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- Mechanical Engineering 382
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- AshbyJonesV1 Solutions.pdf

pete w.

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SOLUTIONS MANUAL Engineering Materials I An Introduction to Properties, Applications and Design, 3rd edn Solutions to Examples 2.1. (a) For commodity A Plparenoritrparenori=C A exp r A 100 tcommaori and for commodity B Qlparenoritrparenori=C B exp r B 100 tcommaori where C A and C B are the current rates of consumption lparenorit =t 0 rparenori and Plparenoritrparenori and Qlparenoritrparenori are the values at t =t. Equating and solving for t gives t = 100 r B −r A ln parenleftbigg C A C B parenrightbigg periodori (b) The doubling time, t D , is calculated by setting Clparenorit =trparenori= 2C 0 , giving t D = 100 r ln2 ≈ 70 r periodori Substitution of the values given for r in the table into this equation gives the doubling times as 35, 23 and 18 years respectively. (c) Using the equation of Answer (a) we find that aluminium overtakes steel in 201 years; polymers overtake steel in 55 years. 2.2. Principal conservation measures (see Section 2.7): Substitution Examples: aluminium for copper as a conductor; reinforced concrete for wood, stone or cast-iron in construction; plastics for glass or metals as containers. For many applications, substitutes are easily found at small penalty of cost. But in certain specific uses, most elements are not easily replaced. Examples: tungsten in cutting tools and lighting (a fluorescent tube contains more tungsten, as a starter filament, than an incandescent bulb!); lead in lead-acid batteries; platinum as a catalyst in chemical processing; etc. A long development time (up to 25 years) may be needed to find a replacement. 1 2 Solutions Manual: Engineering Materials I Recycling The fraction of material recycled is obviously important. Products may be re-designed to make recycling easier, and new recycling processes developed, but development time is again important. More Economic Design Design to use proportionally smaller amounts of scarce materials, for example, by building large plant (economy of scale); using high-strength materials; use of surface coatings to prevent metal loss by corrosion (e.g. in motor cars). 2.3. (a) If the current rate of consumption in tonnes per year is C then exponential growth means that dC dt = r 100 Ccommaori where r is the fractional rate of growth in % per year. Integrating gives C =C 0 exp braceleftbigg rlparenorit−t 0 rparenori 100 bracerightbigg commaori where C 0 was the consumption rate at time t =t 0 . (b) Set Q 2 = integraldisplay t 1/2 0 Cdtcommaori where C =C 0 exp braceleftBig rt 100 bracerightBig periodori Then Q 2 bracketleftbigg C 0 100 r exp braceleftBig rt 100 bracerightBig bracketrightbigg t 1/2 0 commaori which gives the desired result. 2.4. See Chapter 2 for discussion with examples. 3.1. (a) Poisson’s ratio, SLnu, can be defined as the negative of the ratio of the lateral strain to the tensile strain in a tensile test. SLnu =− lateral strain tensile strain periodori (Note that the lateral strain, here, is a negative quantity so that SLnu is positive.) The dilatation, UPDelta, is the change of volume over the whole volume. Solutions Manual: Engineering Materials I 3 (b) Volume change per unit volume (small strains) is SLepsilon 1 +SLepsilon 2 +SLepsilon 3 . But in a uniaxial extension SLepsilon 2 =−SLnuSLepsilon 1 and SLepsilon 3 =−SLnuSLepsilon 1 . Hence UPDelta= SLdeltaV V =lparenori1−2SLnurparenoriSLepsiloncommaori