is: test or deformation temperature and the usual practice is to report the percentage of each of these zones in the fracture Su ¼ Pmax surface. A0 1.15 Solved Problems ¼ 200 Â 103 N mÀ2 490:874 Â 10À6 1.15.1. A round tensile specimen with an initial gage length of 100 mm, fixed as per ASTM specification, shows after ¼ 407:44 Â 106Pa ¼ 407:44 MPa: fracture a gage length of 130 mm and a minimum diameter of 21 mm at fracture. If the load corresponding to the 0.2% (c) (i) Fracture strength is: offset yield is 160 kN, that at the point of the onset of necking is 200 kN and that at the point of fracture is 170 kN, Sf ¼ Pf ¼ 170 Â 103 N mÀ2 determine the following: A0 490:874 Â 10À6 ¼ 346:32 Â 106 Pa ¼ 346:32 MPa: (ii) Applied true fracture strength is: rf ¼ Pf ¼ 170 Â 103 N mÀ2 Af 346:36 Â 10À6 ¼ 490:82 Â 106 Pa ¼ 490:82 MPa: (d) (i) Elongation is: ef ¼ Lf À L0 ¼ 0:13 À 0:1 ¼ 0:3 or 30%: L0 0:1

86 1 Tension (ii) True strain at fracture is: ef ¼ ln A0 ¼ 490:874 Â 10À6 A0 ¼ pD02 ¼ pð12Þ2 ¼ 113:1 mm2: Af ln 346:36 Â 10À6 4 4 ¼ 0:3487 or 34:87%: (e) (i) Reduction of area at fracture is: Hence, the corresponding instantaneous cross-sectional areas are: rf ¼ A0 À Af ¼ ð490:874 À 346:36Þ Â 10À 6 A15 ¼ 1 A0 ¼ 113:1 ¼ 98:3 mm2; A0 490:874 Â 10À6 þ e15 1:15 ¼ 0:2944 or 29:44%: and A25 ¼ A0 ¼ 113:1 ¼ 90:5 mm2: þ e25 1:25 (ii) True reduction of area at fracture is: 1 490:874 Â 10À6 Therefore, the corresponding true stresses from (1.25) are: ln 346:36 Â 10À6 rf0 ¼ ln A0 ¼ P15 3 Â 103 Nmm2; Af A15 98:3 r15 ¼ ¼ ¼ 30:5 ¼ 0:3487 or 34:87%: 3:3 Â 103 Nmm2: 90:5 (f) From (1.131), zero-gage-length elongation is: and r25 ¼ P25 ¼ ¼ 36:5 A25 ÀÁ 1 rf ¼ 1 0:2944 ¼ 0:4172 or 41:72%: Alternative determination of r15 and r35 ef 0¼ À rf À 0:2944 1.15.2. In a tensile test of a cylindrical specimen with an Combining (1.28), and (1.24), we get the true stress r ¼ initial diameter of 12 mm, the loads at 15 and 25% engi- Sð1 þ eÞ ¼ ðP=A0Þð1 þ eÞ; so the corresponding true stresses neering strain are recorded to be 3 and 3.3 kN, respectively. are: If the flow curve of the above specimen is represented by the Hollomon parabolic stress–strain relation, neglect elastic r15 ¼ P3A0105:5ð1Nþme1m5Þ2;¼an3d1Â131:103 Â ð1 þ 0:15Þ strain and calculate the following: ¼ (a) % Uniform elongation. r25 ¼ 3PA6205:5ð1Nþme2m5Þ2;¼ 3:3 Â 103 Â ð1 þ 0:25Þ (b) UTS. ¼ 113:1 (c) 0.2% Offset yield strength. (d) (i) % Uniform reduction of area and (ii) % uniform true Following (1.90b), we can write reduction of area. log r15 ¼ log K þ n log e15 and log r25 ¼ log K þ n log e25 (e) True toughness in the plastic range up to the point of Or; log r25 À log r15 ¼ nðlog e25 À log e15Þ tensile instability. Solution Therefore, strain-hardening exponent is: Given that the initial diameter of specimen, D0 ¼ 12 mm;  0    0   the load at engineering strain e15 ¼ 0:15 is P15 ¼ 3 Â r25 e25 36:5 0:2231 103 N; and that at engineering strain e25 ¼ 0:25 is n ¼ log r15 log e15 ¼ log 30:5 log 0:1398 P25 ¼ 3:3 Â 103 N: ¼ 0:384: From (1.10), the true strains corresponding to engineering strains e15 and e25 are, respectively: Strength coefficient, e15 ¼ lnð1 þ e15Þ ¼ lnð1:15Þ ¼ 0:1398; r25 !! and e25 ¼ lnð1 þ e25Þ ¼ lnð1:25Þ ¼ 0:2231: en25 Nor, mre1n1m55 2 36:5 30:5 K ¼ ¼ 0:22310:384 or, 0:13980:384 ¼ 64:9 : Since from (1.9), L=L0 ¼ 1 þ e and from constancy of volume, (a) From (1.97), we know that uniform true strain is eu ¼ A ¼ ðL0=LÞA0; where L0 and L are initial and instantaneous n ¼ 0:384: From (1.134), uniform elongation or uni- lengths and A0 and A are initial and instantaneous form engineering strain is given by cross-sectional areas of specimen, then A ¼ A0=ð1 þ eÞ: Now,

1.15 Solved Problems 87 eu ¼ expðeuÞ À 1 ¼ expðnÞ À 1 Alternatively, to determine uniform reduction of area: ¼ expð0:384Þ À 1 ¼ 0:468; or; 46:8% uniform elongation: Since eu ¼ Lu À L0 ¼ Lu À 1 ¼ A0 À 1; or; (b) Since the flow curve represented by Hollomon parabolic L0 L0 Au stress–strain relation, i.e. (1.90a) is valid from the onset of yielding to the point of maximum load, so at the point Au ¼ 1 ¼ 1 1 ¼ 0:681: of maximum load the true tensile strength will be: A0 1 þ eu þ 0:468 ru ¼ K44e:nu94¼NKnmn m¼26o4r:;9MÂPa0::3840:384 ) ru ¼ 1 À ðAu=A0Þ ¼ 1 À 0:681 ¼ 0:319; ¼ Or; 31:9% uniform reduction of area: From (1.121), UTS is given by (ii) Uniform true reduction of area is given by Su ¼ ru expðÀeuÞ ¼ 44:94  expðÀ0:384Þ ¼ 30:61 MPa: ru0 ¼ lnðA0=AuÞ ¼ uniform true strain; eu ¼ 0:384; Or; 38:4% uniform true reduction of area: Alternatively, determination of UTS from (1.120): (e) From the true stress–strain curve, true toughness in the Su ¼ 1 ru ¼ 1 44:94 ¼ 30:61 MPa: plastic range up to the point of tensile instability is þ eu þ 0:468 given by ÀÁ Zeu Zn UToughnessu True (c) Given that offset yield strain, i.e. offset engineering ¼ rde ¼ Kende strain at the yield point, is e0 ¼ 0:002: From (1.10), the corresponding true strain is: 0 en þ 1 !n0 K Ânn þ 1à nþ ¼ þ ¼K n 1 10 e0 ¼ lnð1 þ e0Þ ¼ lnð1:002Þ ¼ 1:998  10À3: 64:9 Â0:3840:384 þ 1à mm3 MJm3 ¼ 0:384 þ 1  MN or From (1.91b), the true stress at the yield point, where ¼ 12:47 MJ mÀ3: e0 ¼ 1:998  10À3; is: ½r0Še0 ¼ 1:998  10À 3 ¼ 5K:9e60n 6¼N64m:9mÂ2 oÀr,1M:99P8a  10À 3Á0:384 1.15.3. For a material showing uniform strain distribution ¼ : and obeying the Hollomon parabolic true stress–strain rela- tion up to the point of fracture, From (1.28), 0.2% offset yield strength is given by (a) Prove that in the plastic range, the true toughness is the same as the engineering toughness. ½S0Še0¼0:002¼ ½r0 Še0 ¼1:998Â10À3 ¼ 5:966 ¼ 5:954 MPa: 1 þ e0 1 þ 0:002 (b) Derive an expression for the tensile strength in terms of strain-hardening exponent and strength coefficient. Using (d) If Lu is the length and Au is the cross-sectional area of the data from problem 1.15.2, verify the value of UTS. specimen at the point of tensile instability, then (c) Derive an expression for the tensile-to-yield strength (i) Uniform reduction of area is given by ratio in terms of engineering as well as true yield strain and strain-hardening exponent. Using the data from ru ¼ A0 À Au ¼ 1 À Au : problem 1.15.2, verify the value of the tensile to yield A0 A0 strength ratio from the derived expression for 0.2% offset yield strength. Neglect elastic strain. Since from (1.135b) we can write Au=A0 ¼ expðÀeuÞ; (d) Derive an expression between the yield strength and the ) ru ¼ 1 À ðAu=A0Þ ¼ 1 À expðÀeuÞ reduction in cross-sectional area at the yield point. ¼ 1 À expðÀ0:384Þ ¼ 0:319; Using the data from problem 1.15.2, verify the value of 0.2% offset yield strength corresponding to the reduc- Or; 31:9% uniform reduction of area: tion of area at 0.2% offset engineering yield strain. Neglect elastic strain.

88 1 Tension Solution (b) At the point of maximum load, ru ¼ Kenu: Equation (1.90a), i.e. r ¼ Ken, is given in the problem. Let Using (1.121) and (1.97), we can write us take the definitions of all terms to be the same as those of problem 1.15.2. Su Seuxp¼ðeKuÞ¼neKn:eun ; or, Su expðnÞ ¼ Knn; ) (a) Let us assume ef is the true plastic strain at fracture and ef is the engineering plastic strain at fracture. Since In problem 1.15.2, it was found that K ¼ strain distribution is uniform, so according to (1.10), 64:9 N mm2 or MPa; and n ¼ 0:384: ÀÁ ÀÁ ef ¼ ln 1 þ ef ; or; ef ¼ exp ef À 1: ) UTS; Su ¼ 64:920::3718840:384MPa = 30:61 MPa: The true toughness in the plastic range is the area under the true plastic stress–strain curve up to the point of fracture Hence, the value of UTS is the same as that found in and will be given by problem 1.15.2. ÀÁ Zef Zef K Âen Ãef UToughness True þ ¼ r de ¼ K ende ¼ þ 1 0 n 1 (c) At the yield point, r0 ¼ Ken0: 00 ¼ n K 1 enf þ 1 ¼ n K 1 ÂÀ ef ÁÃn þ 1: Using (1.28) and (1.10), we can write þ þ ln 1 þ r S0ð1 þ e0Þ ¼ K½lnð1 þ e0ފn; From (1.28), we know that S ¼ 1 þ e : The engineering toughness in the plastic range is the area Or, S0 K¼Kne½lnn0ð11Kþþ½elen00ðÞ11Šnþþ¼ee00Kފenee0n0 ) Su ¼ under the engineering plastic stress–strain curve up to the S0 point of fracture and will be given by ÀÁ Zef Zef r Zef K en ¼ nn 1 þ e0 : UToughness Engg: 1þ 1þe e ½lnð1 þ e0ފn ¼ S de ¼ e de ¼ de 00 0 In problem 1.15.2, it was found that n ¼ 0:384; and Su=S0 ¼ 30:61=5:954 ¼ 5.141. Zef K½lnð1 þ eފn 1þe For 0.2% offset yield strength, since offset engineering ¼ de: 0 yield strain e0 ¼ 0.002, Let lnð1 þ eÞ ¼ z; then de ¼ dz: ) Su ¼ 0:3840:384 1 þ 0:002 ¼ 5:141: 1þe S0 2:718 ½lnð1 þ 0:002ފ0:384 ÀÁ Zef K½lnð1 þ eފn UToughness Engg:¼ 1þe ) de Again, using (1.29), we can write 0 S0 expðe0Þ ¼ K en0; lnðZ1 þ ef Þ ¼ K Âzn þ 1 Ãlnð1 þ ef Þ )Or,SSu0S¼0 ¼¼KKneneexnnpe0ðn0eeee0Kn00Þe:ee0n¼0 ¼ K zndz þ n 1 0 0 ¼ K ÂÀ þ ef ÁÃn þ 1:  n neðe0ÀnÞ þ ln 1 e0 n 1 Now, it is proved that ÀÁ ¼ ÀÁ n K 1 ÂÀ þ ef ÁÃn þ 1 For 0.2% offset yield strength, since offset true yield UToughness True UToughness Engg:¼ þ ln 1 strain e0 ¼ lnð1 þ e0Þ ¼ lnð1:002Þ ¼ 1:998  10À3; ¼ K enf þ 1: þ n 1

1.15 Solved Problems 89 ) Su ¼ 1:9908:3Â8410À30:384expÀ1:998  10À3 Á (a) Uniform engineering strain. S0 À 0:384 (b) UTS. (c) Modulus of elasticity based on the true yield stress in ¼ 5:141: the Ludwik equation. Hence, the value of the tensile to yield strength ratio for (d) Plastic work done per unit volume up to the point of offset engineering as well as true yield strain is the same as that found in problem 1.15.2. uniform true strain. (e) Elastic work done per unit volume based on the true (d) Let Ly is the length and Ay is the cross-sectional area of specimen at the yield point, then from constancy of yield stress in the Ludwik equation. volume, L0 Ly ¼ Ay A0: (f) Percentage of work done for elastic deformation with respect to the uniform plastic work done. Solution Hence, reduction of area at the yield point is given by From (1.95), we know that at the point of maximum load up to which true or engineering strain is uniform, the condition ry ¼ A0 À Ay ¼ 1 À Ay : is dr=de¼r: A0 A0 Hence from the given Ludwik equation, we can write the From (1.7), true yield strain e0 ¼ ln Ly ¼ ln A0 ¼ ln 1 1 ry ; condition at the point of maximum load as: L0 Ay À ð507  0:35Þe0:35À1 ¼ 28 þ 507 e0:35; so, expðe0Þ ¼ 1 1 ry : Or, 28 þ 507e0:35 À 177:45 eÀ0:65 ¼ 0: À Let 28 þ 507 e0:35 À 177:45 eÀ0:65 ¼ y; when y ¼ 0; then At the yield point, r0 ¼ K en0; so from (1.29), we get e ¼ eu (uniform true strain). S0 expðe0Þ ¼ K 1 !n S0 ! Let us take the starting value of e ¼ 0:35 and then with ) ln 1 À or; 1 the change of the values of e, find by trial and error the 1 !n 1 À ry values of y which will be either zero or very close to zero, ln 1 À ry ry with a positive as well as a negative value. ¼K e 0.35 0.30 0.32 0.33 0.325 0.323 0.324 0.3235 0.3234 y 28 −27.4 −3.9 7.15 1.684 −0.536 0.576 0.02078 −0.09044 ÀÁ 1 !n S0 ¼ K 1 À ry ln À ry : 1 Reduction of area corresponding to 0.2% offset engi- neering yield strain is: ry ¼ 1 À Ay ¼ 1 À L0 ¼ 1 À 1 1 ¼ 1 e0 It will not be unreasonable to assume linearity between A0 Ly þ e0 þ e0 the two coordinate points of ð0:3235; 0:02078Þ and ð0:3234; À0:09044Þ; which are very close to each other but ¼ 0:002 ¼ 1:996  10À3: opposite sides of y ¼ 0: Equation of this straight line will be: 1:002 In problem 1.15.2, it was found that K ¼ y À 0:02078 ¼ e À 0:3235 ; 64:9 N mm2 or MPa; and n ¼ 0:384; so 0.2% offset yield 0:02078 þ 0:09044 0:3235 À 0:3234 strength from the derived relation can be obtained as follows: Or, e ¼ 0:3235 þ y À 0:02078  0:0001 0:11122 À 10À3Á 1 !0:384 ¼ 0:3235 þ ðy À 0:02078Þ Â 0:899  10À3: 64:9 1 ln 1 À 1:996  10À3 ½S0 Še0 ¼0:002 ¼ À 1:996  It is obvious that the straight line joining these above two ¼ 5:954 MPa: coordinate points will intersect y ¼ 0; and at this intersection point, the value of e is the uniform true strain, i.e. e ¼ eu: Hence, the value of 0.2% offset yield strength is the same Hence putting y ¼ 0 and e ¼ eu in the above straight line as that found in problem 1.15.2. equation, we get 1.15.4. The flow curve of a metal is described by the Ludwik Uniform true strain, eu ¼ 0:3235 þ ð0 À 0:02078Þ equation of the form rðMPaÞ ¼ 28 þ 507e0:35: Determine 0:899  10À3 ¼ 0:32348: the following:

90 1 Tension (a) From (1.134), uniform elongation or uniform engi- (f) Percentage work done used for elastic deformation with neering strain is: respect to the uniform plastic work done is eu ¼ expðeuÞ À 1 ¼ expð0:32348Þ À 1 ¼ 0:3819; 3:57  100 ¼ 0:0039%: Or; 38:19% uniform elongation: 90:89  103 (b) True tensile strength, Exercise ru ¼ 28 þ 507 e0u:35 1.Ex.1. A round isotropic metal specimen of 100 mm length ¼ 28 þ 507 ð0:32348Þ0:35 and 20 mm diameter is elastically deformed in tension to a ¼ 369:55 MPa : length of 100.1 mm and a diameter of 19.994 mm. Calculate its Poisson’s ratio and volume strain. From (1.121), UTS, Su ¼ ru expðÀeuÞ ¼ 369:55 expðÀ0:32348Þ ¼ 267:42 MPa: 1.Ex.2. The principal strains are measured to be 0.003 and 0.002 by strain gages on the free surface of a thin copper (c) The modulus of elasticity E can be computed from (1.92), sheet. The modulus of elasticity for copper is 129.8 GPa, in which we can put r0 ¼ 28 MPa, K ¼ 507 MPa and its Poisson’s ratio is 0.343. Calculate the corresponding and n ¼ 0:35; from the given Ludwik equation. principal stresses. Hence, 1.Ex.3. Calculate the average elastic strain energy stored by an edge dislocation per unit length of the dislocation in an  1   1 annealed crystal of a metal for which the shear modulus is K 1Àn 507 1À0:35 70 GPa and Burgers vector is 0.3 nm. r0 ¼ 28 ¼ En E0:35 or; ¼ ; 1.Ex.4. Show that for a uniaxial tensile test the equations for significant stress and strain reduce to the tensile stress and 1 strain values. ð507Þ0:35 E ¼ ;1À0:35 1À0:35 28 )E¼ 5070:135 MPa ¼ 109904:4 MPa 1À0:35 ð28Þ 0:35 ¼ 109:9 GPa: (d) Plastic work done per unit volume up to the point of 1.Ex.5. When the mean grain diameter of a polycrystalline uniform true strain will be given by material is reduced from 0.05 to 0.01 mm, it is found that the Zeu 0:Z32348 À 507e0:35Áde yield strength increases from 125 to 250 MPa. Assuming r de ¼ 28 grains to be spherical, find the yield strength of that material þ 00 for an ASTM grain-size index number of 8. ¼ 28  0:32348 þ 507 1 ð0:32348Þ0:35 þ 1 MN mm3 or, MJ mÀ3 0:35 þ 1.Ex.6. Assuming constancy in volume, calculate the ¼ 90:89 MJ mÀ3: engineering strain, true strain, reduction in cross-sectional area and true reduction in cross-sectional area for a rod (e) Elastic work done per unit volume based on the true which is yield stress in the Ludwik equation can be written as (a) stretched to 2 times of its initial length. r20 ð2  EÞ from (1.108b), where r0 is the true yield (b) compressed to half of its initial length. stress and E is the modulus of elasticity. From the given Ludwik equation, it is seen that r0 ¼ 28 MPa: 1.Ex.7. If a sheet specimen with an initial gage length of Elastic work done per unit volume ¼ 2 r02 100 mm, fixed as per current specification in Great Britain, ÂE shows a cross-sectional area of 190 mm2 at fracture, what will be the value of zero-gage-length elongation for that ¼ 282 MPa or MJm3 sheet specimen? 2ð109:9  103Þ ¼ 3:57  10À 3 MJ m3 ¼ 3:57 kJ mÀ3: 1.Ex.8. A 250-mm-long rod with an initial diameter of 25 mm is stretched under a tensile load of 4 kN. If the

1.15 Solved Problems 91 diameter of the rod reduces to 22 mm at this load, assuming (a) modulus of elasticity; and constancy in volume, calculate the following: (b) r0: (a) The corresponding final length of the rod. 1.Ex.14. The flow curve of an isotropic homogeneous duc- (b) The corresponding true stress and strain. tile metal is described by Hollomon parabolic stress–strain (c) The corresponding engineering stress and strain. relation with strength coefficient value of 300 MPa and strain-hardening exponent value of 0.4. If the initial 1.Ex.9. A 100-mm-long member made of an aluminium cross-sectional area of the metal is 9 mm2, calculate the alloy with a rectangular cross-section of 20 mm by 15 mm is following: subjected to a tensile load of 32 kN. Assuming linear elastic behaviour with Young’s modulus of 72 GPa and Poisson’s (a) Rate of strain hardening of that metal at the point of ratio of 0.3, determine the total change in (a) length, (b) cross-sectional dimensions, and (c) volume. (i) 0.2% offset yield strength, neglecting elastic strain; and (ii) tensile instability. 1.Ex.10. A member whose diameter is 30 mm shows a (b) Maximum load at the point of tensile instability. linear elastic deformation of 0.03 mm in gage length of (c) Cross-sectional area of the metal at the point of maxi- 150 mm under an applied tensile load of 12 kN. Compute mum load. the following: 1.Ex.15. Assume that the flow curve of copper in the region (a) Modulus of elasticity. of uniform plastic deformation is described by r ðMPaÞ ¼ (b) Strain energy per unit volume at the applied load. 317 Á e0:54: Compute the following: (c) Modulus of resilience, if the load experienced by the material of the member at the proportional limit is 55 kN. (a) Value of true strain rate corresponding to the engineering strain rate value of 0.1 s−1 at the point of maximum load for 1.Ex.11. A member of 400 mm length is to be designed to copper, neglecting the effect of elastic deformation of the withstand a tensile load of 27 kN without exceeding the testing machine on flow properties. yield strain. Three choices are available: (b) Amount of work done per unit volume on copper in the uniform plastic deformation range. (i) Medium-carbon steel, for which elastic modulus, E ¼ 207 GPa and yield strength, S0 ¼ 310 MPa: 1.Ex.16. Assume that the flow curve of steel in the region of (ii) A magnesium alloy, for which E ¼ 45 GPa and uniform plastic deformation is described by rðMPaÞ ¼ S0 ¼ 196:5 MPa: 1270 Á e0:15: Determine the (iii) An acrylic polymer, for which E ¼ 3:4 GPa and S0 ¼ 14 MPa: (a) UTS; (b) Uniform elongation; Using a safety factor of 1.5, (c) Uniform reduction of area. (d) Toughness in the plastic range up to the point of the (a) Determine the necessary cross-section for each material. onset of necking. (b) Compute the total amount of strain energy stored by each at the 27 kN load. 1.Ex.17. If the flow stress of a material at 200 and 1000 °C are, respectively, 200 and 70 MPa at constant strain and 1.Ex.12. If a sheet specimen with an initial cross-sectional strain rate, compute the activation energy for plastic flow in area of 400 mm2, on unloading from the point of tensile J mol−1. instability, shows a permanent extension of 59 mm mea- sured over the initial gage length fixed as per ASTM Stan- 1.Ex.18. At a true strain value of 0.25, the strain-rate sen- dard and obeys Hollomon parabolic stress–strain relation sitivity of a metal at 25 °C is 0.06 and that at 450 °C is 0.2. with strength coefficient of 317 MPa, calculate the UTS of At the same true strain value at a strain rate of 10À2 s-1, if the that material. flow stress is found to be 53.3 MPa at 25 °C and 5.8 MPa at 450 °C, determine the factor by which the flow stress will 1.Ex.13. The flow curve of a metal in the form of a Ludwik change at each of the temperatures when the strain rate is equation is described by rðMPaÞ ¼ r0 þ 640e0:15; and its increased to 102 s−1. modulus of resilience based on the value of r0 in the above Ludwik equation is 128:69 kJ m3: Calculate the value of

92 1 Tension 1.Ex.19. Indicate the correct or most appropriate answer(s) (A) e ¼ pn;ffiffi (B) e ¼ n2; from the following multiple choices: (C) e ¼ n; (D) e ¼ 2n; (a) In general, strain-hardening coefficient increases with where e ¼ true plastic strain and n ¼ strain-hardening (A) increasing both strength level ðrÞ and stacking fault exponent. energy (SFE); (B) decreasing both r and SFE; (i) If a round plastic metal with diameter of 20 mm and (C) increasing r and decreasing SFE; length of 100 mm is uniformly elongated to 400 mm, the (D) decreasing r and increasing SFE. diameter of that metal after deformation will be: (b) The strain-rate sensitivity of flow stress for the occur- (A) 5 mm; (B) 10 mm; rence of superplasticity is in the range: (C) 15 mm; (D) 20 mm. (A) Less than 0.1; (B) 0.1–0.2; (j) For superplastic deformation in an alloy, the necessary conditions are: (C) 0.4–0.9; (D) 1.5 to 2.0. (A) an extremely fine and uniform grain size; (c) Same elongation will result for 14-mm-wide two sheet (B) high strain rate; specimens of same metal, if one specimen has thickness of (C) coarse and non-uniform grains; 2 mm and gage length of 30 mm and another specimen has (D) high homologous temperature. thickness of 8 mm and gage length of (k) The characteristic mechanical behaviour of an elastomer (A) 15 mm; (B) 30 mm; or rubber is: (C) 60 mm; (D) 120 mm; (E) none of the above. (A) large recoverable strains; (B) an adiabatic decrease in temperature on stretching; (d) If zero gage length elongation of a material is 0.25, the (C) an increase in elastic modulus with increasing corresponding value of engineering reduction of area at temperature; fracture is (D) a decrease in elastic modulus with increasing temperature. (A) 0.2; (B) 0.25; (C) 0.3; (D) none of the above (l) Which of the following statements are true about edge dislocation? (e) The elastic strain energy of a dislocation is related to its Burgers vector as follows: (A) edge dislocations have an incomplete extra plane of atoms associated with them; (A) directly proportional; (B) edge dislocations lie parallel to the Burgers vector; (B) proportional to the square of the Burgers vector; (C) edge dislocations can avoid obstacles by cross-slip; (C) proportional to the square root of the Burgers vector; (D) depending on the geometry, parallel edge dislocations of (D) not related at all. opposite sign can attract or repel each other. (f) Von Mises’ yielding criterion for ductile metals predicts (m) According to Hall–Petch relation, the yield strength of a that the yield stress in uniaxial tension is related to that in metal is related to its grain size d as pure shear as:  pffiffiffi pffiffiffi (A) equal to each other; (B) 2ptffiffiimes; (A) d2; (B) 1 d2; (C) d; (D) 1 d: (C) one half; (D) 3 times. (n) As compared to the engineering stress–strain curve, the (g) Critical resolved shear stress in single crystal is calcu- portion of the true stress–strain curve beyond the necking of lated by applying a ductile specimen in tension is: (A) Hooke’s law; (B) Braggs’ law; (A) below and to the left; (B) below and to the right; (C) Schmid’s law; (D) Coulomb’s law. (C) above and to the right; (D) above and to the left. (h) If a strain-hardening metal obeys Hollomon true stress– Answer to Exercise Problems strain relation, then necking in a tensile specimen occurs 1.Ex.1. 0.3; and 4 Â 10À4: when

1.15 Solved Problems 93 1.Ex.2. 542.2 MPa; 445.6 MPa. Backofen, W.A., Turner, I.R., Avery, D.H.: Trans. ASM 57, 981 1.Ex.3. 3:15 Â 10À9J=m: (1964) 1.Ex.5. 165.9 MPa. 1.Ex.6. (a) e ¼ 1; e ¼ 0:693; r ¼ 0:5; r0 ¼ 0:693; Bain, E., Paxton, H.W.: Alloying Elements in Steel, 2nd edn, p. 37. (b) e ¼ À0:5; e ¼ À0:693; r ¼ À1; r0 ¼ À0:693: American Society for Metals, Metals Park, Ohio (1961) 1.Ex.7. 64.87%. 1.Ex.8. (a) 322.83 mm; (b) 10.5 MPa, and 25.6%; Barba M.J.: Résistance à Ia traction et allongements des métaux après (c) 8.15 MPa, and 29.1%. rupture. Mem. Soc. Ing. Civils, 682, pt. 1, France, ler partie, (1880) 1.Ex.9. (a) þ 0:148 mm (increase); (b) longer dimension: À8:89 Â 10À3mm (decrease) and shorter dimension: Bechtold, J.H.: Acta Metall. 3, 253 (1955) À6:67 Â 10À3mm (decrease); (c) þ 17:78 mm (increase). Bridgman, P.W.: Trans. ASM 32, 553 (1944) 1.Ex.10. (a) 84.9 GPa; (b) 1.698 kJ m−3; (c) 35.66 kJ m−3. Bridgman, P.W.: Effects of high hydrostatic pressure on the plastic 1.Ex.11. (a) (i) 130.6 mm2, (ii) 206.1 mm2, (iii) 2892.86 mm2; (b) (i) 5.39 J, (ii) 15.72 J, (iii) 14.82 J. properties of metals. Rev. Mod. Phys. 17, 3–14 (1945) 1.Ex.12. 135.6 MPa. Cahn, R.W.: Adv. Phys. 3, 363–445 (1954) 1.Ex.13. (a) 207 GPa; (b) 230.82 MPa. Chokski, A.H., Rosen, A., Karch, J., Gleiter, H.: On the validity of the 1.Ex.14. (a) (i) 4998.3 MPa, (ii) 207.94 MPa; (b) 1254.5 N; (c) 6.03 mm2. Hall-Petch relationship in nanocrystalline materials. Scr. Metall. 23, 1.Ex.15. (a) 0.058 s−1; (b) 79.7 MJ m−3. 1679 (1989) 1.Ex.16. (a) 822.4 MPa; (b) 16.18%; (c) 13.93%; Conrad, H.: J. Iron and Steel Inst. 198, 364 (1961) (d) 124.63 MJ m−3. Considère, A.: Ann. ponts et chausses 9(6), 574–775 (1885) 1.Ex.17. 6.57 kJ mol−1. Cotrell, A.H.: Dislocation and Plastic Flow in Crystals. Clarendon 1.Ex.18. At 25 °C: increase by a factor of 1.74; and at Press, Oxford (1953) 450 °C: increase by a factor of 6.31. Cotrell, A.H.: Vacancies and Other Point Defects in Metals and Alloys, 1.Ex.19. (a) (B) decreasing both r and SFE. (b) (C) 0.4 to p. 1. Institute of Metals, London (1958) 0.9. (c) (C) 60 mm. (d) (A) 0.2. (e) (B) pprffioffi portional to the Cottrell, A.H., Hunter, S.C., Nabarro, F.R.N.: Phil. Mag. 44, 1064 square of the Burgers vector. (f) (D) 3 times. (g) (C) (1953) Schmid’s law. (h) (A) e ¼ n: (i) (B) 10 mm. (j) (A) an Datsko, J.: Material Properties and Manufacturing Processes, pp. 18– extremely fine and uniform grain size; (D) high homologous 20. Wiley, New York (1966) temperature. (k) (A) large recoverable strains; (C) an Davis, H.E., Troxell, G.E., Wiskocil, C.T.: The testing and inspection increase in elastic modulus with increasing temperature. of engineering materials, 3rd edn, p. 119. McGraw-Hill Book (l) (A) edge dislocations have an incomplete extra plane of Company, New York (1964) atoms associated with them; (D) depending on the geometry, Decker, R.F.: Metall. 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Compression 2 Chapter Objectives • Standard specimen, elastic and plastic range, stress, strain and strain rate, brittle and ductile materials in compression. Advantages over tension test. • Bauschinger effect. • Barreling and buckling in compression test and remedial measures against them. Critical column stress for buckling. • Various compressive failures. • Problems and solutions. 2.1 Introduction cohesive strength is exceeded during tensile loading, the material will fail in tension. On the contrary, no such limi- Compression test is simply the reverse of the tension test tation exists in the increase of interatomic repulsive stress with reference to the sense or direction of applied stress. within the material; as a result, it might be concluded that the There are many similarities between static compression and actual compressive strength of such a material is limitless and tension test as far as the behaviour of materials is concerned. the applied normal compressive stress always produces an However, static compression differs adequately from static elastic action. Therefore, the resolved shear stress acting on tension in some important matters, such as (1) manner of inclined planes must be responsible for inelastic action in loading, (2) test specimen geometry, (3) frequent differences compression. In compression of homogeneous non-porous in stress–strain diagrams with a pronounced difference in the body, slip on the inclined planes produced by the resolved plastic range, (4) the extent of ductility shown by a material, shear stress follows the same laws and equations as in tension. (5) the mode of failure and fracture surface appearance. Compression test is preferable to tension test for brittle Although this chapter involves the discussions on simple materials. Compression test becomes essential for ductile uniaxial compression, it is worthy to point out the effect of materials which undergo more than 50% plastic strain during hydrostatic compression in the inelastic range. Hydrostatic mechanical working processes. compression is a state of stress where equal compressive stresses are applied from all directions on a material, such as Pure compressive stress causes to push the atoms in a a body submerged in water that experiences equal pressures material closer together, which obviously cannot produce from all directions. Hydrostatic compression pushes the failure of a homogeneous non-porous body. The atomic bond atoms and molecules of the material from all directions and curve in Fig. 1.39 shows that there is a definite peak in the brings them closer and closer, but their atomic structure is interatomic attractive binding stress that corresponds to the not distorted, i.e. they maintain absolutely identical config- theoretical cohesive strength of the material. Once this urations as before. Hence, there is a simple reduction in volume of the material but no tendency for plastic © Springer Nature Singapore Pte Ltd. 2018 95 A. Bhaduri, Mechanical Properties and Working of Metals and Alloys, Springer Series in Materials Science 264, https://doi.org/10.1007/978-981-10-7209-3_2

96 2 Compression deformation. If an ideal state of hydrostatic compression (i.e. that the test piece is cantered with respect to the centre of the exactly equal compressive stresses acting from all directions) spherical surface of this block. is maintained throughout, the above condition may extend to The standard compression specimens recommended by any level of stress and failure of material becomes mostly the ASTM E9 for metallic materials fall into the following impossible because failure cannot occur simply due to three categories: gathering together of atoms. 1. ASTM short specimen—It is either a 2-in. (50.8 mm) cube or a cylinder of 1 in. (25.4 mm) long and 1 1 in: 8 2.2 Standard Specimen (28.575 mm) in diameter, i.e. L0C ðinitial lengthÞ=D0C ðinitial diameterÞ ¼ 0:889 ’ 0:9: It is proposed to use For standard compression test method of metals at room for brittle metallic materials. temperature, the reader is referred to ASTM Standard 2. ASTM medium length specimen—It is usually a (ASTM E9 2009). The specimen employed in compression cylindrical specimen used for general purpose, whose test is generally a plain right cylinder or prism with proper initial length L0C to initial diameter D0C ratio is 3, that is dimensions in accordance with the ASTM Standards. A L0C =D0C ¼ 3: Different initial lengths and initial diame- cylindrical specimen is usually preferred over other shapes ters are proposed, which are shown in Table 2.1 (Davis because a circular cross-section can be uniformly stressed. et al. 1964). However, a square or rectangular cross-section may also be (3) ASTM long cylindrical specimen—It is used for more used. To avoid bending of the specimen due to eccentricity accurate test results and to determine the modulus of in loading, the specimen is precisely machined so that the elasticity. The long specimens have initial lengths to straightness of the specimen and the flatness and parallelism initial diameters ratios of 8 or 10, as shown in Table 2.2 of both ends of the specimen are maintained as far as pos- (Davis et al. 1964). sible. The specimen is kept between two parallel compres- sion flat plates, also known as anvils or bearing blocks, For compression tests of sheet metals, specimens are attached to a universal testing machine that applies com- loaded in a jig, and the details of which along with the pressive load to the specimen. The lower compression plate specimen are described in ASTM E9. The jig is used as remains fixed on the frame of the machine, whereas the support to prevent bending of the specimen, but it must not upper one is fitted with the movable cross-head of the obstruct the axial deformation of the specimen. machine. Prior to placing the specimen between the anvils, care must be taken to observe that the axis of the specimen For concrete and mortars, the standard specimen is a matches with the centre-line of bearing blocks of machine to cylinder whose initial length L0C to initial diameter D0C ratio avoid the effect of eccentricity which is more noticeable in is 2, that is L0C =D0C ¼ 2; although ASTM has specified a compression than in tension. In order to overcome the effect 2 in. (50.8 mm) cube as specimen for mortar. Further, a 6-in. of lack of perfect parallelism of both ends of the specimen, it (152.4 mm) cube is commonly used as a specimen for is desirable to place a spherically seated bearing block at the concrete in England and in Europe. Dimensions of cylin- upper end of the test piece so that an uniform stress is drical specimens for concrete depending on the size of their applied on each end of the test piece. It must be observed constituent aggregates as well as for mortars are shown in Table 2.3 (Davis et al. 1964). Table 2.1 Dimensions of Initial length L0C Initial diameter D0C L0C ASTM medium length cylindrical D0C specimens (Davis et al. 1964) 1 1 in: ð38:1 mmÞ 1 in: ð12:7 mmÞ 2 2 3 2.976 2 3 in: ð60:325 mmÞ 0.798 in. (20.2692 mm) 3 8 3 3 in. (76.2 mm) 1 in. (25.4 mm) L0C D0C 3 3 in: ð85:725 mmÞ 1 1 in: ð28:575 mmÞ 8 8 7.9887 ’ 8 10 Table 2.2 Dimensions of Initial length L0C Initial diameter D0C ASTM long cylindrical specimens (Davis et al. 1964) 6 3 in: ð161:925 mmÞ 0.798 in. (20.2692 mm) 8 12 1 in: ð317:5 mmÞ 1 1 in: ð31:75 mmÞ 2 4