where SLepsilonlparenori≡SLepsilon 1 rparenori is the tensile strain. Clearly UPDelta is zero if SLnu = 0periodori5. (c) The dilatations are: Most metals ≈ 0periodori4SLepsilon Cork = SLepsilon Rubber = 0 3.2. The solid rubber sole is very resistant to being compressed, because it is restrained against lateral Poisson’s ratio expansion by being glued to the rela- tively stiff sole. However, the moulded surface has a much lower resistance to being compressed, because the lateral Poisson’s ratio expansion of each separate rubber cube can occur without constraint (provided the gaps between adjacent cubes do not close-up completely). So your colleague is correct. 3.3. The axial force applied to the cork to push it into the bottle results in a zero lateral Poisson’s ratio expansion, so it does not become any harder to push the cork into the neck of the bottle. However, the axial force applied to the rubber bung results in a large lateral Poisson’s ratio expansion, which can make it almost impossible to force the bung into the neck of the bottle. 4.1. Refer to Fig. 4.11. Force F between atoms = dU dr periodori 4 Solutions Manual: Engineering Materials I At the equilibrium distance, r o , the energy U is a minimum (i.e. F is zero, and U is the “dissociation energy” U o ). dU dr = mA r m+1 − nB r n+1 = 0commaori or B = m n r n−m o Aperiodori U o =− A r m o + 1 r n o · m n r n−m o A =− A r m o parenleftBig 1− m n parenrightBig periodori Now, for r o = 0periodori3nmcommaoriU o =−4eV. A= 4· 5 4 lparenori0periodori3rparenori 2 = 0periodori45eV nm 2 = 7periodori2×10 −20 Jnm 2 periodori B = 1 5 lparenori0periodori3rparenori 8 ×0periodori45 = 0periodori59×10 −5 eV nm 10 = 9periodori4×10 −25 Jnm 10 . Max force is at d 2 U dr 2 = 0. i.e. at value of r given by − mlparenorim+1rparenoriA r m+2 + nlparenorin+1rparenoriB r n+2 = 0 which is r = braceleftbigg B A n m lparenorin+1rparenori lparenorim+1rparenori bracerightbigg 1 n−m = braceleftbigg n+1 m+1 bracerightbigg 1 n−m r o = parenleftbigg 11 3 parenrightbigg1 8 ×0periodori3= 0periodori352nmperiodori and Force = dU dr = mA r m+1 braceleftbigg 1− r n−m o r n−m bracerightbigg = 2×0periodori45 lparenori0periodori352rparenori 3 braceleftbigg 1− lparenori0periodori3rparenori 8 lparenori0periodori352rparenori 8 bracerightbigg = 14periodori9eV nm −1 = 14periodori9×1periodori602×10 −19 10 −9 Jm −1 = 2periodori39×10 −9 Nperiodori 4.2. The term −A/r m is an attractive potential which depends on the type of bonding. The B/r n term is a repulsive potential due to charge-cloud overlap and diminished screening of the nuclei (see Section 4.2). 4.3. The values of ˜ A are shown below. The mean is 88. The calculated values of the moduli are Solutions Manual: Engineering Materials I 5 Material Calculated from 88 kT M /UPOmega Measured Ice 1periodori0×10 10 Nm −2 7periodori7×10 9 Nm −2 Diamond 9periodori0×10 11 Nm −2 1periodori0×10 12 Nm −2 The calculated values, for these extremes of elastic behaviour, are close to the measured values. The important point is that the moduli are roughly propor- tional to absolute melting temperatures. ˜ A values: Ni, 98; Cu, 78; Ag, 76; Al, 89; Pb, 51; Fe, 96; V, 61; Cr, 116; Nb, 48; Mo, 138; Ta, 72; W, 127. 5.1. (a) Let the spheres have a diameter of 1. Then (referring to Fig. 5.3) the unit cell has an edge length √ 2, and a volume 2 √ 2. It contains 4 atoms, with a total volume 4SLpi/6. Hence the density, SLrho, is given by SLrho= 4SLpi 6periodori2 √ 2 = 0periodori740periodori (b) Glassy nickel is less dense than crystalline nickel by the factor 0.636/0.740. The density is therefore 8periodori90lparenori0periodori636/0periodori740rparenori= 7periodori65Mg m −3 . 