2.2 Standard Specimen Material Initial length 97 L0C Table 2.3 Dimensions of Concrete containing aggregates of maximum size up to 0.75 in. Initial diameter cylindrical compression (19.05 mm) 6 in. D0C specimens for concrete and Concrete containing aggregates of maximum size up to 2 in. (152.4 mm) 3 in. (76.2 mm) mortars (50.8 mm) Concrete containing aggregates of maximum size up to 2.5 in. 12 in. 6 in. (63.5 mm) (304.8 mm) (152.4 mm) Concrete containing aggregates of maximum size up to 6 in. 8 in. (152.4 mm) 16 in. (203.2 mm) Mortars (406.4 mm) 18 in. (457.2 mm) 36 in. 2 in. (50.8 mm) (914.4 mm) 4 in. (101.6 mm) Compression specimen for wood is generally a rectan- Elastic stress gular prism whose dimensions are 2 by 2 by 8 in. (50.8 by in tension 50.8 by 203.2 mm) for tests performed parallel to the grains of wood or 2 by 2 by 6 in. (50.8 by 50.8 by 152.4 mm) Elastic strain Tension when tests are conducted perpendicular to the grains of in compression wood. 0 2.3 Elastic Range The elastic action for non-porous crystalline solids in com- Elastic strain pression is absolutely same as that in tension but in the in tension opposite direction. For a given applied compressive stress, the squeezing of atoms from their equilibrium interatomic Compression Elastic stress spacing causes to develop the necessary internal interatomic in compression repulsive force to balance the applied compressive stress at roughly the same atomic displacements as in case of tensile Fig. 2.1 Elastic stress–strain diagram in tension and compression for a loading. The reason is that the slope or curvature on either material obeying Hooke’s law side of the equilibrium point in the curve of interatomic bond stress versus interatomic spacing does not vary sharply; C rather Fig. 1.38 reveals the continuity of the curve through its equilibrium. Thus, for non-porous homogenous and iso- Rubber tropic crystalline materials including polycrystals obeying Hook’s law, the linear elastic stress–strain diagram in ten- Elastic stress in compression Cork sion is extended in the reverse direction to get the elastic curve in compression as shown in Fig. 2.1, because the B stress and strain in tension are assumed to be positive while A those in compression are negative during mechanical testing of materials. But usually the elastic strain in compression is 0 more than that in tension, and hence, the only difference Elastic strain in compression between the two elastic curves lies in the amount of their elastic strains. Fig. 2.2 Elastic stress–strain curves in compression for rubber and cork Under compression, long chain polymers like rubber or fibrous and cellular materials like wood and cork exhibit nonlinear elastic stress–strain diagrams. Elastic stress–strain diagrams in compression for rubber and cork are shown in Fig. 2.2. The structure of rubber having a random coiled configuration in the unstressed condition stiffens more and more quickly with increase in compressive stress on it,

98 2 Compression exhibiting a rise of stress with strain from the beginning but instantaneous cross-sectional area AC of the member con- the rate of increase in stress with strain becomes gradually tinues to increase due to Poisson effect and the strain hard- more and more, as evident in its stress–strain curve. The ening always increases the true stress rC required to behaviour of plastic foam in compression is quite similar to compress the member. Therefore, the axially applied com- that of rubber in compression. Wood and cork show an pressive load PC always increases with reduction in length, initial rise of stress with strain because of the deformation DLC; of the member in the plastic range because PC ¼ resistance produced by the stiffness of their cell walls. When rC Â AC: During compressive deformation of a member, the the compressive stress increases to such high level that their spreading of the material over the anvils to increase cell walls undergo lateral instability, i.e. they buckle, there is an abrupt reduction in their stiffness. Under such condition, cross-sectional area of the member is resisted by the fric- there is a slight rise of stress with strain producing a segment from A to B in the stress–strain curve as shown in Fig. 2.2. tional stresses developed at the contact area between the As the cells are compressed further, the compaction causes the stiffness to increase once more and the stress–strain member and the anvils because the frictional stresses act in a curve rises rapidly upwards to point C (Fig. 2.2). The stress– strain diagram for cork or wood in compression looks very direction opposite to the direction of flow of material, i.e. in similar to that for rubber in tension. the inward direction and will thus be compressive in nature. The elastic properties in compression test, such as mod- ulus of elasticity, proportional limit, elastic strength or limit, So, the region of the member in contact with as well as near modulus of resilience, and Poisson’s ratio for linear elastic materials and secant modulus, tangent modulus and resi- the anvils will be in a state of triaxial compressive stresses. lience for nonlinear elastic materials, are determined abso- lutely in the same manner as in the tensile test. Just like This triaxial state is developed by the axial compressive true tension, the volume of the object subjected to compression stress rC applied in the longitudinal direction and the increases and always shows a value of Poisson’s ratio less imposed compressive stresses sC acting inward in the other than 1=2 in the elastic range. two lateral directions, as shown in Fig. 2.3. With the help of Von Mises’ yielding criterion given by (1.64), it can be proved as follows that the longitudinal compressive true stress, rC; required to deform the material near the ends of the member is higher than the flow stress under uniaxial state of compression r0C . As all stresses of the triaxial state are compressive, hence it is assumed that these compressive stresses are positive quantities. So, r0 ¼ r0C ; r1 ¼ rC and r2 ¼ r3 ¼ sC: 2.4 Plastic Range Hence, Beyond the elastic range, the plastic range starts and con- r0C ¼ p1ffiffi h À sC Þ2 þ ðsC À sC Þ2 þ ðsC À rC Þ2 i1=2 tinues till the point of final failure of the material. The most 2 ðrC commonly used compressive properties in the plastic range are yield strength, compressive strength, secant modulus, ¼ p1ffiffi h ðrC À sC Þ2 i1=2 ; tangent modulus and toughness. Definitions of most of the 2 2 properties are same as those of the corresponding tensile properties, and they are determined in the same way as in Or, r0C ¼ rC À sC; or rC ¼ r0C þ sC; i:e: rC [ r0C tension. ð2:1Þ Since necking of the specimen does not occur in com- pression, the problem of plastic instability as encountered in The above (2.1) shows that the material near the anvil the tension test does not arise in the compression test. So, the surfaces is strengthened, i.e. a zone of undeformed material plastic range is likely to continue over a much larger strain is created internally near the ends of the member. The value in compression than in tension. In compression, strain transverse compressive forces that restrain the outward flow hardening takes place after the initiation of yielding and of material are the greatest at the end surfaces in contact with volume remains constant in plastic range, just like tension. If the anvils and decrease gradually towards the mid-length of a body is compressed along its longitudinal axis, it will the member. It has been found that for a cylindrical speci- expand laterally and the lateral strain of the body will be half men of initial diameter D0C , these restraining forces become of its longitudinal strain because the value of Poisson’s ratio zero at a distance of approximately equal to D0C from any is again 1=2 during compressive plastic deformation. Hence end of the specimen. Hence, the region of stronger unde- with the progress of compressive plastic deformation, the formed material existing internally near the ends of the member takes the shape of a cone as shown by the shaded region in Fig. 2.4. The axial compressive load versus

2.4 Plastic Range Friction forces 99 Fig. 2.3 Triaxial state of compressive stresses at the ends Anvil Direct axial of a compression specimen in PC compressive stress, σC contact with anvils—the diagram shows only one end of the Lateral specimen compressive stress, τC Fig. 2.4 Conical undeformed regions (shaded) due to interface Lateral compressive friction at the ends of compression specimens with stress, τ due to interface length to diameter ratios being C larger in (a) and smaller in (b) friction acting inward Element at centre of end of specimen Anvil (a) (b) Pc Specimen before compression Pc Compressed specimen Conical undeformed regions deformation curve is shown in Fig. 2.5. With decrease in 2.4.1 Stress and Strain length of the compression specimen, the above-mentioned cone-shaped region of stronger material from each end of the The manner of determining conventional and true stress and specimen comes near to each other and tends to overlay. strain in the compression test is similar to that in the tension This action causes a rapid increase in the axial compressive test. All the equations for stress and strain and their relation load required for further reduction in length of the specimen developed in tension can be used for compression provided resulting in an upward bend in the load–deformation curve, all the values of compression are substituted in those equa- as can be seen from Fig. 2.5. For constant diameters of tions with negative signs, otherwise the following equations, cylindrical compression specimens, the shorter the specimen obtained by making all vector quantities of the tensile the larger will be the proportion of stronger undeformed equations negative, can be used for compression keeping material region in the specimen and the greater will be the their values positive. Any quantity of the tensile equation axial compressive load to produce the equal percentage of with subscript ‘Ç’ represents the same quantity in com- reduction in length of the specimen. Therefore, the load– pression. Hence from (1.5) and (1.7), we get deformation curves in compression shift upwards at any given value of reduction in length of the specimen, when the Conventional or engineering compressive strain, ratio of initial diameter D0C to initial length L0C , i.e. D0C =L0C of the specimen increases, as exhibited in Fig. 2.5. eC ¼ DLC ¼ ÀDL ¼ L0C À LC ð2:2Þ L0C L0C L0C

100 2 Compression Natural or true compressive stress, rC ¼ PC ¼ 4PC ¼ 4PC LC ð2:8Þ AC pDC2 pD20C L0C D Further from (2.8), (2.7), (2.4) and (2.3), the following 0C increasing (2.9) and (2.10) are obtained, either of which can be used to determine the true compressive stress. L0C True compressive stress, Axial compressiveload rC ¼ 4PC LC ¼ SC LC ¼ SCð1 À eC Þ ð2:9Þ pD20C L0C L0C True compressive stress, rC ¼ 4PC LC ¼ SC LC ¼ SC expðÀeCÞ ð2:10Þ pD20C L0C L0C 0 The axial compressive load versus reduction in length, Reduction in length of specimen i.e. PC versus DLC curve for any D0C =L0C ratio of the specimen in Fig. 2.5 can be converted to the corresponding Fig. 2.5 Load–deformation curves in compression with increasing engineering compressive stress–strain, i.e. SC versus eC ratio of initial diameter D0C to initial length L0C i.e. D0C =L0C of the diagram with the help of (2.7) and (2.2). The shape of the specimen resulting SC versus eC diagram will be the same as that of the load–deformation, i.e. PC versus DLC diagram from which Natural or true compressive strain, SC À eC curve is drawn, because SC / PC and eC / DLC: ZLC dL ln L0C When a crystalline material is strained in compression L LC beyond its linear elastic part of stress–strain curve and then eC ¼ À ¼ ð2:3Þ unloaded from its plastic range, the result will be same as that described in tension. The total strain consists of elastic, L0C anelastic and plastic part as in tension and the unloading path will also be similar. On unloading, the elastic strain is From (2.2), recovered immediately, the anelastic strain disappears with time and the plastic strain remains permanently in the eC ¼ 1 À LC ; or LC ¼ 1 À eC ð2:4Þ specimen. When the material is reloaded in compression L0C L0C after unloading from its plastic range, the compressive stress required for yielding increases, i.e. its compressive yield From (2.3) and (2.4), natural or true compressive strain, strength is raised to a higher value, which is similar to the increase in tensile yield strength observed on reloading in eC ¼ À ln LC ¼ À lnð1 À eC Þ ¼ ln ð1 1 ð2:5Þ tension. L0C À eCÞ Let us assume that the compression specimen is cylin- 2.4.2 Strain Rate drical whose initial diameter = D0C and instantaneous diameter = DC at any moment during homogeneous com- In homogeneous compression, in which the specimen is pressive deformation. From constancy of volume during plastic deformation, one can write pD02C pDC2 reduced in length uniformly without any barreling, the 4 4 L0C ¼ LC ; or, D2C ¼ D20C L0C ð2:6Þ conventional or engineering strain rate, denoted by e_C, and LC the true strain rate, denoted by e_C, are defined as follows: Hence from (1.24), (1.25) and (2.6), we get Compressive engineering strain rate, Conventional or engineering compressive stress, deC ÀdL 0 1 dL dt L0C dt ¼ À L0C dt e_ C ¼ ¼ ð2:11aÞ SC ¼ PC ¼ 4PC ð2:7Þ A0C pD02C

2.4 Plastic Range 101 Compressive true strain rate, ÀdL 0 LC 0 Since e_C ¼ dt and e_C ¼ constant, e_C = deC ¼ ÀdL dt ¼ À 1 dL ð2:11bÞ ZLnC dL Ztn dt LC LC dt LC À ¼ e_C dt; so, where L0C and LC are, respectively, the initial length and L0C 0 instantaneous length of the compressive specimen and ‘ÀdL’ is the compression of the specimen in the time interval Or; ln LnC ¼ Àe_ C tn ; dt. Since one end of the compression specimen remains fixed L0C and its other end is attached to a movable cross-head of the testing machine, the compression, ‘ÀdL’, of the specimen Or, LnC ¼ L0C expðÀe_CtnÞ ð2:13Þ per unit time is equal to the cross-head velocity. The ) vnC ¼ e_CLnC ¼ e_CL0C expðÀe_CtnÞ ð2:14Þ cross-head velocity of the compression testing machine, denoted by vC, is: The relation between the true strain rate, e_C; and the engineering strain rate, e_C, during uniform deformation of Cross-head velocity in compression, the compression specimen is given by the following (2.15): dL Since engineering compressive strain, dt vC ¼ À ð2:11cÞ L0C À LC LC L0C L0C eC ¼ ¼ 1 À ; The compressive strain rate is related the cross-head velocity in compression, vC, by the following (2.12a) and Or, L0C ¼ 1 ; (2.12b). From (2.11a) and (2.11c), LC 1 À eC Compressive engineering strain rate, ) e_C ¼ vC ¼ vC L0C ¼ e_ C ð2:15Þ LC L0C LC À eC ÀdL 0 ÀdL=dt vC 1 L0C L0C L0C e_ C ¼ dt = = ð2:12aÞ It is clear from (2.12a) that e_C / vC since L0C ¼ constant. 2.4.3 Brittle Materials Hence, the compression test can easily be carried out at a constant compressive engineering strain rate e_C, if the The compressive strength, SuCB , of a brittle material is the cross-head velocity in compression, vC, is kept constant. maximum engineering stress that the material can withstand. From (2.11b) and (2.11c), It is defined as the maximum compressive load, PmaxC , Compressive true strain rate, divided by the average original cross-sectional area of the ÀdL 0 ÀdL=dt vC compressive specimen A0C , as given below. For brittle LC LC LC materials, the compressive strength SuCB is the same as the e_C ¼ dt = = ð2:12bÞ breaking or fracture strength SfCB . Hence, As the specimen is compressed, its instantaneous length, Compressive strength, LC, decreases. Therefore, the true strain rate e_C increases when the cross-head velocity vC is maintained constant in SuCB ¼ maximum compressive load compression, as evident from (2.12b). To achieve a constant average original cross-sectional area of specimen compressive true strain rate e_C, during uniform compression of the specimen, the decrease in the cross-head velocity must ¼ PmaxCB A0C be proportional to the decrease in the instantaneous length of ¼ compressive load at fracture; PfCB the specimen. At the beginning of the test, i.e. at initial time A0C t ¼ 0; the initial length of the specimen is LC ¼ L0C . Let, the reduced length of the specimen after test duration of t ¼ tn; ¼ fracture strength; SfCB is LC ¼ LnC : At time t ¼ tn; the cross-head velocity vnC must decrease during uniform deformation of the compression ð2:16Þ specimen according to the following (2.14): Compressive strengths of brittle materials, such as cast iron, ceramics and concrete, are higher than their tensile

102 2 Compression Table 2.4 Comparison of Brittle materials Tensile Compressive Ratio of compressive tensile and compressive strengths strength, Su strength, SuC in to tensile strength, i.e. of brittle materials in MPa MPa SuC Grey cast iron 165.48 827.4 Su Dense, polycrystalline Al2O3 (ceramics) 280 2100 5 2.76 Concrete (1:3 mix by volume) at 34.48 7.5 water/cement ratio of 0.64 by volume (after 28 days) 12.5 Acrylic plastic like Plexiglas, Lucite 73.09 117.22* 1.6 *Stress at which excessive deformation or rupture occurs strengths, sometimes by a large factor, as shown in Table 2.4 stress. Such attainment of higher compressive stress without (Richards 1961; Shackelford 1992), the reason for which is causing brittle material to fail makes the material stronger in explained below. In tension, the fracture of a brittle material compression than in tension. The compressive strengths of is initiated by some flaws or imperfections in the material, brittle materials like concrete, soils and other granular such as microscopic cracks or holes. These flaws act as stress materials are further increased due to the internal frictions raisers and increase the stresses locally at their tips to high acting along the slip planes between macroscopic or values so that the local stresses are equal to the theoretical microscopic particles (Taylor 1948). The engineering stress– cohesive strength of the material and cause breaking of the strain diagram in tension as well as in compression for grey interatomic bonds in those areas. Thus, tensile fracture takes cast iron and acrylic plastic (Plexiglas) are, respectively, place normally by complete separation of the member along shown in Figs. 2.6 and 2.7. a crystallographic cleavage plane lying normal to the applied uniaxial tensile stress. Under such situation, the applied 2.4.4 Ductile Materials stress remains well below that required to produce slip on the slip planes oriented at 45° to the applied stress axis. The typical engineering stress–strain curve (SC vs. eC dia- Conversely, under the applied compressive stress, any cracks gram) and the corresponding true stress–strain curve (rC vs. or holes present in brittle materials tend to close up causing eC curve) for a ductile material subjected to compression are their harmful effect to disappear. Thus, the imperfections fail shown in Fig. 2.8 for the purpose of comparison. Only to produce stress concentration in compression. Hence, the during uniform plastic compression, one can determine the applied compressive stress can be raised to such higher value value of true compressive stress rC, from the engineering that the resolved shear stress can initiate slip along a plane compressive stress SC, according to (2.9) or (2.10) and the usually making an angle of 45° with the axis of applied Fig. 2.6 Engineering stress– Stress in tension 165.48 MPa strain diagram in tension as well as in compression for grey cast Break iron Tension 0 < 5× 10-3 0.065 Strain in compression Strain in tension Compression Stress in compression Break 827.4 MPa

2.4 Plastic Range 103 0.2 73.09 MPa Break Stress in tension Tension Strain in compression 0 0.04 Strain in tension Compression Stress in compression 117.22 MPa No break Fig. 2.7 Engineering stress–strain diagram in tension as well as in compression for an acrylic plastic (Plexiglas) Engineering and true stresses in compressionEngineering stress – strain curvecompressive stress rC and true compressive strain eC; (2.8) and (2.3) are, respectively, used. Further, the frictional True stress – strain curve stresses at the ends of the specimen create a triaxial state of 0 Engineering and true strains in compression compression, which increases the axially applied compres- Fig. 2.8 Comparison of stress–strain curves in compression for a sive load required for deformation of the material. Hence, ductile metal in the plastic range, using engineering and true stresses the higher load will increase the compressive flow stress rC; and strains which will approach the correct value with decrease in friction. To obtain the correct compressive flow stress rC value of true compressive strain eC, from the engineering without friction (Cooke and Larke 1945), the axial com- compressive strain eC, using (2.5). Due to the presence of pressive load PC versus D0C =L0C for various values of friction at the contact surfaces between the specimen and the reduction in length DLC of the specimen is plotted with the anvils, there is often non-uniform plastic deformation in the help of Fig. 2.5. When each curve of the plot is extrapolated compression specimen. Then to determine the true to ðD0C =L0C Þ ¼ 0; the correct value of axial compressive load PC for each value of reduction in length DLC of the specimen is obtained. From this correct value of PC; the correct compressive flow stress rC without friction can be calculated using (2.8). On the two curves in Fig. 2.8, the corresponding points are joined to show their relations one to the other. Since the instantaneous cross-sectional area is greater than the original cross-sectional area of the specimen, i.e. AC [ A0C ; or the instantaneous length is lower than the original length of the specimen, i.e. LC \\ L0C ; the values of compressive flow stresses will be less than their corresponding compressive

104 2 Compression engineering stresses, i.e. rC \\ SC; in accordance with (2.8). Bauschinger effect, discovered by Bauschinger in 1881. This On the other hand, the values of compressive true strains will can be explained as follows by considering a pure stress– be higher than their corresponding compressive engineering strain diagram without any reversal of load in both tension strains, i.e. eC [ eC; as shown in Fig. 1.2. Hence in Fig. 2.8, and compression for a ductile metal, as shown by BAOA0B0 the points on the true-stress–true-strain curve are to the right in Fig. 2.9. In this figure, the magnitude of initial yield and below of those on the engineering curve. For ductile strength in tension given by point A is the same as that in metals, the flow curve in compression (rC vs. eC curve) is compression represented by A0. Now, the same metal in usually taken identical to that in tension (r vs. e curve) up to virgin condition is strained in tension past its original yield the point that corresponds to the maximum load in tension strength A to the point B following the path O-A-B and then and as a result, the compressive properties of ductile metals unloaded along the path B-C, neglecting small elastic hys- are generally deduced from their tensile properties. The teresis effects, to zero stress represented by the point C in engineering stress–strain curve in compression certainly Fig. 2.9. If the metal is now reloaded in compression from differs from that in tension; but in the early part of the plastic the unloaded point C, the metal will start yielding at the range, they do not indeed differ much and are generally stress corresponding to the point D in Fig. 2.9, which is assumed to be same for convenience in engineering appli- substantially lower than the original compressive yield cations. For example, slip in ductile material causes yielding strength A0 of the metal. Hence, the new tensile yield at a compressive engineering stress that is roughly identical strength of the metal was increased from the point A to the to the tensile yield strength. Further, upper and lower yield point B because of its strain hardening, while its new com- points of mild steel in compression are normally observed to pressive yield strength has been decreased from A0 to D. This be the same as those in tension. So, the yield strength for phenomenon is called the Bauschinger effect, which is continuous as well as discontinuous yielding in compression reversible. This means that if the metal would have been is usually quite similar to that in tension. Since for ductile initially strained in the plastic range under compression materials, the plastic range in compression is theoretically followed by stressing plastically in tension instead of the unlimited and no fracture takes place under compressive above situation of tensile deformation followed by com- stress, the compressive strength of ductile materials bears pressive one, then the tensile yield strength of the metal very little sense and is sometimes arbitrarily defined as would decrease instead of the compressive one. The mech- follows: anism of the Bauschinger effect is explained below: The compressive strength, SuCD ; of a ductile material is 1. During plastic deformation, dislocations move on the slip the maximum engineering stress required to distort the plane through the crystal but grain boundaries in poly- material heavily so that the material is treated as to constitute crystals or other lattice defects in single-as well as complete failure. In fact, this is a yield stress of a ductile poly-crystalline materials act as barriers to the movement material at large plastic strain and the value of the strength of dislocations and dislocations are thus piled-up at such depends on the degree of distortion. Therefore, barriers during the initial loading cycle. This accumula- tion of dislocations at barriers produces a back stress, Compressive strength, because of the repulsive force created by the dislocations of same sign lying on the same slip plane when they are SuCD ¼ load required to distort heavily so as to constitute complete failure brought together during their pile-ups. When the slip average original cross-sectional area of specimen direction is made opposite by 180° with respect to the initial slip direction, the back stress created during the ¼ PHeavy DistotionCD initial loading cycle will assist to move the dislocations A0C in the reverse slip direction and thus decreases the stress required to initiate slip, i.e. the yield strength of the ð2:17Þ material is decreased. 2.5 Bauschinger Effect 2. The sources that have generated the dislocations to cause plastic flow in the first slip direction can produce dislo- If a material is strained either in tension or in compression, cations of opposite sign when the direction of loading is beyond its yield strength, unloaded to zero stress and then reversed. Since dislocations of opposite sign lying on the reloaded in the direction completely opposite to the initial same slip plane attract and cancel or annihilate each slip direction, the yield strength of the material on reloading other, the strength of the lattice will further decrease. will be lower than that observed during the initial loading Hence, a lower stress will be required to initiate yielding cycle. This phenomenon of lowering of the yield strength in the reverse slip direction. when plastic deformation in one direction is followed by deformation in the reverse direction is called the

2.5 Bauschinger Effect Engineering stress in tension 105 B Fig. 2.9 Bauschinger effect A showing the lowering of yield stress on complete reversal of slip direction. The virgin engineering stress–strain curve in tension and compression for a ductile material is represented by BAOA0B0; in which yield stress in tension represented by point A and that in compression represented by point A0 are of equal magnitude O C Engineering strain in tension Engineering strain Engineering stress in compression in compression D Aꞌ Bꞌ 3. It is expected that the obstacles behind the dislocations elsewhere compressive. When the material is reloaded for are not so strong and closely spaced as they are directly compressive deformation, those crystals already having ahead of the dislocations. When the slip direction is compressive residual stresses will start to yield at a stress reversed, dislocations experience less obstructions on lower than the virgin yield strength of the material and their path and can therefore move a considerable distance therefore there will be decrease in the total yield strength easily requiring a lower stress to initiate yielding. of the material. 4. The above factors are responsible for Bauschinger effect The Bauschinger effect can have important consequences in single-as well as in poly-crystalline materials but there in cold working of metals. For example, the Bauschinger is another additional factor only for polycrystals. This effect can be seen in the bending of steel plates (Rolfe et al. factor involves the non-uniformity of yielding in poly- 1968) and can result in work-softening (Polakowski 1963) crystals because of random orientations of the crystals. when heavily cold-worked metals are subjected to stresses of Suppose, the material is first plastically strained in ten- opposite sign. This can take place when rolled sheets or sion after which it will be unloaded and followed by drawn bars are subjected to alternating bending stresses compressive deformation. During the initial straining in during their straightening operations by passing through tension, the randomly orientated crystals yield by dif- roller-levelling machine. The yield strength of work-piece ferent amounts. As a result, there will be slight variation can be reduced resulting in an increase in the elongation in stress from one crystal to another on a microscopic value compared to the cold-worked state during such scale but macroscopically average yield strength of the roller-levelling operations. material will be observed. During unloading, the material contracts till the point of zero value of average stress but 2.6 Advantages of Compression Over this amount of contraction will microscopically vary for Tension Test differently orientated crystals. On unloading, the stresses in the least yielded crystals do not completely come back All brittle materials are weaker in tension than in compres- to zero value, rather stay above zero, i.e. are under ten- sion. This fact coupled with the problems related to brittle sion, whereas the stresses in the most yielded ones cross materials mentioned below makes them less useful in tensile zero and remain in compression. Thus, microscopic applications than in compression. Moreover, compression residual stresses known as Heyn stresses (Seitz 1943) or test is also performed for ductile materials particularly when textural stresses (Freudenthal 1950) exist throughout the polycrystalline material, somewhere tensile and

106 2 Compression the test involves a much larger plastic strain of the material, not often used in tension because of the difficulty in because the plastic range of ductile materials under com- applying the tensile load as explained below. pression is theoretically unlimited. Hence, it is obvious that the compression test has several following advantages over (a) When the ends of a member made of wood are the tension test, although there are some problems associated gripped with clamps for application of tensile load in with compression test that will be discussed subsequently in the longitudinal direction of grains, the transverse this chapter. compressive pressure on the ends of this member caused by the clamping leads to compressive failure 1. In compression test, necking of the specimen does not of the wood in the grip regions, because wood is occur. Hence, the problem with necking as encountered weak in the transverse direction. in the tension test does not arise. So, the compression test of a ductile material can be carried out to measure the (b) When tensile load is applied through hook inserted flow stress up to true strains in excess of 2.0. Since into the ends of the wooden member, the ends shear mechanical working of material involves very large off easily under tensile force. plastic strain in the order of true strain value 2–4, it is Other porous or cellular materials also experience the very much essential to determine the flow curve at that above-mentioned same difficulties when they are strain level. The application of tension test to determine loaded in tension. the flow curve is of limited use in the fields of mechanical working, because the neck formation does not allow 4. In compression test, an axial compressive force is applied uniform deformation of the specimen to exceed a true on a cylindrical or prismatic specimen, which is kept strain value of about 0.5. On the contrary, the flow stress between two parallel flat plates, known as anvils. So of a ductile material at the desired strain level required there is no problem related to gripping of brittle or cel- for mechanical forming applications can easily be lular materials unlike the tensile test. Further, the com- determined by compression test. However, since many pression test is not only easy to perform but the specimen metalworking operations involve high temperatures required for the test is also of simple shape having no where the flow stress is nearly independent of strain but change in cross-section, while the tension test requires a strongly dependent on strain rate, so tests to determine laborious preparation of specimen. This preparation flow stress must be conducted under controlled condi- involves the formation of thicker grip regions with or tions of temperature and constant true strain rate without threads at both ends of specimen, a reduced (McQueen and Jonas 1971). cross-section at its middle zone and a fillet of suitable radius in between the grip and middle regions. 2. It is much easier to apply a load to brittle material in compression test than in tension test, because the grip- 2.7 Problems in Compression Test ping of brittle materials in tension test is a serious problem, as narrated below. Two main problems of the compression test are (a) If screw threaded grips are used, the fracture in the 1. Buckling or lateral deflection and thread regions or at the fillet sections of the specimen 2. Barreling or barreled specimen profile or end restraint. is difficult to prevent because brittle materials are strongly influenced by stress concentration devel- 2.7.1 Buckling oped in the threads or at the fillets. If a member subjected to a compressive load applied along (b) If clamp-type grips with wedge blocks that provide its longitudinal axis has its length larger than any of its automatic clamping are used, the transverse com- transverse dimensions, it is termed as column. Experience pressive pressure on the specimen caused by the shows that when the axially applied compressive load is clamping leads to compressive fracture (shear frac- small, the column is stable. However, as the compressive ture or fragmentation) of the specimen in the grip load is gradually increased it is observed that at certain value regions. of this load, the column becomes laterally instable and this lateral instability of column persists even after the applied 3. Wood has cellular structure, and its grains have direc- compressive load is removed. This lateral deflection is called tional property, i.e. the cell structures of wood are buckling, which consequently develops non-uniform stress stronger in the longitudinal than in the transverse direc- tion. Note that wood does not show brittle characteristics under tensile loading and surprisingly, it is considerably stronger in tension than compression. However, wood is

2.7 Problems in Compression Test 107 in the column due to its bending. This condition may lead to ‘strength’ of a long column that behaves elastically is not sudden complete failure at an applied compressive stress which will be lower than the true compressive strength of the dependent upon the compressive strength of the material, but column material. Therefore, care must be taken in the design of column so that it can bear safely the axially applied only on its geometry and the stiffness of the column material. compressive load without acquiring a bent configuration. Equation (2.18) further shows that if the moment of inertia I of the cross-section of a long column is increased the column strength may be increased. Increase in I without increasing the cross-sectional area may be accomplished by distributing the 2.7.1.1 Critical Column Stress material as far as possible from the principal axes of the Due to imperfections, no column is really straight and starts to buckle at some critical compressive load applied along its cross-section. Hence, it is more economical to use tubular longitudinal axis. The magnitude of the maximum applied compressive load at which the column begins to buckle is sections as compression members than solid sections. The called the buckling load or critical load for buckling, which has been denoted by Pcr. So, it is important to know the stability of the column may be increased by increasing the critical load for a column to resist its buckling. Let us con- sider that buckling has occurred in a slender pin-ended transverse dimensions and reducing the wall thickness of such column, which is shown in Fig. 2.10. Euler’s column for- mula showing critical load for buckling of a slender tubular sections. However, if the wall thickness is diminished pin-ended long column, assuming linear elasticity, is given as (Timoshenko and Young 1968): below a certain limit, the column as a whole may not buckle, Euler buckling load, but the wall itself may be unstable leading to the buckling of the longitudinal elements of the column. This results in a corrugation of the tube-wall, called local buckling. The critical compressive engineering stress, Scr; at which the long column with pinned ends begins to buckle can be p2 EC I obtained from (2.18) by dividing with the initial L02C cross-sectional area A0C of the column. Critical column stress, Pcr ¼ ð2:18Þ  p2 EC I L02C Àpp2ffiAEffiffi0ffiCffiCffiffi=ffiffiffiIffiÁ2 P L02C A0C Scr ¼ A0C cr ¼ ¼ where L02C =Àpp2EffiIffi=ffiCffiffiAffiffiffi0ffiCffiffiÁ2 À p2EC L0C =q0C Pcr maximum or critical axial load on the column just ¼ ¼ Á2 ð2:19Þ prior to the onset of buckling; where pffiffiffiffiffiffiffiffiffiffiffi EC Young’s modulus of elasticity in compression; q0C radius of gyration of the cross-section = I=A0C ; I the minimum value of moment of inertia of the and L0C =q0C slenderness ratio of the column. It is a measure of column’s cross-section about an axis lying on its the column’s flexibility. cross-section; L0C unsupported initial length of the long column, whose ends are pinned Equation (2.18) shows that Pcr for a given column is If (2.19) is plotted with slenderness ratio as abscissa and directly proportional to ECI; called flexural rigidity, and inversely proportional to the square of the length L0C of the the critical column stress as ordinate, it gives the curve ABC, column, but independent of the compressive strength of as shown in Fig. 2.11, which is called Euler’s curve. For a column material. Thus, two geometrically identical slender column of any given slenderness ratio, this curve provides the corresponding value of the critical column stress Scr; for columns made of the same metal, one of high strength and which the column becomes laterally unstable. A column can the other of low strength, will fail by buckling at nearly the fail by buckling, yielding or a combination of both. To same value of the load since their moduli of elasticity remain prevent such failure, the compressive stress applied to the relatively unchanged. It must therefore be noted that the column must remain underneath the curve ABSplC in Fig. 2.11, where SplC is the compressive proportional limit of P P the column material. L0C /2 L0C /2 Equations (2.18) and (2.19) are not applicable to com- Longitudinal axis of column pressive deformation beyond the linear elastic region of the Fig. 2.10 Deflection of a slender pin-ended column due to applied column. Since the elastic behaviour of material was assumed compressive load at the beginning of lateral buckling, we conclude that (2.19) is valid only if the critical column stress Scr is equal to or lower than the compressive yield strength S0C , or truly speaking, compressive proportional limit SplC of the column

108 2 Compression Scr Critical buckling load for short column, Splc C PcrShort ¼ p2ETC I ð2:21Þ B L02C Critical buckling stress for short column, Scr= 2Ec ScrShort ¼ À p2ETC Á2 ð2:22Þ (L0c / ρ0c )2 L0C =q0C A Equation (2.22) may also be applied to the material L0c exhibiting nonlinear elastic region as there is no proportional ρ0c limit, and Young’s modulus does not exist in such case. Use L0c EC of (2.22) to resist buckling of short column is associated with ρ0c Splc the estimation of ETC ; which itself depends on the critical stress ScrShort for short column, whereas for a given value of Fig. 2.11 Euler’s curve showing critical column stress versus slen- ETC ; ScrShort is a function of the slenderness ratio L0C =q0C ; as derness ratio shown by (2.22). Now the following procedure, which is an material, i.e. for engineering purpose, Scr S0C ; or more indirect approach, explains how to obtain the magnitudes of correctly, Scr SplC . For a given column, the lower the ScrShort from (2.22) for different values of L0C =q0C . slenderness ratio L0C =q0C ; the higher is the critical column stress Scr; given by (2.19). Hence, the limiting value of the First, a stress–strain curve of the desired material in slenderness ratio below which Euler’s column formula does tension, assuming that the curve holds for compression also, not apply and the column undergoes inelastic compressive is taken. This curve is shown schematically in Fig. 2.12. deformation is obtained by setting Scr ¼ SplC as follows: Then the slope of this curve at each point of stress on the À p2EC Á2 ¼ SplC curve is computed. This slope is the tangent modulus ETC of L0C =q0C sffiffiffiffiffiffiffi the material in tension as well as in compression and a curve of ETC versus the corresponding stress is drawn on the same plot of stress–strain diagram, as shown in Fig. 2.12. From the tangent modulus versus stress curve, several values of ETC and its corresponding stresses are selected. From each pair of values of ETC and stress, the corresponding L0C EC ð2:20Þ q0C SplC ) ¼p 0 Tangent modulus pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi Column, for which L0C =q0C \\p EC=SplC is called short column that requires critical column stress for buckling higher than the proportional limit. This value of L0C =q0C ; Stress – strain curve Tangent modulus given by (2.20), is represented by point B on the curve in vs. stress curve Fig. 2.11 and marks the boundary beptwffiffieffiffieffiffiffinffiffiffiffitffiffihffiffie long and Stress short columns. Thus for L0C =q0C \\p EC=SplC ; the com- pressive stress in a pin-ended short column will reach the proportional limit before lateral buckling can occur and (2.19) becomes inapplicable. For this reason, the part BC of the Euler’s curve in Fig. 2.11 is represented by a dotted line; only the part AB is valid. Since Hook’s law cannot be applied to the buckling of Linear elastic region short column, as well as to the material exhibiting nonlinear 0 0 Strain elastic region, (2.18) and (2.19) have to be modified by Fig. 2.12 Schematic plot of tangent modulus versus stress, computed substituting instantaneous stiffness, measured by tangent from stress–strain curve for a desired material in tension as well as in compression modulus ETC ; in place of Young’s modulus EC. Thus for short columns, (2.18) and (2.19) take the following forms that agree well with the experimental findings.