5.2. (a) 6 Solutions Manual: Engineering Materials I (b) (c) (d) 5.3. (a) If the atom diameter is d, then the lattice constant for the f.c.c. structure is a 1 = 2d 1 √ 2 = 0periodori3524nmperiodori Solutions Manual: Engineering Materials I 7 (b) The weight of one atom is Atomic wtperiodori Avogadro’s number = 58periodori71 6periodori022×10 26 = 9periodori752×10 −26 kgperiodori There are 4 atoms per unit cell, so density = 4lparenori9periodori752×10 −26 rparenori lparenori0periodori3524×10 −9 rparenori 3 = 8periodori91Mg m −3 periodori (c) If the atom diameter is d 2 , then the lattice constant for the b.c.c. structure is a 2 = 2d 2 √ 3 = 0periodori2866nmperiodori (d) One atom weighs 55periodori85 6periodori022×10 26 = 9periodori274×10 −26 kgperiodori There are 2 atoms per unit cell. So density = 2lparenori9periodori274×10 −26 rparenori lparenori0periodori2866×10 −9 rparenori 3 = 7periodori88Mg m −3 periodori 8 Solutions Manual: Engineering Materials I 5.4. (a) Copper Have 4 atoms per unit cell lparenori8× 1 8 from cube corners = 1+6× 1 2 from cube faces = 3rparenori Atoms touch along cube-face diagonal: This gives atom radius r = √ 2 4 a. Required percentage = 4× 4 3 SLpir 3 ×100 a 3 = 16SLpi 3a 3 · √ 22a 3 ×100 16×4 = SLpi √ 2 6 ×100 = 74%periodori Answer is same for magnesium — both are close-packed structures. (b) Copper Density = 4×m a 3 where m is atomic mass. Atomic weight for Cu = 63periodori54. m= 63periodori54kg 6periodori023×10 26 = 10periodori55×10 −26 kgperiodori 8periodori96Mg m −3 = 4×10periodori55×10 −26 a 3 kgperiodori a 3 = 4periodori71×10 −29 m 3 periodori a = 3periodori61×10 −10 mcommaori or 0periodori361nmperiodori Solutions Manual: Engineering Materials I 9 Magnesium Have 6 atoms per unit cell lparenori12 × 1 6 from corners = 2 + 2 × 1 2 from end faces = 1+3 insiderparenori Volume of unit cell = 3 braceleftbigg a 2 √ 3 2 c bracerightbigg periodori Density = 2×6m 3 √ 3a 2 c = 4m √ 3a 2 c periodori m = 24periodori31kg 6periodori023×10 25 = 4periodori04×10 −26 kgperiodori 1periodori74 Mg m −3 = 4×4periodori04×10 −26 kg √ 3a 2 c periodori a 2 c = 5periodori36×10 −29 m 3 periodori In face-centred structure, plane spacing = √ 3a 3 commaori where r = √ 2a 4 periodori Spacing = √ 3 3 4r √ 2 periodori 10 Solutions Manual: Engineering Materials I In close-packed-hexagonal structure, plane spacing = c 2 = √ 3 3 4r √ 2 = √ 3 3 4 √ 2 a 2 periodori c a = 1periodori633commaori c = 1periodori633aperiodori Using value for a 2 c of 5periodori36×10 −29 m 3 we obtain: a = 3periodori20×10 −10 mor0periodori320nmsemicolonori c = 5periodori23×10 −10 mor0periodori523nmperiodori 5.5. See Section 5.10. 5.6. Let V be the volume fraction of the polyethylene which is crystalline. Then, (a) 1periodori014V +0periodori84lparenori1−Vrparenori= 0periodori92, giving V = 0periodori46, or 46%. (b) 1periodori014V +0periodori84lparenori1−Vrparenori= 0periodori97, giving V = 0periodori75, or 75%. 6.1. The two sets of values for the moduli are calculated from the formulae E composite =V f E f +lparenori1−V f rparenoriE m (upper values); E composite = 1 V f E f + 1−V f E m (lower values); where V f is the volume fraction of glass, E m the modulus of epoxy, and E f that of glass. Values are given in the table and plotted in the figure. The data lie near the lower level. This is because the approximation from which the lower values are derived (that the stress is equal in glass and epoxy) is nearer reality than the approximation from which the upper values are