2.7 Problems in Compression Test 109 slenderness ratio L0C =q0C is calculated by putting the stress stresses. Therefore, compression of a short length spec- value in place of the critical stress ScrShort and its corre- sponding value of ETC in (2.22). Now for various values of imen causes an increase in the axial compressive load L0C =q0C ; their corresponding critical stresses [the stresses, which have been substituted in (2.22)] are available and if required for deformation, which in turn requires a testing required, a plot of ScrShort versus L0C =q0C can be made from machine of a relatively higher capacity. The fact that the which the critical stress can be determined for any given axial load for compression increases with increasing the value of slenderness ratio. ratio of initial diameter D0C to initial length L0C i.e. D0C =L0C of the specimen has been explained earlier in Sect. 2.4 and also shown in Fig. 2.5. 2.7.1.2 Remedial Measure Against Buckling 2.7.2 Barreling To resist buckling failure of the specimen during its com- pressive deformation, the critical column stress in the linear Although lateral expansion at both ends of the specimen is elastic region or the critical buckling stress in the inelastic or opposed by the frictional forces developed at the contact area in the nonlinear elastic region must be greater than the axial between the specimen and the compression plates during its compressive stresses to be applied during the test. To ensure compressive deformation as discussed earlier in Sect. 2.4, higher value of Scr and S ;crShort the slenderness ratio L0C =q0C the material at the mid-length of specimen can uninterrupt- must be decreased to a quite low value as evident from edly flow outward that leads to lateral expansion of the (2.19) and (2.22). For a round cross-section, if D0C = the central portion of specimen. This condition, known as end initial diameter of the specimen, the radius of gyration of the restraint, tries to retain both ends of a compression specimen cross-section, near their initial dimensions while the middle part expands, resulting in a barreled specimen profile, as shown in qffiffiffiffiffiffiffiffiffiffiffi rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffi Fig. 2.13. Sometimes, both ends of the specimen may be q0C ¼ I=A0C ¼ pD40C =64  4=pD20C ¼ D20C =16 completely prevented from lateral expansion. During longi- tudinal compression of a cylindrical specimen with initial ¼ D0C =4; diameter of D0C ; the expansion in the transverse direction, i.e. the increase in diameter DDC is: Àwhich meÁans q0C / D0C . Therefore, the slenderness ratio is L0C =q0C / ðL0C =D0C Þ: So to avoid buckling, a cylindrical DDC ¼ DC À D0C ¼ D0C eTC ¼ D0C ðÀm eLC Þ ¼ jD0C m eLC j; specimen must have a low value of the ratio of L0C ðinitial lengthÞ=D0C ðinitial diameterÞ: Hence to eliminate where DC = instantaneous diameter of the specimen, the probability of lateral instability, ASTM short specimen eTC = transverse engineering strain, m = Poisson’s ratio and (see Sect. 2.2) can be used provided barreling is ignored. eLC = longitudinal engineering strain. This short length specimen is used in the standard quality control tests of brittle materials when a comparison of results on different specimens is only required. Hence, tests on short length specimens can provide useful comparison between materials. Although a short length specimen provides a better stability of the test piece under compression, it has the following limitations: 1. It is difficult to measure strain with precision from such a Fig. 2.13 Barreled specimen profile resulting from end restraint test piece of smaller length because the circumferential surface of a short length specimen under compression will take a barrel-shaped profile. 2. The capacity of the testing machine for compressing a short length specimen must be large, the reason of which is explained below. The ASTM short specimen is a cylinder with L0C ’ 0:9 Á D0C or a 2-in. (50.8 mm) cube. So, the cross-sectional area to length ratio of a short specimen is relatively larger than that of a medium length or a long specimen (see Sect. 2.2). Hence, a short length specimen undergoing reduction in length under uniaxial compression will be in a state of triaxial compressive

110 2 Compression 2.7.2.1 Remedial Measures Against Barreling of tensile in nature constituting triaxial state of stress with the axially applied compressive stress rC. As yielding of 1. Use of Specimen with Higher L0C ðInitial LengthÞ=D0C material takes place under compressive stress, in the present ðInitial DiameterÞ Ratio context it is assumed that the compressive stress is positive Recalling the fact that the transverse restraining compressive and the tensile stress is negative. Hence, r0 ¼ r0C ; r1 ¼ stresses developed due to friction at both ends of the spec- rC and r2 ¼ r3 ¼ ÀsT ; because sT is the shear stress in imen vanish at a distance of roughly equal to the initial tension. With the help of Von Mises’ yielding criterion given diameter D0C ; from any end of the specimen, the lower limit for a cylindrical specimen is L0C =D0C ¼ 2; which is the by (1.64), we get minimum requirement to avoid the end restraint. The spec- imen with this minimum dimensional ratio is used for con- r0C ¼ p1ffiffi h þ sT Þ2 þ ðÀsT þ sT Þ2 þ ðÀsT À rC Þ2 i1=2 crete and mortar. For metallic materials, ASTM 2 ðrC medium-length specimen with L0C =D0C ¼ 3 (see Sect. 2.2) can be used. There will be a region in the middle of this ¼ p1ffiffi h ðrC þ sT Þ2i1=2; specimen, which is free from the influence of end restraint 2 2 and subjected to uniaxial compression. The length of this region ’ L0C À 2 Á D0C ’ D0C : Hence, about one diameter Or, r0C ¼ rC þ sT ; or; rC ¼ r0C À sT ; i:e: rC\\r0C long central region of this specimen is experiencing simple compression over which strains can satisfactorily be mea- ð2:23Þ sured. For more accurate results in strain measurements, ASTM long specimens (see Sect. 2.2) may be used. As the The above (2.23) shows that there is a weakening of the ratio L0C =D0C of the specimen is increased, the tendency material near the anvil surfaces instead of a strengthening, as towards buckling increases. To avoid the danger from shown earlier by (2.1). Consequently, this lateral tensile bending of long specimen, it is required to machine the stress sT causes the compression specimen to split longitu- specimen more precisely and reduce the eccentricity in dinally. This longitudinal splitting may be avoided by loading to the minimum level. Again, as the ratio L0C =D0C of making fine concentric grooves in both ends of the speci- the specimen is decreased, the effect of the end restraint men. The grooves must uniformly be spread over the entire becomes predominant. The right selection of the ratio contact surfaces of the specimen. The grooves are introduced L0C =D0C for a compression specimen thus becomes a com- to retain the lubricant applied at the contact surfaces so that promise between high L0C =D0C to avoid the effects of end the outward flow of lubricant is prevented. In spite of this, restraint and low L0C =D0C to resist buckling. It may be the test is carried out in increments so that the lubricant can concluded that the most satisfactory compromise is to adopt be replaced at intervals (Hsu 1969). The kind of lubricant the ASTM medium length specimen with L0C =D0C ¼ 3. applied and the size and shape of the grooves have to be established through experiment. This above approach can 2. Proper Lubricant Applied in the Fine Concentric almost eliminate the barreling effect in ductile specimens. Grooves Made in Both Ends of Specimen Using the above technique, one can reach a compressive strain value of about e ¼ 1:0 with a slight amount of The friction at the specimen–anvil contact surfaces can be barreling. reduced (van Rooyen and Backofen 1960; Pearsall and Backofen 1963) by applying lubricant at both ends of the 3. Shaping the Anvils and the Ends of the Specimen in specimen and using hardened anvils with smooth surface the Form of a Cone finish. Usually, lubricants used are Teflon sheet or graphite dispersed in paraffin oil for cold deformation and molten Shaping the anvils and the ends of the specimen introduces glass or mixture of molybdenum disulphide and graphite for hot deformation, although proper lubricant to be used must an outward force which will just balance the inward friction be decided experimentally. But the problem is that under force if the contact surface is shaped in the form of a cone in high contact pressure, the lubricant tends to flow outward at the ends of the compression specimen and develops lateral such a way that the normal to the conical surface makes an stresses acting outward, i.e. in a direction opposite to that of the friction stresses. Hence, this lateral outward stresses are angle equal to the angle of friction f with the longitudinal axis of the specimen. The angle of friction f is related to the coefficient of friction l at the anvil–specimen contact sur- faces as f ¼ tanÀ1 l: Figure 2.14 shows the forces exerted by the anvil on an elemental initial area DA0C at the specimen–anvil interfaces for two specimens made of same material—one with plane ends and another with conical ends. Let, SC = compressive stress applied along the longitu- dinal axis of the specimen;

2.7 Problems in Compression Test 111 (a) (b) Compressive stress, SC Compressive stress, SC Anvil ΔRc f Anvil NC ΔA0 f 0 ΔRC ΔA τ C ΔA 0 τC ΔA0 ΔA0 C Specimen N f = angle of friction NC = SC τC = μ=NNCτCC =μ ΔA0 ΔRC = tan f Specimen SC ΔA0 Fig. 2.14 Method to get rid of barreling of specimen in compression normal to the conical surface makes an angle of friction f with the by eliminating lateral compressive stresses. a Specimen with flat ends longitudinal axis of the specimen, barreling will not occur where barreling may occur. b Specimen with conical ends, where if the NC = normal stress or pressure exerted by the applied of material at both ends of specimen. Hence, barreling is compressive stress SC on the elemental initial area DA0C of developed by unhindered lateral expansion at the central the specimen; portion of specimen having plane ends. sC = tangential shearing friction stress developed at the For specimen having conical ends, the normal stress NC is anvil–specimen interface if it is assumed that the outward inclined at an angle of f (angle of friction) to the direction of thrust for lateral expansion of the specimen is enough to applied compressive stress SC and therefore, produce some slipping between the end of specimen and the anvil; NC ¼ SC cos f : According to Coulomb’s law of sliding friction, So from (2.25), the resultant force for specimen with conical ends is given by frction stress ¼ coefficien of friction  normal stress; Or; sC ¼ l NC ¼ tan f NC DRC ¼ SC cos f DA0C ¼ SC DA0C ð2:27Þ cos f ð2:24Þ Let, DRC = resultant force developed due to the action of Hence, (2.27) shows that the magnitude of the resultant normal force, NCDA0C ; and friction force, sCDA0C ; on the force DRC is same as that of the applied compressive force. elemental initial area DA0C of the specimen. For specimens Further, the direction of resultant force DRC is parallel to that with plane ends as well as conical ends, the resultant force DRC of the applied compressive stress SC; because both DRC and is inclined to the direction of the normal stress NC ; at an angle SC make the same angle of f (angle of friction) with the direction of the normal stress for specimen with conical of f (angle of friction). Hence, we get the resultant force as ends. Since the magnitude and direction of the resultant force are the same as those of the applied compressive force, qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi so the action of the resultant force will be the same as that of the applied uniaxial compressive force and the material at DRC ¼ NC2 DA20C þ sC2 DA02C ¼ DA0C NC2 þ ðlNCÞ2 ½using ð2:24ފ both conical ends of the specimen will be subjected to homogeneous compression, i.e. uniaxial state of compres- ¼ DA0C pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ¼ NC DA0C sion. As a result, there will be unhindered lateral uniform NC 1 þ tan2 f cos f expansion throughout the length of the specimen during its compression and barreling of the specimen will be com- ð2:25Þ pletely eliminated. For specimen with plane ends, the normal stress NC is parallel to the longitudinal axis of the specimen, i.e. parallel to the direction of applied compressive stress SC and NC ¼ SC: So from (2.25), the resultant force for specimen with plane ends is given by DRC ¼ ðNCDA0C Þ= cos f ¼ ðSCDA0C Þ= cos f ð2:26Þ 2.8 Compressive Failure of Materials Since DRC is inclined at an angle of f (angle of friction) to A material subjected to compressive load may usually fail in the direction of applied compressive stress SC; so the any one of the following ways: resultant force acts inward to resist the lateral outward flow

112 2 Compression 1. If the applied compressive load exceeds the Euler Strees buckling load given by (2.18) in the linear elastic range of the member or the critical buckling load for short θ θ = 45° – φ column shown by (2.21) in the nonlinear elastic range or 2 in the inelastic region of the member, there will be failure due to lateral deflection of the member and it is called α = 45° + φ buckling failure. It may happen in any material whether it 2 is ductile or brittle and porous or cellular. α 2. When the applied compressive stress exceeds the yield strength of the material but remains below the critical buckling stress, there will be slip in the material. (a) Failure in Ductile Materials In a ductile material, slip leads to the formation of a barreled Strees specimen profile during its compression and barreling increases with the applied compressive stress, provided that Fig. 2.15 Relation between the angle of break, h, made by a shearing the specimen must not buckle or bend. Excessive barreling plane and the angle of internal friction, u causes to develop secondary tensile stresses in the circum- ferential direction of the specimen. These circumferential deviate from the theoretical value due to the effect of end tensile stresses produce surface cracks parallel to the loading restraint, which is more pronounced in a short specimen. For axis and thus tensile failure instead of compressive one will specimens of adequate length, the values of a (see Fig. 2.15) occur in a ductile material. But if the ductile material could are observed to vary approximately between 50° and 60° for be compressed homogeneously without any friction at the several brittle materials, such as cast iron, concrete, sand- contact surfaces, it would not bulge laterally even after a stone and brick (Withey and Aston 1939). For discussions of large reduction of its length. In such case, no tensile stress failure of granular materials whose internal friction resists would develop in the circumferential direction and the failures, the reader may see textbooks on soil mechanics specimen would not crack. (Taylor 1948) or other literature (Endersby 1940). (b) Failure in Brittle Materials The end restraint causes a cone-shaped region of stronger undeformed material to remain internally near the ends of the In a brittle material, slip leads to permanent rupture of specimen, around which a weak zone of material exists at the atomic or molecular bonds and ultimately causes fracture. periphery of the specimen. The shear crack initiates at this Fracture may occur completely along a single shear plane weak region and propagates from the edge at each end of the known as shear fracture, or in the form of fragmentation, specimen and ultimately meets in the centre of the specimen. which consists of a lot of small fractures on shear planes in different directions. Usually, fracture occurs somewhat in • Usually, cast iron breaks along an inclined diagonal between these above two extremities. Further, if there is plane and this shear fracture is shown in Fig. 2.16a, unhindered lateral expansion at the ends of a brittle material, where the fracture occurs along the dashed line. it generally fails by longitudinal splitting into columnar fragments called splitting failure or columnar fracture, (a) Stress (b) Stress (c) Stress which is pronounced in short specimens. Stress Stress Stress In brittle materials like cast iron and concrete, shear fracture occurs along the planes that extend from corner to Fig. 2.16 Types of compressive fractures of brittle materials, such as corner of the specimen. Slip does not occur on planes of cast iron, concrete and mortar or stone cubes. a Shear fracture, e.g. cast maximum shear stress whose normal is oriented at 45° to the iron and concrete. b Hourglass or shear cone fracture, e.g. concrete, loading axis but planes of shear leading to fracture are near mortar or stone cubes. c Shear cone fracture with splitting above, e.g. those of maximum shear. Since internal friction as well as concrete cohesion resists failures in these materials, shearing planes along which fractures take place are function of the angle of internal friction u. The relation between the angle of fracture h made by a shearing plane and the angle of internal friction u is given in Fig. 2.15. Further, the angle of fracture h may

2.8 Compressive Failure of Materials 113 • Failures in cylindrical-shaped specimens of concrete, Solution mortar or stone cubes commonly occur along conical shear surface forming the shear cone fracture, sometimes UTS, Su = 414 MPa, and uniform engineering tensile strain, called an hourglass fracture, as shown in Fig. 2.16b, eu = 0.16. where the curved dashed line indicates the fracture path. If the cross-section of specimen is of square or rectan- From (1.120), the true tensile strength is: gular shape, the hourglass fracture will have a pyramidal-shaped fracture instead of a cone-shaped ru ¼ Suð1 þ euÞ ¼ 414ð1 þ 0:16ÞMPa ¼ 480:24 MPa: fracture as observed for a cylindrical specimen. From (1.134), the true uniform tensile strain is: • In addition to the hourglass fracture, concrete may either fracture along an inclined diagonal plane showing shear eu ¼ lnð1 þ euÞ ¼ ln ð1 þ 0:16Þ ¼ 0:14842: fracture like cast iron, as shown in Fig. 2.16a or show the shear cone fracture with longitudinal splitting above, Since as per given condition, the true stress and true strain as shown in Fig. 2.16c, where the fracture path is dis- in compression are identical to those in tension, so com- played by the dashed line. pressive true stress rC ¼ ru ¼ 480:24 MPa and compressive true strain eC ¼ eu ¼ 0:14842: 3. Failure in Wood Hence, the desired engineering stress in compression corresponding to the given UTS value can be obtained from (2.10) and is given below: Wood is a fibrous and cellular material where the constituent SC ¼ rC exp ðeCÞ ¼ 480:24 Â exp ð0:14842Þ MPa cells align themselves forming a series of columns along the ¼ 557:078 MPa: longitudinal direction of the grain of wood. The rupture of the cell walls causes individual collapse of columnar fibres. Similarly, the desired engineering strain in compression On applying compressive load parallel to the grain of wood, corresponding to the given uniform engineering tensile strain various types of failure are observed in wood as described value can be obtained from (2.5) and is given below: below: eC ¼ 1 À 1 ¼ 1 À exp 1 • The plane of fracture is nearly horizontal, known as exp ðeCÞ ð0:14842Þ crushing failure. ¼ 0:1379 or 13:79%: • The plane of rupture makes an acute angle, usually about 45°, with horizontal, called shear failure. 2.9.2. When a uniaxial compressive load of 6 Â 105 kgf is applied to the top surface of a well-lubricated solid cube of • Fracture into columnar pieces occurs roughly parallel to side 100 mm, the cube just yields. What load would be the grain of wood, and this longitudinal separation is required to produce yielding if the surfaces of the cube other called splitting failure. than the top surface were constrained by compressive loads 1:5 Â 105 kgf and 3 Â 105 kgf? Neglect the frictional • If the direction of splitting is radial or tangential, it is effects. known as wedge split failure. Solution • Fracture path initially takes a shear plane followed by longitudinal separation parallel to the grain of wood and Given that the uniaxial yield load in compression, PyC ¼ finally again changes to a shear plane causing complete 6 Â 105 Â 9:807 N ¼ 5;884;200 N; the area of each face of separation of the member. This may be termed as the cube AC ¼ 100 Â 100 mm2 ¼ 104 mm2: shearing and splitting failure. Hence, yield stress in compression, 2.9 Solved Problems r0C ¼ PyC ¼ 5;884;200 N mmÀ2 ¼ 588:42 N mmÀ2: AC 104 2.9.1. Assume that the flow curve of certain ductile steel in tension is identical to that in compression. If UTS of the steel The problem states the cube is subjected to a triaxial state is 414 MPa and uniform engineering tensile strain is 0.16, of stress, and it is asked to determine the compressive load compute the engineering stress and strain in compression required for yielding under such condition. Let this load be corresponding to the values given in tension, assuming P1C and the corresponding compressive stress is r1C . Other constancy in volume. constraining compressive loads given are:

114 2 Compression P2C ¼ 1:5  105  9:807 N ¼ 1;471;050 N and Alternative Computation of r1C in case of Von Mises’ P3C ¼ 3  105  9:807 N ¼ 2;942;100 N: yielding criterion: Hence, other constraining compressive stresses are: Here, calculations of r2C and r3C are not needed. Only calculation of yield stress in compression, i.e. r0C is required r2C ¼ P2C ¼ 1;471;050 N mmÀ2 ¼ 147:105 N mmÀ2: to determine r1C from Von Mises’ yielding criterion, AC 104 because loads given in the problem are such that r3C ¼ P3C ¼ 2;942;100 N mmÀ2 ¼ 294:21 N mmÀ2: P2C :P3C :PyC ¼ 1:2:4; and the area AC of each face of the AC 104 cube is constant, which means r2C :r3C :r0C ¼ 1:2:4; i.e. r2C ¼ r0C =4; and r3C ¼ r0C =2: For yielding under triaxial stress condition, yielding cri- terion is to be considered and the value of the required load Hence, from Von Mises’ yielding criterion (1.64): for yielding depends on the yielding criterion selected. According to Tresca yielding criterion (1.68a), yielding in  À r0C 2 þ r0C À r0C 2 compression will occur when r1C þ4r0C 4 2 2 2 r1C ¼ 2r02C : À r1C À r2C ¼ r0C : or, r21C À r0C r1C þ r20C þ r20C þ r20C Hence according to Tresca yielding criterion, 2 16 16 4 À r0C r1C þ r12C À 2r02C ¼ 0; r1C ¼ r0C þ r2C ¼ 588:42 þ 147:105 ¼ 735:525 N mmÀ2: or, 2r12C À 3 r0C r1C À 13 r02C ¼ 0; 2 8 Therefore, load required to cause yielding is: Hence, 735:525  104 rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 104 9:807 ÀðÀ1:5r0C Þ Æ ðÀ1:5r0C Þ2À4  2 À1:625r20C P1C ¼ r1C  AC ¼ 735:525  N ¼ kgf ¼ 7:5  105 kgf: r1C ¼ 2Â2 pffiffiffiffiffiffiffiffiffiffiffi ¼ 1:5 Æ 4 15:25 Á r0C According to Von Mises’ yielding criterion (1.64), yielding in compression will occur when ¼ 1:35128 Á r0C ¼ 1:35128  588:42 N mmÀ2 ðr1C À r2C Þ2 þ ðr2C À r3C Þ2 þ ðr3C À r1C Þ2¼ 2 r02C : ¼ 795:12 N mmÀ2; ðneglecting negative valueÞ: Hence according to Von Mises’ yielding criterion, Therefore, load required to cause yielding is: ðr1C À 147:105Þ2 þ ð147:105 À 294:21Þ2 P1C ¼ r1C  AC ¼ 795:12  104 N ¼ 795:12  104 kgf þ ð294:21 À r1C Þ2¼ 2  588:422; 9:807 ¼ 8:108  105 kgf: or, r12C À 294:21r1C þ 21639:88 þ 21639:88 þ 86559:52 À 588:42r1C þ r12C ¼ 692476:193; 2.9.3. An 18 m long pin-ended column of square cross- section, made of copper metal, can withstand an application of or, 2r21C À 882:63r1C À 562636:913 ¼ 0; a maximum compressive stress of 11 MPa acting normal to its cross-section. Young’s modulus of copper is 110 GPa. Using Hence, a factor of safety of 2.5 in computing Euler’s critical load for buckling, determine the cross-sectional dimension of the qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi column if the column can safely sustain the following com- pressive loads applied along its longitudinal axis: r1C ¼ ÀðÀ882:63Þ Æ ðÀ882:63Þ2 À 4  2 ðÀ562636:913Þ 2Â2 ¼ 795:12 N/mm2; ðneglecting negative valueÞ: Therefore, load required to cause yielding is: (a) 900 kN and (b) 1800 kN. P1C ¼ r1C  AC ¼ 795:12  104 N ¼ 795:12  104 kgf Solution 9:807 ¼ 8:108  105 kgf: Let the dimension of the square cross-section is a. Hence, the minimum value of moment of inertia about an axis lying on its cross-section is I ¼ a4=12:

2.9 Solved Problems 115 (a) Buckling Criterion rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 4:5  106  182 or; a! 4 p2  110  109  12 m According to given condition, critical load for buckling, ¼ 0:3563 m ¼ 356:3 mm: Pcr ! factor of safety  applied compressive load Stress Criterion ¼ 2:5  900 kN ¼ 2:25  106 N: From Euler’s column formula for critical load, given by Cross-sectional area ¼ a2 ! applied compressive load (2.18) allowable compressive stress ¼ 1800 kN ¼ 1800  103 m2: 11 MPa 11  106 p2 EC I L20C ! 2:25  106 N: rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi where EC ¼ 110 GPa ¼ 110  109 N=m2 and L0C ¼ 18 m: ) a! 1800  103 m ¼ 0:4045 m ¼ 404:5 mm: 11  106 ÀÁ We must select the maximum value of a from the results ) I ¼ a4 ! 2:25  106 L02C 2:25  106  182 obtained by these above two criteria. Hence, the dimension 12 p2EC ¼ p2  110  109 m4: of square cross-section of the column will be a ! 404.5 mm. In this case, the stress criterion is considered for the rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi design of the column. 2:25  106  182 or, a! 4 p2  110  109  12 m ¼ 0:2996 m ¼ 299:6 mm: Exercise Stress Criterion Cross-sectional area ¼ a2 ! applied compressive load 2.Ex.1. If the engineering stress and engineering strain in allowable compressive stress compression are, respectively, 300 MPa and 30%, calculate the corresponding values of compressive true stress and ¼ 900 kN ¼ 900  103 m2: compressive true strain. 11 MPa 11  106 2.Ex.2. Calculate the true stress (in MPa) and true strain for rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi a cylindrical specimen of 10 mm diameter and 15 mm 900  103 height, which is homogeneously compressed to one-third of ) a! 11  106 m ¼ 0:286 m ¼ 286 mm: its original height by applying a maximum load of 150 kN. Assume that volume remains constant. We must select the maximum value of a from the results obtained by these above two criteria. Hence, the dimension 2.Ex.3. A rod of 25 mm diameter is homogeneously com- of square cross-section of the column will be a ! 299.6 pressed within its linear elastic range to reduce its length by mm. In this case, the buckling criterion is considered for the 1 mm at an engineering stress of 50 MPa. If the speed of the design of the column. machine is 5 mm s−1, what will be the power required for the test? (b) Buckling Criterion 2.Ex.4. If a cylindrical specimen of initial length 20 mm is According to given condition, critical load for buckling, homogeneously compressed with a constant true strain rate of 0:01 sÀ1; what will be the length of the above specimen Pcr ! factor of safety  applied compressive load after a period of 1 min from the start of the compression ¼ 2:5  1800 kN ¼ 4:5  106 N: test? Neglect the effect of elastic deformation of the testing machine on flow properties. From Euler’s column formula for critical load, given by (2.18) 2.Ex.5. A cylindrical specimen of initial length 70 mm is homogeneously compressed maintaining a constant true p2 EC I ! 4:5  106 N: strain rate till the end of compression test. If the initial L20C cross-head velocity of the machine is 10À1 mm=s; what will be the cross-head velocity after a period of 5 min from the ÀÁ start of the compression test? Neglect the effect of elastic ) I ¼ a4 ! 4:5  106 L02C 4:5  106  182 deformation of the testing machine on flow properties. 12 p2EC ¼ p2  110  109 m4:

116 2 Compression 2.Ex.6. If a homogeneous compression test with a constant (c) Even when length to diameter ratio of a cylindrical cross-head speed of the machine requires 5 min to produce compression sample is nearly one, purely compressive fail- 40% compressive engineering strain in a specimen with an ure that is buckling occurs for initial length of 100 mm, neglecting the effect of elastic deformation of the testing machine on flow properties what (A) concrete; (B) gray cast iron; will be the (C) stone cube; (D) wooden cork. (a) Engineering strain rate in the test? (d) If a plastic specimen with cross-sectional area of 6 mm2 (b) True strain rate at 40% compressive engineering strain? and height of 20 mm is uniformly compressed to 12 mm (c) Cross-head velocity of the machine? without any lateral buckling and barreling of the specimen, the cross-sectional area of that specimen after compression 2.Ex.7. Assume that the flow curve of certain ductile metal will be: in tension is identical to that in compression. If the metal is found to follow Hollomon parabolic true stress–strain rela- (A) 6 mm2; (B) 8 mm2; (C) 10 mm2; (D) 12 mm2. tion in tension with strain-hardening exponent of 0.35 and strength coefficient of 450 MPa, compute the engineering (e) The flow stress of a ductile metal is 200 MPa in tension stress and strain in compression corresponding to the true as well as in compression. If the metal is subjected to tensile strength and the uniform true tensile strain of the compressive stress of r2 ¼ r3 ¼ 50 MPa laterally along the metal, assuming constancy in volume. second and third principle directions, the metal yields when the longitudinal stress in the first principal direction is: 2.Ex.8. A steel bar of rectangular cross-section, with breadth of 50 mm and height of 25 mm, is used as a pin-ended long (A) r1 = 200 MPa in tension, or, r1 = 150 MPa in column. If Young’s modulus of steel is 207 GPa, and the compression; proportional limit in compression is 207 MPa, compute the (B) r1 = 150 MPa in tension, or, r1 = 200 MPa in following: compression; (C) r1 = 150 MPa in tension, or, r1 = 250 MPa in (a) Critical compressive stress for buckling of the long col- compression; umn, if it is 1.2 m long. (D) r1 = 250 MPa in tension, or, r1 = 150 MPa in (b) Minimum length of the long column for which Euler’s compression. column formula can be applied. (f) Bauschinger effect is related to 2.Ex.9. A worked alloy steel tube with outer diameter of 50 mm and wall thickness of 3 mm is to be used as a long (A) Decrease in yield strength when plastic deformation in column with pinned ends. If Young’s modulus of the alloy one direction is followed by deformation in the reverse steel is 207 GPa, and the proportional limit in compression direction; is 1100 MPa, determine the following: (B) Barreling under compression; (C) Increase in yield strength when plastic deformation in (a) Critical compressive stress for buckling of the long col- one direction is followed by deformation in the reverse umn, if it is 3.6 m long. direction; (b) Minimum length of the long column for which Euler’s (D) Buckling failure in compression. equation applies. Answer to Exercise Problems 2.Ex.10. Indicate the correct or most appropriate answer 2.Ex.1. 210 MPa; 35.67%. from the following multiple choices: 2.Ex.2. 636.62 MPa; 109.86%. (a) The failure of concrete under compression test will show a 2.Ex.3. 61.36 W. (A) Cup and cone fracture; (B) Helical fracture; 2.Ex.4. 10.976 mm. (C) Hourglass fracture; (D) Flat fracture. 2.Ex.5. 0.065 mm s−1. 2.Ex.6. (a) 1:33 Â 10À3 sÀ1; (b) 2:22 Â 10À3 sÀ1; (b) As compared to the engineering stress–strain curve in (c) 0:133 mm sÀ1: compression, the true stress–strain curve in compression is 2.Ex.7. 442.22 MPa; 29.53%. (A) below and to the left; (B) below and to the right; 2.Ex.8. (a) 73.89 MPa; (b) 717 mm. (C) above and to the right; (D) above and to the left. 2.Ex.9. (a) 43.7 MPa; (b) 717.6 mm. 2.Ex.10. (a) (C) Hourglass fracture. (b) (B) below and to the right. (c) (D) wooden cork. (d) (C) 10 mm2. (e) (C) r1 = 150

2.9 Solved Problems 117 MPa in tension, or, r1 = 250 MPa in compression. (f) (A) McQueen, H.J., Jonas, J.J.: Hot workability testing techniques. In: Decrease in yield strength when plastic deformation in one Hoffmanner, A.L. (ed.) Metal Forming: Interaction between Theory direction is followed by deformation in the reverse direction. and Practice. Plenum Publishing Corporation, New York (1971) References Pearsall, G.W., Backofen, W.A.: Trans. ASME, Ser. D J. Basic Eng. 85, 68 (1963) ASTM E9: Standard test methods of compression testing of metallic materials at room temperature. Designation: E9–09, ASTM Inter- Polakowski, N.H.: Proc. ASTM 63, 535 (1963) national, West Conshohocken, Pa. (2009). https://doi.org/10.1520/ Richards, C.W.: Engineering Materials Science, p. 151. Wadsworth E0009-09 Publishing Company Inc., Belmont, California (1961) Bauschinger, J.: Zivilingur 27, 289–347 (1881) Rolfe, S.T., Haak, R.P., Gross, J.H.: Trans. Amer. Soc. Mech. Eng. Cooke, M., Larke, E.C.: J. Inst. Met. 71, 371–390 (1945) Davis, H.E., Troxell, G.E., Wiskocil, C.T.: The Testing and Inspection J. Basic Eng. 90, 403–408 (1968) Seitz, F.: The Physics of Metals, 1st edn (Metallurgy and Metallurgical of Engineering Materials, 3rd edn, pp. 140–141. McGraw-Hill Book Company, New York (1964) Engineering Series). McGraw-Hill Book Company, New York, Endersby, V.A.: The mechanics of granular and granular plastic London, p. 147 (1943) materials. Proc. ASTM 40, 1154–1173 (1940) Shackelford, J.F.: Introduction to Materials Science for Engineers, 3rd Freudenthal, A.M.: The Inelastic Behavior of Engineering Materials edn, p. 390. Macmillan Publishing Company, New York (1992) and Structures. Wiley, New York (1950) Taylor, D.W.: Fundamentals of Soil Mechanics. Wiley, New York Hsu, T.C.: Mater. Res. Stand. 9(20–25), 47–53 (1969) (1948). (Chaps. 13–15) Timoshenko, S., Young, D.H.: Elements of Strength of Materials, 5th edn (an east-west edn.). Litton Educational Publishing Inc., New York, p. 272 (1968) van Rooyen, G.T., Backofen, W.A.: Int. J. Mech. Sci. 1, 1–27 (1960) Withey, M.O., Aston, J.: Johnston’s Materials of Construction, 8th edn, p. 867. Wiley, New York (1939)

Hardness 3 Chapter Objectives • Classification of hardness measurement and precautions to avoid erratic measurement. • Measurement of hardness with Mohs’ scale, file test and Herbert pendulum. • Brinell hardness: principle, method with loading period and derivation of expression showing hardness. Indenters and loads for standard and non-standard tests. Condition to obtain a constant hardness for the same material. Advantages and disadvantages. • Meyer hardness: definition, expression and its load sensitivity. Meyer’s law and condition to obtain the same hardness. • Rockwell hardness: principle, loads, indenters, direct-reading hardness dial, hardness scale, method of testing, expression showing hardness, and advantages. • Rockwell superficial hardness: principle and expression showing hardness, hardness scale, merits and demerits. • Vickers hardness: principle, indenters and loads, derivation of expression showing hardness, operational method, and minimum test piece thickness. Advantages and disadvantages, and comparison with Brinell hardness. • Microhardness (Knoop hardness): principle and derivation of expression showing hardness, penetrators and loads, advantages and disadvantages. • Monotron hardness: principle and expression, indenters and hardness scales, advantages and disadvantages. • Shore scleroscope: principle, mass effect of test piece and advantages. • Poldi impact hardness: principle and expression, determination of Brinell hardness with supplied table, advantages and disadvantages. • Nanohardness: indenters, derivation of expression showing hardness with Berkovich indenter, determination of contact depth of penetration, correction for machine compliance, correction due to imperfect geometry of indenter tip, and errors due to pile-up. Consideration of Martens hardness. • Relationship to flow curve and prediction of tensile properties. • Problems and solutions. © Springer Nature Singapore Pte Ltd. 2018 119 A. Bhaduri, Mechanical Properties and Working of Metals and Alloys, Springer Series in Materials Science 264, https://doi.org/10.1007/978-981-10-7209-3_3

120 3 Hardness 3.1 Introduction 3.2 Classification of Hardness The term ‘hardness’ is a structure-sensitive mechanical Tool of material which causes deformation or indentation on property of materials, primarily associated with the surface. the surface of the specimen to be tested must be harder than If a material is uniform in composition and structure, the the test piece. This tool is usually known as indenter, and the hardness measured on the surface layer will represent the deformation mark or impression on the surface of the test hardness of the bulk of the material. In general, the hardness piece is called indentation. Depending on the manner in is defined as the resistance of a material to permanent or which the hardness test is conducted, hardness may be plastic deformation of its surface, usually by indentation, classified as follows: under static or dynamic load. (1) Indentation hardness—It is the resistance of a material From the hardness measurement, one can determine the to permanent indentation under static or dynamic load. tensile or the compressive flow curve of a ductile material up Most of the hardness tests operate on indentation to certain plasticity, if that ductile material does not show principle. Names of such indentation hardness tests are high anisotropy of deformation. Empirical relation between given below: hardness and tensile strength of steel or yield strength of a metal has also been developed. Shear stress–shear strain (i) Brinell; (ii) Meyer; (iii) Vickers (macro- and micro- diagram, as obtainable by the torsion test (Chap. 5), can be hardness); (iv) Rockwell (regular and superficial); achieved without carrying out the test if the tensile flow curve (v) Knoop (microhardness); (vi) Nanohardness (mostly of a material is known which may be obtained from the by Vickers and Berkovich indenters); (vii) Monotron; indentation hardness data. The hardness tester can also be (viii) Poldi; (ix) Herbert pendulum; etc. used to generate constant-load creep curve of a material (Chap. 7) at different stress levels, from which creep strength (2) Rebound hardness—It is the resistance offered by a can be determined. Endurance limit of a material found out material to strike and absorb energy for plastic defor- by the fatigue test (Chap. 8) is related to its tensile strength by mation under impact loads, causing the hammer to some material’s dependent known factor. Thus, hardness rebound. Most common example is the ‘Shore sclero- measurement also helps to get the endurance limit. Hardness scope hardness test’ which measures the hardness in tester is relatively inexpensive, and hardness measurement terms of the rebound height of the indenter. It is vir- requires relatively little experience in comparison to machi- tually an indentation test. nes and test procedures for other types of test mentioned above. Although hardness test cannot substitute the different (3) Scratch hardness—It is the resistance of a material to types of test mentioned above, it may give a rough idea about scratch by another material, for example Mohs’ scale of the aforementioned different mechanical properties of a hardness. material. Thus, the hardness test becomes the easiest way of measuring strength property. Moreover, the quality level of (4) Wear or abrasion hardness—It is the resistance of materials or products may be checked or controlled by the a material to abrasion and wear, when subjected to hardness measurements. Often results of such measurements rotational or sliding motion, for example file hardness become a determining factor in the acceptance or rejection of test. products, especially in any heat-treating or hardening pro- cess. Same materials are employed for different applications (5) Cutting hardness—It is the resistance of a material to depending on their hardness values. Thus, hardness mea- various cutting or drilling operations. This hardness is a surement plays an important role in both the fields of research measure of machinability of a material. ‘Bauer drill and routine shop or laboratory inspection. The hardness and test’ is one of the various tests employed to determine its measurement have been covered in details in a number of the cutting hardness or machinability of materials. In publications (O’Neill 1935, 1967; Williams 1942). this test, a special drill running at a constant speed and pressure is used to make hole in the test piece, and the depth of the hole made in a given time is reported as the hardness of the material.

3.2 Classification of Hardness 121 Table 3.1 Classification of hardness tests Direction of load Fixed load; variable indentation Variable load; fixed indentation application Normal to surface of Static load Dynamic load Static load Dynamic specimen load (a) Brinell (a) Shore (a) Monotron; (b) Meyer scleroscope (b) Wood-hardness tool (c) Vickers (macro- and microhardness) (d) Rockwell regular (b) Poldi (e) Rockwell superficial (f) Knoop (microhardness) (g) Nanohardness (Vickers and Berkovich indenters) Parallel to surface of (a) Herbert pendulum (a) Mohs’ scale of hardness specimen (b) Bauer drill (machinability test) (c) File test Hardness usually means indentation hardness, unless which permits a free plastic flow of material around the specified otherwise. Indentation and rebound-type tests, indenter owing to lack of sufficiently available supporting or because of their simplicity, are most widely used as one of constraining material. the important quality control tests for metal. Depending on the mode of load application during hardness measurement, Decarburized or re-carburized areas from the test surface hardness tests can be classified as shown in Table 3.1. should first be removed by grinding, unless they themselves are of primary interest, as hardness measurements on such 3.3 Precautions to Avoid Erratic Hardness surfaces will not be representative of the entire test piece. In Measurement doubtful cases, it is recommended that two or more tests be made under different loading conditions until the results check one another. To avoid erratic hardness determination, a number of pre- 5. The bottom surface of test piece should be free from cautions, listed below, must be observed for the hardness scale, dirt or other foreign material that might crush or tests, to be discussed in the subsequent articles, particularly flow under the test load, for which a portion of the where the direction of load application is normal to the applied load might be consumed. So, the less amount of surface of specimen. load available for indentation will reduce the size of the indentation for fixed load tests, or more amount of load 1. The indenter and anvil should be clean and well seated. is required to obtain a fixed indentation size for constant 2. To prevent injury to the anvil and indenter, they should depth tests, anyone of which would result in a higher hardness value of the specimen. So careful grinding of not be brought into contact without a test specimen the back side of the specimen or specimen mount is between them. required. 3. The surface to be tested should be flat, smooth and per- pendicular to the vertical axis of the indenter. So the 6. Specimens that have sufficient overhang so that they do surface of the specimen must be prepared by belt not balance themselves on the anvil should be properly grinding and by polishing with emery papers. Diamond supported. polishing may be required particularly for very small-sized indentation to facilitate its measurement. 7. Minimum thickness of the test piece and anvil effect—In 4. The test surface should be dry and clean, which means it general, the necessary minimum thickness of the test should be free from oil, oxide scale, grit particles or foreign piece will vary depending upon the relative hardness of materials and pits, particularly at the point of indentation. the metal and the magnitude of the applied load. It is observed that the material of the test piece is plastically Oiled surfaces generally give slightly lower hardness deformed over an appreciable distance below the sur- readings than dry ones because of the reduced friction under face of the indentation. From tests on a large variety of the indenting tool. materials, where the thickness of the test piece was less than the ‘critical’ value, the National Bureau of Stan- A pitted surface results in a low hardness reading because dards noted that for each indentation on the specimen, a some indentations form near the edge of a pit (a depression), spot of altered surface was visible on the surface

122 3 Hardness opposite to the indentation surface. The ASTM speci- formed indentation may be either too large due to lack fies that no hardness measurements on thin sections or of availability of sufficient constraining material near soft materials can be considered reliable if the com- the previous indentation or too small owing to pression or indenter mark or bulge appears on the strain-hardening effect of the material caused by the reverse side of the test piece during indentation. It is previous indentation. However, tests have shown that recommended that the thickness of the test piece should one can get rid of the erroneous hardness measurement be at least ten times the depth of the indentation. The if the spacing between the centres of the adjacent reason for such recommendation is that the hardness of indentations is equal to or greater than three times the the test piece is influenced not only by the hardness of diameter of the previous indentation. the material at the point of indentation but also by the 11. Curved or cylindrical surfaces—Test on the convex hardness of the material below the point of indentation side of a curved surface or on the outer surface of a up to at least ten times the depth of the indentation. hollow cylindrical body or on the surface of a solid Consequently, if a soft layer or a hard surface is present cylinder causes a free plastic flow of material around within this depth, the size of the indentation will, the indenter due to the lack of availability of con- respectively, be larger or smaller indicating a lower or a straining material at the point of indentation and will higher hardness value of the specimen. Obviously if the produce a larger indentation resulting in a lower specimen supporting hard anvil is situated within the ten hardness value than that obtained from test on a flat times the depth of the indentation during testing of a thin surface. On the other hand, in the absence of any sample, the influence of the hard anvil, called ‘anvil proper support below the curved convex surface or effect’, will cause noticeable reduction in the depth of the inside the hollow cylinder, a portion of the applied load indentation for fixed load tests, or increase the amount of will be consumed for the compression of the curved load required to obtain a fixed indentation depth for surface causing the amount of load available for the constant depth tests, and result in a hardness higher than indentation to become less for constant load tests, when a thick sample of the same material is tested. or requiring higher amount of load for constant depth 8. A single thickness of material should only be used for tests. Any one of these will result in a higher hardness testing. To increase the thickness of test piece if several value. thin pieces of the same material are combined or clamped for loading in a test, the tendency for relative movement Test on the concave side of a curved surface or on the on the surfaces between the various pieces and lack of inner surface of a hollow cylinder will produce a smaller flatness of the separate pieces will consume a portion of indentation and result in a higher hardness value due to the the applied load for the sliding action as well as for the restricted lateral plastic flow of the material caused by the flattening of the pieces on the anvil. Thus, the less amount constraining material surrounding the point of indentation. of load available for indentation will reduce the size of Indentation on the concave surface is preferable to that on the indentation for fixed load tests, or more amount of the convex surface if no proper support is provided below load is required to obtain a fixed indentation size for the convex surface or inside the hollow cylinder to avoid constant depth tests, anyone of which would result in a compression of the curved surface. higher hardness value of the specimen. So when thick sample is not available, thin specimens to be clamped If comparative results are not desired in the testing of must be completely flat and care must be taken during curved surfaces or round sections, a flat spot for taking clamping so that no sliding takes place under load. indentation should be made on the curved or round surface 9. Edge effect—If an indentation is made too near the edge by either filing or grinding in order to avoid the above error of the specimen, the lateral plastic flow of the material in the hardness measurement. Further, the test piece must be along the specimen surface will be easier during loading appropriately clamped and a mandrel must be used as a owing to lack of sufficiently available supporting or support wherever necessary. constraining material near the edge and the indentation may be both too large and unsymmetrical, called ‘edge 12. Effect of vibration—If the base on which the hardness effect’, resulting in a lower hardness value of the speci- machine is mounted is subjected to vibration during men. So it is recommended that the distance from the edge moving the operating handle for application of load or of the test piece to the centre of the indentation should be otherwise, the hardness value will be too low, since the at least 2.5 times the diameter of the indentation. indenter will sink farther into the material than when 10. Effect of previous indentation—If the test piece is such vibrations are absent. In such case, felt or rubber indented too close to a previous indentation, the newly washers should be placed beneath the machine in order to absorb the vibration.

3.3 Precautions to Avoid Erratic Hardness Measurement 123 13. Speed of load application—The speed of application of corundum with HM = 9, the assigned Mohs’ hardness value the load should always be standardized through of tungsten carbide will be between 9 and 10. Similarly, the adjustment of an oil-dashpot arrangement in the Mohs’ hardness of mica scratched by fluorite and not by machine. A rapid rate of applying the load affects the calcite will be between 3 and 4. size of the indentation in the following two ways: Mohs’ scale of hardness is widely used in the field of (i) The inertia effect of the weights due to sudden mineralogy but rarely applied for the testing of metals and application of the entire load causes a momentary alloys, because the intervals are not widely spaced in the high rise of the load above the selected or preset load hardness range of Mohs’ scale and metals of similar hardness and consequently enlarges the indentation. cannot be distinguished clearly. Most hard metals fall in the Mohs’ hardness range of 4–8. Hardened steel forming (ii) A rapid rate of indentation allows less time for the martensite has approximately 7 Mohs. However, annealed plastic flow of the material, resulting in a decreased copper and most pure metals have hardness below 4 Mohs. size of the indentation. Soft lead cannot be included in the Mohs’ scale as it lies below 1 Mohs. Table 3.2 lists ten standard minerals of the Tests have shown that the error due to inertia effect is scale, together with their formulas and types of bonding much more than that due to an insufficient period of loading (Zwikker 1954). for plastic flow to occur. Thus, rapid rate of load application would cause an increased indentation size and lower the With the development of extremely hard abrasives, the hardness value. The rate of load application may be stan- need to distinguish between materials in the range of hard- dardized with the help of standard specimen of known ness between that of quartz and that of diamond is felt. hardness. Table 3.3 (Davis et al. 1964) shows an extension of Mohs’ scale devised for the above purpose. 3.4 Mohs’ Scale of Hardness 3.5 File Hardness Test One of the first systematic hardness scales is the Mohs’ File hardness test is a simple qualitative method to determine hardness scale, proposed by a German mineralogist, Frie- the abrasion hardness of metals and widely used in many drich Mohs, in 1832. The scale consists of ten standard plants for inspection and quality control purposes. It is per- minerals with a series of numbers from 1 to 10, arranged in formed by subjecting a test piece to the sliding action of a order of increasing hardness. Diamond, the hardest material, predesigned standard file and noting whether a visible cut is is assigned the Mohs’ hardness number HM = 10. Talc, one produced or not. of the softest minerals, is assigned the Mohs’ number HM = 1. The other integers, 2 through 9, are assigned to The test piece is either clamped in a vise or firmly gripped other natural minerals of corresponding relative hardness. in one hand, and a standard file of appropriate hardness is slowly but firmly drawn for a single time over surface of a Mohs’ hardness of a material is determined quickly and test piece. If that file slides over the surface without pro- qualitatively by surface scratching with the standard miner- ducing any visible cut on it, the material under test is als listed in the Mohs’ scale. For example, as tungsten car- acceptable. But if the file bites into the test piece, a visible cut bide is scratched by diamond with HM = 10 and not by Table 3.2 Standard Mohs’ scale Mohs’ hardness Minerals Formula Types of bonding of hardness number, HM 1 Talc 3MgOÁ4SiO2ÁH2O van der Waals (between layers) 2 Gypsum CaSO4Á2H2O van der Waals (between layers) 3 Calcite CaCO3 van der Waals (between layers) 4 Fluorite CaF2 Ionic 5 Apatite CaF2Á3Ca3P2O5 Ionic 6 Orthoclase feldspar K2OÁAl2O3Á6SiO2 Mixed ionic-covalent 7 Quartz SiO2 Mixed ionic-covalent 8 Topaz (AlF)2SiO4 Mixed ionic-covalent 9 Corundum Al2O3 Mixed ionic-covalent 10 Diamond C Covalent

124 3 Hardness Table 3.3 Extension of Mohs’ scale of hardness (Davis et al. 1964) angle of contact between the file and the surface of the test piece should be maintained uniformly for all sam- Extended Mohs’ hardness number Minerals ples of the same material for each test. This is important because the file may produce a cut on projections and 1 Talc sharp edges more easily than on fiat surfaces. 2 Gypsum Advantages 3 Calcite • The test is speedy in the hands of a skilled operator. Due to the rapidity of the test, file testing provides a simple 4 Fluorite method for routine checking to control the hardness level of materials or products. For example, the hardness of 5 Apatite each tooth of a gear may be determined in a very short time by the file testing, compared to other testing meth- 6 Orthoclase feldspar ods that require much longer time. 7 Vitreous pure silica • Filing is carried out for each piece in the lot, and file is drawn for a single time over the surface of the test piece, 8 Quartz whereas due to the requirement of longer time, more accurate conventional methods of testing are carried out 9 Topaz for only representative samples from each lot. 10 Garnet • The test allows determining the hardness of a material in inaccessible places, whereas with a table-type or con- 11 Fused zirconia ventional hardness testing machine, it would be usually impossible to obtain hardness measurements in such 12 Fused alumina places. 13 Silicon carbide • Surface preparation of the test piece is not required, because enough adhering surface scale will normally be 14 Boron carbide removed during filing. 15 Diamond Disadvantages will be noted on the surface and the material will be rejected, • The accuracy of the file hardness testing depends on the as it proves to be softer than the desired one. skill of the technician performing the test. As the test compares the hardness between the material in • The test results are generally not reproducible enough. question and a predesigned standard file, obviously, the test This may be due to the flexibility of the inevitable vari- method depends upon the following factors: ations in characteristics of ordinary files. (1) Type of file—The type of file that will be used for all • It is impossible to make fine distinctions in hardness. samples of the same product throughout subsequent • Quantitative measurements of hardness that are given by tests must be the same with respect to the following parameters. table-type or conventional hardness testing machines are not possible. (a) Size of the file—It is usually either six inch or eight inch. 3.6 Brinell Hardness (b) Shape of the file—It is usually either pillar or In 1900, Dr. J. A. Brinell, of Sweden, proposed and stan- three-square file. Pillar file is used on fiat, oblong dardized a method for quantitative determination of inden- or square test pieces, and three-square file is used tation hardness, which was the first widely accepted in testing objects of irregular shapes having cre- indentation hardness test, known as ‘Brinell hardness’ test vices and grooves. (Wahlberg 1901). Figure 3.1a shows a pump-type hydraulic Brinell hardness tester machine, and Fig. 3.1b shows (c) Hardness of the file—It is generally standardized schematically its salient features. by noting whether or not it will cut the material under test that has the proper hardness, as deter- mined quantitatively by more precise hardness testing method. (2) Speed, pressure and angle of filing during the test—In order to achieve more precise and valid hardness indi- cation of the test piece, the speed of filling over the test surface must be slow, because rapid filling will remove material, not only from the test surface, but also from the file, and thereby indicates the test piece to be softer than is actually the case. The amount of applied pres- sure during filing must be the same for all samples of the same material throughout subsequent tests. The

3.6 Brinell Hardness (a) (b) 125 Bourdon gage Fig. 3.1 a A pump-type Weights which Weight to balance hydraulic Brinell hardness tester regulate load Pump handle oil pressure at machine. b Schematic diagram desired load showing the salient features of a Plunger hydraulic Brinell hardness tester Ball indenter Small (Davis et al. 1964) Specimen piston Brinell ball Anvil Oil pressure (spherical seat) applied by Table for specimen Hand wheel pump for elevating anvil Elevating screw 3.6.1 Principle of Testing diameter. Moreover, the curved surface area of the indentation will be more accurate when it is calculated from the mea- The Brinell hardness test consists of forming an indentation surement of the larger dimension that is the diameter rather by forcing a standard spherical ball indenter into the surface than the depth of indentation. So, the most common method is of the material under a definite static load applied for a to measure the diameter of indentation as its measurement standard period of time. The standard Brinell hardness tester does not in general influence the Brinell hardness. operates usually under hydraulic pressure that applies force. 3.6.2 Derivation for BHN The Brinell hardness number (BHN), expressed in units of kilograms per square millimetre, is defined as the ratio of It is assumed in the following derivation for BHN that there the applied load, P in kilograms, to the curved surface area is no elastic deformation or distortion of the spherical ball of the elastically recovered indentation, AS in square mil- indenter and no elastic recovery of the indentation takes limetres. That is, place when the applied load is released. That means the diameter of the indentation after the removal of load will Brinell hardness number or BHN ¼ P kgmm2 ð3:1Þ remain the same as that of the indentation formed just before AS the release of the applied load, and the vertical centre plane of the indentation coincides with that of the spherical ball In order to determine the Brinell hardness number of a indenter. material, it is necessary to know the curved surface area of the indentation in the test piece. This can be calculated in terms Figure 3.2 shows that an indentation of diameter ‘d’ with of the spherical ball diameter, D; and the diameter of the depth ‘h’ is formed after the indenter ball with diameter ‘D’ indentation, d; or the depth of the indentation, h; as shown has penetrated into the surface of test piece under an applied below. The diameter of the indentation means the diameter of load, ‘P’. Let ‘2a’ be a solid angle of cone formed at the centre the circular impression along the surface of the test piece. of the indenter by the circular edge of the indentation on the surface of the test piece. On the curved surface of the inden- To calculate the curved surface area AS of the indentation, tation, let us arbitrarily choose an infinitesimally thin circular the diameter or the depth of the indentation has to be measured section of average radius ‘r’ that makes an arbitrary solid after the release of applied load. But after the removal of angle of ‘2h’ at the centre of the indenter. The arbitrary applied load, the indentation is recovered elastically and if semisolid angle ‘h’ formed at the centre of the indenter varies there is no or insignificant distortion of the spherical indenter from ‘0’ along the centre plane of the indentation to ‘a’ at the during loading the indentation will take the shape of a semicircular edge of the indentation on the surface of the test spherical cap but with a radius of curvature higher than that of piece, i.e. hMinimum ¼ 0 and hMaximum ¼ a: It is further the spherical ball indenter due to the elastic recovery of assumed that the thickness of the above section makes an indentation. The harder the material, the greater is the elastic angle dh at the centre of the indenter, as shown in Fig. 3.2. recovery. Elastic recovery affects the depth of indentation greatly but its effect on the diameter of indentation is negli- gible, owing to the larger dimension associated with the

126 3 Hardness P Again from Fig. 3.2 ðd=2Þ=ðD=2Þ ¼ sin a; or; d=D ¼ sin a; pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi rffiffiffiffiffiffiffiffiffiffiffiffiffiffi rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 À sin2 a d2 D2 À d2 Vertical centre plane of ) cos a ¼ ¼ 1 À D2 ¼ D2 indenter and indentation pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ¼ D2 À d2 D 2α So from (3.2a), total curved surface area of indentation is: D pD2 pD2 \" pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi# θ dθ 2 2 D2 À d2 D AS ¼ ½1 À cos aŠ ¼ 1À D 2 \" pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi# ¼ pD2 D À D2 À d2 2D ¼ pD h À pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffii ð3:2cÞ D D2 À d2 h Surface of 2 test piece r d Hence from (3.1), (3.2b) and (3.2c), Brinell hardness φ number can be given by the following (3.3a) and (3.3b), 2φ although the working formula for BHN is given by (3.3a) because it does not involve the depth of indentation h; which Fig. 3.2 Figure for derivation of BHN, showing that an indentation of is largely affected by elastic recovery on the release of diameter ‘d’ with depth ‘h’ is formed after the indenter ball with applied load. diameter ‘D’ has penetrated into the surface of test piece under an applied load, ‘P’. The included angle between the tangents drawn to the From (3.1) and (3.2c), edge of a Brinell indentation is ‘2u’ BHN ¼ P ¼ h 2pPffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffii ð3:3aÞ AS pD D À D2 À d2 Now, the curved surface area of the infinitesimally thin From (3.1) and (3.2b), circular section is: BHN ¼ P ¼ P ð3:3bÞ AS pDh D dAS ¼ 2pr 2 dh ¼ prDdh It is evident from the above (3.2b) and (3.2c) that ) Total curved surface area of indentation will be pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi Depth of indentation; h ¼ D À D2 À d2 Z Zh¼a Zh¼a ð3:4Þ 2 AS ¼ dAS ¼ pD rdh ¼ pD D sin hdh When the load is released, the resulting indentation is not 2 perfectly spherical if the spherical ball indenter deforms elastically under load. Although the derivation of equation AS h¼0 h¼0 for BHN has assumed a perfectly spherical indentation with no elastic recovery of the indentation after removal of load, Zh¼a the BHN, as computed by (3.3a), is nevertheless accepted as ¼ pD D sin hdh valid for all practical purposes, provided standard conditions of testing are maintained. 2 h¼0 ¼ pD2 ½À cos hŠa0¼ pD2 ½À cos a þ 1Š 2 2 ¼ pD2 ½1 À cos aŠ ð3:2aÞ 2 Since from Fig. 3.2, cos a ¼ fðD=2Þ À hg=ðD=2Þ; it fol- 3.6.3 Indenters, Loads and Loading Periods lows from (3.2a) that The diameters of spherical steel ball indenters used in the standard Brinell hardness test are either 5 or 10 mm. AS ¼ pD2 1 À ðD=2Þ À ! ¼ pD2 2h ¼ pDh ð3:2bÞ Although odd-size ball indenters having diameters such as 2 D=2 h 2 D 1.25, 2.50 and 7.00 mm are also available but the use of

3.6 Brinell Hardness 127 these non-standard indenters is not considered to be a stan- above standard test specifications. For such cases, as dis- dard Brinell test. The ball indenter normally used is made cussed below, the material may be tested with either smaller from heat-treated hard high-carbon steel, known as ‘Hult- loads or with indenters of smaller diameter. However, such gren ball’ or made from tungsten carbide. When the indenter Brinell hardness tests should not be considered as standard is forced into the test piece, the indenter will be subjected to one. With a non-standard load or ball diameter, it is neces- varying amount of elastic deformation depending on the sary to produce geometrically similar indentations so that magnitude of the applied load and the hardness of the test the same BHN can be obtained, at least theoretically. Geo- piece and the indenter. In case of a softer material having metrically similar indentations will be achieved provided BHN less than or equal to 300, there will be practically no or that the solid angle ‘2a’ in Fig. 3.2 remains constant, where insignificant distortion of the spherical indenter under load. ‘2a’ is a solid angle of the cone formed at the centre of the If the spherical indenter undergoes some elastic deformation indenter by the circular edge of the indentation on the sur- under load, the surface of the indentation does not remain face of the test piece. Now the question arises—what should truly spherical. Hardness beyond a Brinell number of about be the constant value of ‘a’? For the same material, it is 600, the distortion of the Hultgren ball indenter during aimed to achieve the Brinell hardness number identical with loading period becomes substantially high causing the the more precise and accurate Vickers hardness number curved surface area of the indentation to increase so that the (VHN), to be discussed in Sect. 3.10, because both the BHN of the material will show a lower value than it should above hardness numbers are defined in a similar way and be. So the Hultgren ball is used till the BHN of the material have the same theoretical hardness scale. The included angle does not exceed a value of 600. But if instead of a Hultgren between the two opposite faces of the square-based ball, a tungsten carbide ball indenter is used to minimize the pyramidal-shaped Vickers indenter is 136: To maintain distortion of the indenter, this upper limit of 600 Brinell may the similarity between BHN and VHN, the included angle, be extended to approximately 725 Brinell. ‘2u’, as shown in Fig. 3.2, between the tangents drawn to the edge of a Brinell indentation must be 136: Hence, the 3.6.3.1 Standard Test solid angle, 2a, formed at the centre of the indenter will be: For the Brinell hardness test, the American Society for Testing Materials (ASTM) and the British Standards Insti- 2a ¼ 180 À 2u ¼ 180 À 136 ¼ 44; or; tution specify the application of either a load of 3000 kg on a semi-solid angle; a ¼ 22: 10-mm-diameter ball or a load of 750 kg on a 5-mm- diameter ball for at least 10 s when testing iron, steel and It is possible to keep ‘a’ at the desired value of 22; if the alloys having hardness similar to steel. For non-ferrous ratio of the diameter of indentation, d, to that of the ball metals and alloys, a 10-mm-diameter ball is specified under indenter, D, is 0.375, because, a load of either 500 kg with a loading period of at least 30 s in case of copper, annealed brass, magnesium alloys, etc., or d=D ¼ sin a ¼ sin 22 ¼ 0:375: 1000 kg for at least 15 s in case of gun metal/bronze, cold-worked brass, etc. With the help of (3.1) and (3.2a), the Brinell hardness number can also be expressed as follows: The loading in the Brinell hardness test should be con- tinued for such a time period that the material under test BHN ¼ P ¼ P À cos aŠ ¼ 2P cos aŠ becomes fully strain hardened. The softer the material the AS ðpD2=2Þ½1 pD2½1 À longer is the time required for its strain hardening. That is why the minimum time of full load application for harder ð3:3cÞ materials is less than that for softer materials. From tests at the National Bureau of Standards, it was found that the Equation (3.3c) shows that for a and BHN to remain plastic flow for most materials is quite rapid during the first constant, the applied load, P, must be proportional to the 30 s under the applied load; it is much less rapid in the square of the ball diameter, D2, that is the load and ball interval from 30 to 120 s. For most metals, the Brinell number varies less than 1% for loading intervals between 30 diameter must be varied in the ratio and 120 s. This is due to the insignificant enlargement of the indentation in a fully strain-hardened material that produces P1 ¼ P2 ¼ P3 ¼ ÁÁ Á ¼ C ðAn arbitrary constantÞ: practically no plastic flow. D21 D22 D23 3.6.3.2 Non-Standard Test The values of C may vary from 1 to 30 with intermediate Frequently, either for thinner or for softer specimens, the values as 2.5, 5, 10, 15, 20 and 25, depending on the Brinell hardness test cannot be conducted according to the hardness and the thickness of test piece. Unless precautions are taken to maintain P D2 constant, the BHN will generally vary with load. If value of C ¼ 1 is considered for an indenter diameter of 2.5 mm, the magnitude of applied load

128 3 Hardness in Brinell test may be as low as 6.25 kgf, which is occa-  sionally used. guideline to select P D2 ratio depending on the material under test as well as its hardness, so that 0:25 d=D 0:5; 3.6.3.3 Selection of Ratio of Load to Square is satisfied. Again, the diameter of the indenter D is selected depending on the thickness of the test specimen so that the of Indenter Diameter thickness of the test piece must be greater than or equal to ten times the depth of indentation. It is experimentally inconvenient to maintain a constant value of 0.375 for the ratio of d=D: Hence, P D2 ratio 3.6.4 Method of Testing should always be so adjusted that the ratio of d=D would lie Initially, the specimen to be tested is made flat by grinding and then polished so that the indentation formed by loading on the between 0.25 and 0.50, average of which is 0.375. However surface of the specimen is clearly visible under the Brinell microscope, used for measurement of indentation diameter. in many instances, deviation from the standard testing Then depending on the diameter of Brinell indenter, the load to be applied is selected maintaining the condition of testing, as specifications will be necessary to maintain the above range discussed in Sect. 3.6.3.3. The specimen is placed on the anvil of the testing machine, and the anvil is raised by rotating the of d=D ratio. If the value of d=D is found to be lower in hand wheel so that the specimen surface is brought in tight contact with the apex of the indenter. Then, depending on the Brinell hardness test, it indicates that the test piece is material under test, the time of load application is set and the selected load is applied to the specimen for the preset duration. becoming harder, which will cause more distortion of the After the application of load, the specimen is released by lowering the anvil. Now, the specimen is ready for measure- indenter during the loading period. So, the minimum value ment of the diameter of the indentation, which has been formed on the specimen surface. To achieve the proper Brinell of d=D is kept at 0.25 to minimize as far as possible the hardness number, the indentation diameter must be read accurately. Errors in reading the diameter of the indentation elastic deformation of the indenter and its adverse effect on may arise from the following two causes: the hardness value. The maximum value of d=D is limited to (i) Error in reading from the microscope and (ii) The boundary of the indentation is not distinct. 0.5 to avoid the damage of the cap of the indenter, that holds the ball in place, due to the contact of the cap with the surface of the specimen because the indenter will sink more into the test piece as it is becoming softer. In order to avoid the anvil effect, the thickness of a uni- formly hard test piece should be at least 10 times the depth of the indentation, i.e. ‘10 Á h;’ as discussed in Sect. 3.3. Thus, the magnitude of the applied load is selected by the size of the indenter and the hardness as well as the thickness of the material under test.  To obtain a constant BHN for the same material, P D2 ratio must be selected in such a way that both the conditions —(i) 0:25 d=D 0:5; and (ii) thickness of the test piece ! 10 Á h; are satisfied. The following Table 3.4 will help as a  Table 3.4 Guidelines for selecting P D2 ratio P Recommended materials BrinellÀ hardnessÁ Recommended Diameter of the Applied Time of load D2 range kg mm2 thickness range of ball indenter, load, P (kg) application (s) test piece (mm) D (mm) 10 ± 2 30 Iron, steel and similarly hard 142.6–601.5 10 3000 ferrous and other alloys More than 4 5 750 15 ± 2 4–2 2.5 187.5 10 Harder non-ferrous metals and 47.5–200.5 Less than 2 10 1000 30 ± 4 alloys like gun metal/bronze, 23.8–100.2 More than 6 5 250 cold-worked brass 11.9–50.1 6–3 2.5 62.5 60 ± 6 Less than 3 10 500 5 Non-ferrous metals and alloys More than 4 5 125 like copper, annealed brass, 4–2 2.5 31.25 magnesium alloys Less than 2 10 250 More than 6 5 62.5 2.5 Softer non-ferrous metals and 6–3 2.5 15.625 alloys like aluminium, lead, tin Less than 3 and their alloys

3.6 Brinell Hardness 129 The diameter of the indentation is measured accurately Sometimes for measurement of the true diameter of with a low-power microscope. The eyepiece of the micro- indentation accurately, the steel ball indenter is coated with a scope has a scale from 0 to 7 mm graduated in tenths of a dull black or a blue pigment or etched lightly with nitric acid millimetre that allows an accuracy of the diameter mea- before making the indentation so that the indentation surement up to 0.05 mm. Two readings of the diameter, becomes more distinct. d1 and d2; of nearly circular indentation at right angles should usually be taken and the arithmetic average of d1 and 3.6.6 Advantages and Disadvantages d2 must be substituted for the diameter, d; of the indentation in (3.3a). Since P and D are already known, so the Brinell The followings are the advantages of Brinell hardness test: hardness of the specimen can easily be calculated by (3.3a). • The relatively large size of the Brinell indentation will A non-circular indentation will result if the compressive include local heterogeneities in the material having properties of a flat specimen are not uniform due to direction heterogeneous structure and provide its average hard- of rolling or cooling stresses or any other reason. Then to ness. So, for materials like cast iron or heterogeneous obtain the average Brinell hardness of the material, the non-ferrous alloys, measurement of Brinell hardness is diameters, d1; d2; d3 and d4; of the non-circular indentation not only advantageous but also preferable. are measured in four directions, roughly 45° apart, and their arithmetic average must be put in place of the diameter, d; of • The influence of surface scratches and roughness will be the indentation in (3.3a). less in the Brinell test than other hardness tests. 3.6.5 Anomalous Behaviour • Surface polishing of the test piece is not needed due to the larger size of the Brinell indentation. Generally, a Two types of anomalous behaviour, as illustrated schemat- satisfactory surface may be obtained by belt grinding or ically in cross-sections of Brinell indentations in Fig. 3.3, filing with a relatively fine file. are frequently observed owing to the localized deformation of material at the indentation. The disadvantages or limitations of Brinell hardness test are as follows: The sketch in Fig. 3.3a shows ‘ridging’ or ‘piling up’, in which there is a formation of a lip or a raised ridge of • Determination of the entire range of hardness encoun- material around the periphery of the indentation. The mea- tered in commercial metals is not possible under appli- sured diameter d of such indentation is found to be higher cation of a single load. than the true diameter of a perfect indentation and would result in a low hardness value. This condition is observed • Due to the larger size of the indentation, the application with cold-worked materials having little ability to of Brinell hardness test is not possible on small jobs or strain-harden. The sketch in Fig. 3.3b shows ‘sinking in’, in critically stressed portions which would crack during the which a depressed surface of the material is formed around indentation. the periphery of the indentation. The measured diameter d of such indentation is lower than the true diameter of a perfect 3.7 Meyer Hardness indentation and a high hardness value would result. This type of behaviour is frequently observed with annealed A definition of indentation hardness under the application of materials having a high rate of strain hardening. In both a constant static load, proposed by Meyer (1908), known as sketches of Fig. 3.3, h1 is the observed indentation depth, ‘Meyer hardness’ is more rational than that proposed by usually determined from the relative motion of the specimen Brinell. Meyer hardness is defined as the ratio of the load to and the ball plunger, but h1 differs from the actual depth h the projected area of the elastically recovered indentation corresponding to the indentation diameter d : Fig. 3.3 Cross-sections of (a) (b) indentations in Brinell test. a ‘Piling up’ indentation showing d d encircling ridge. b ‘Sinking in’ indentation showing encircling h1 h h h1 depression ‘Piling up’ ‘Sinking in’

130 3 Hardness rather than the surface area of the indentation as defined for If R.H. denotes the rate of work hardening, then R.H.= BHN. The mean pressure, pm; between the surface of the indenter and the indentation is same as the Meyer hardness rate of change of Meyer hardness or mean pressure with having units of kilograms per square millimetre just like the Brinell hardness. indentation diameter d; which means d  0  4K0 ðn0 2Þdn0À3 dðdÞ 4K dn0À2 p R:H: ¼ ¼ À ð3:8Þ p Meyer hardness ¼ mean pressure pm ¼ P kg/mm2 ¼ 4P Since K0 ¼6 0; or; d 6¼ 0; and as R:H: ¼ 0; for a fully Ap pd2 strain-hardened material, so, from (3.8), we get n0 ¼ 2: ð3:5Þ In (3.8) when n0 [ 2; the value of R.H. is positive indi- where cating that the material possesses work-hardening capacity. For a fully annealed material, the value of n0 will be about P applied load, kg, 2.5. The value of the exponent in Meyer’s law, n0, is the Ap elastically recovered projected area of indentation in slope of the straight line obtained when log P is plotted the test piece, sq mm, and d chordal diameter of indentation after unloading, mm. against log d: The intercept of the linear plot of log P versus log d at log d ¼ 0; is log K0; which means K0 is the value of Hence, d is measured after the release of applied load. P; when d ¼ 1 mm: From (3.6), we see that K0 will be expressed in kg=mmn0 : Meyer hardness or mean pressure is determined using the Brinell hardness tester but the Meyer hardness measurement 3.7.2 Load Sensitivity is rarely performed for practical purposes, although it is a more fundamental measure of indentation hardness. 3.7.1 Meyer’s Law (Meyer 1908) The Meyer hardness is less affected by the variation of the Meyer formulated an empirical relationship, usually called applied load in comparison to the Brinell hardness when the ‘Meyer’s law’, between the applied load and the size of the diameter of the ball indenter ‘D’ is kept constant. For a fully indentation, as shown below: strain-hardened material with n0 % 2; the Meyer hardness does not depend on load and remains constant, whereas the Meyer’s law: Brinell hardness decreases with the increase of load. The P ¼ K0dn0 ð3:6Þ reason for constancy of the Meyer hardness for a fully strain-hardened material can be obtained as follows by putting n0 % 2; in (3.7), where pm ¼ 4K0 d n0 À2 ¼ 4K0 d2À2 ¼ 4K0 ¼ constant; p p p P applied load, kg; d chordal diameter of indentation after unloading, mm; because K′ = constant for a given ball diameter and material. K′ a material constant expressing resistance of material to The reason for variation of the Brinell hardness for a fully the initial penetration; cold-worked material is explained below: n′ a material parameter related to strain hardening of BHN ¼ load=contact area ¼ projected area material % n + 2. Hence, n0 is a measure of the effect pm load=projected area contact area of deformation on the hardness of the metal (Hoyt 1924). n0 is often called as Meyer index.  projected area As already discussed in Chap. 1, n is the strain-hardening ) BHN ¼ pm contact area  coefficient in Hollomon parabolic relationship for the curve of true-stress–true-strain. As fully work-hardened material ¼ a constant value  projected area  cannot be further strain hardened, so the value of n will approach towards zero, which means n0 % 2; for fully contact area strain-hardened material that can be proved as follows: For a spherical indenter, when the load is increasing the With the help of (3.5) and (3.6), one can rewrite the contact area is increasing more quickly than the projected area. Meyer hardness or mean pressure as Hence, the ratio of projected area to contact area decreases with the increase in load causing the BHN to decrease. pm ¼ 4P ¼ 4K 0 dn0 ¼ 4K 0 d n0 À2 ð3:7Þ pd2 pd2 p For a fully annealed material with n0 % 2:5; as the load is increased the Meyer hardness increases due to the strain hardening produced by the indentation but the rate of increase in hardness gradually decreases because of the

3.7 Meyer Hardness 131 decrease in n0 that gradually approaches towards the value of For geometrically similar indentations, since d=D is 2. Ultimately, the Meyer hardness becomes constant for still constant (see Sect. 3.6), the Meyer hardness must also be higher loads where the deformation of the material will be so constant, as shown by (3.11b). Therefore, geometrically extensive that it becomes fully cold worked with a value of similar indentations will give the same Meyer hardness n0 ¼ 2: However, as the load increases, the Brinell hardness number. Equation (3.12a) can also be rearranged as follows: will initially increase due to the strain hardening but the hardness-increase-rate gradually decreases till the material P ¼ Cd2 d n0À2¼ C dn0 D2 or; P ¼  d n0 will become fully cold worked and thereafter, the BHN D Dn0 D2 C decreases for still higher loads. D 3.7.3 Influence of P/D2 ð3:13Þ If the same material is indented with ball indenters of different  diameters, the values of K0 and n0 are expected to change. For The above (3.13) shows that if the ratio P D2 iskept con- ball indenters of diameters D1; D2; D3; . . .; producing stant, d=D will remain constant. Hence for same P D2 ratio, indentations of diameters d1; d2; d3; . . .; under the application geometrically similar indentations are obtained that will give of same load P; a series of relation is obtained of the type the same Meyer hardness number. Therefore, when the ratio of load to square of indenter diameter is kept constant, the same Meyer hardness value will be obtained for the same material. Meyer hardness testing has some advantages over Brinell hardness testing as follows: P ¼ K10 d1n01 ¼ K20 d2n20 ¼ K30 d3n30 ¼ Á Á Á ð3:9aÞ • The Meyer hardness is less sensitive to the applied load than the Brinell hardness, as discussed above. From a series of investigations, Meyer found experi- mentally that for different diameters of the ball indenters, • The Meyer hardness is a more fundamental measure of n0 % constant, but that the values of K0 decreased when the indentation hardness because it is based on the projected diameters of the indenters were increased and vice versa. area of the indentation rather than the surface area as in Hence, (3.9a) can be rewritten as the case of the Brinell hardness. P ¼ K10 d1n0 ¼ K20 d2n0 ¼ K30 d3n0 ¼ Á Á Á ð3:9bÞ As Meyer hardness is measured using the Brinell hard- ness tester and the Brinell indenter, so the advantages and The relationship between K0 and D can be expressed disadvantages associated with the Brinell hardness testing, empirically in the following form: mentioned in Sect. 3.6, will also be applicable to the Meyer hardness testing. K10 D1ðn0À2Þ ¼ K20 D2ðn0À2Þ ¼ K30 D3ðn0À2Þ ¼ Á Á Á ¼ C ð3:10Þ where C is a constant. The general expression for Meyer’s 3.8 Rockwell Hardness law involving both d and D is then obtained from (3.9b) and (3.10) as follows: In 1908, Professor Ludwig of Vienna described a method to determine the hardness of a metal by ‘differential depth’ P ¼ Cd1n0 ¼ Cd2n0 ¼ Cd3n0 ¼ ÁÁ Á ð3:11aÞ measurement test. This hardness testing method involved the D1n0À2 D2n0 À2 D3n0À2 measurement of the increment of depth of an indenter forced into the metal by a primary and a secondary load. Based on Or in general form, Ludwig’s principles, Stanley P. Rockwell invented the Rockwell hardness test in 1919 (Rockwell 1922). Today, it is P ¼ Cdn0 ð3:11bÞ the most widely used method to determine quantitatively the Dn0À2 indentation hardness, particularly for routine testing purpose in industry. A photograph of one model of Rockwell hard- Two interesting conclusions originate from (3.11b). This ness tester is displayed in Fig. 3.4. equation may first be rewritten as 3.8.1 Principle of Testing P ¼  d n0À2 ð3:12aÞ d2 C The Rockwell test measures the hardness value of material by utilizing the difference in the depth of indentation D ) The Meyer hardness ¼ 4P ¼ C 4  d n0À2 ð3:12bÞ pd2 p D