derived (that the strains are equal in the two components). Note that the sets of values are widely separated near V f = 0periodori5. Fibreglass tested parallel to the fibres, or wood tested parallel to the grain, lie near the maximum composite modulus. Both materials, tested at right angles to the fibres or grain lie near the lower modulus. They are, therefore, very anisotropic: the ratio of the two moduli can be as much as a factor of 4.5 for fibreglass (see the figure, at V f = 0periodori5); it can be more for woods. Solutions Manual: Engineering Materials I 11 E composite E composite Volume fraction E composite (upper values) (lower values) of glass, V f lparenoriGN m −2 rparenori lparenoriGN m −2 rparenori lparenoriGN m −2 rparenori 0 5.0 5.0 5.0 0.05 5.5 8.8 5.2 0.10 6.4 12.5 5.5 0.15 7.8 16.3 5.8 0.20 9.5 20.0 6.2 0.25 11.5 23.8 6.5 0.30 14.0 27.5 7.0 6.2. E c =E f V f +lparenori1−V f rparenoriE m . SLrho c =SLrho f V f +lparenori1−V f rparenoriSLrho m . (a) SLrho c = 0periodori5×1periodori90Mg m −3 +0periodori5×1periodori15Mg m −3 = 1periodori53Mg m −3 . (b) SLrho c = 0periodori5×2periodori55Mg m −3 +0periodori5×1periodori15Mg m −3 = 1periodori85Mg m −3 . (c) SLrho c = 0periodori02×7periodori90Mg m −3 +0periodori98×2periodori40Mg m −3 = 2periodori51Mg m −3 . (a) E c = 0periodori5×390GN m −2 +0periodori5×3GN m −2 = 197GN m −2 . (b) E c = 0periodori5×72GN m −2 +0periodori5×3GN m −2 = 37periodori5GN m −2 . (c) E c = 0periodori02×200GN m −2 +0periodori98×45GN m −2 = 48periodori1GN m −2 . 12 Solutions Manual: Engineering Materials I 6.3. E ⊥ = braceleftbigg V 1 E 1 + lparenori1−V 1 rparenori E 2 bracerightbigg −1 semicolonori see Section 6.4periodori E dblbarSC = V 1 E 1 +lparenori1−V 1 rparenoriE 2 semicolonori see Section 6.4periodori E dblbarSC E ⊥ = lbraceoriV 1 E 1 +lparenori1−V 1 rparenoriE 2 rbraceori braceleftbigg V 1 E 1 + lparenori1−V 1 rparenori E 2 bracerightbigg = 1−2V 1 +2V 2 1 +lparenori1−V 1 rparenoriV 1 braceleftbigg E 1 E 2 + E 2 E 1 bracerightbigg periodori d dV 1 braceleftbigg E dblbarSC E ⊥ bracerightbigg = lparenori1−2V 1 rparenori braceleftbigg E 1 E 2 + E 2 E 1 −2 bracerightbigg periodori For E 1 negationslash= E 2 commaori d dV braceleftbigg E dblbarSC E ⊥ bracerightbigg = 0 when V 1 = 0periodori5 (the only turning point). 6.4. Refer to Chapters 4, 5, 6 and the appropriate References. 7.1. Following eqn (7.6), mass of beam M = lscriptwdSLrho c . Substituting d using the equation given in the example gives M = parenleftbigg SLrho c E 1/3 c parenrightbiggparenleftbigg Flscript 6 w 2 4SLdelta parenrightbigg 1/3 =K parenleftbig SLrho c /E 1/3 c parenrightbig where K is a constant. Values of SLrho c /E 1/3 c taken from the answers to Example 6.2 are as follows. (a) Carbon fibre-epoxy resin = 0periodori26. (b) Glass fibre-polyester resin = 0periodori55. (c) Steel-concrete = 0periodori69. The lightest beam is (a) carbon fibre-epoxy resin. 7.2. (a) Define the bending stiffness of the tube as F/SLdelta. Then F SLdelta = 3SLpiEr 3 t l 3 periodori The mass, M, is given by M = 2SLpirtlSLrhoperiodori Substituting for t in (1) gives M = 2l 3r 2 parenleftbigg F SLdelta parenrightbigg · parenleftBig SLrho E parenrightBig periodori Solutions Manual: Engineering Materials I 13 The lightest bicycle (for a given stiffness) is that made of the material for which lparenoriSLrho/Erparenori is least. The table shows data for six possible (and quite sensible) materials. Material E lparenoriGN m −2 ) SLrho lparenoriMg m −3 ) ˜ plparenori$ tonne −1 rparenori lparenori SLrho E rparenori ˜ plparenori SLrho E rparenori Mildsteel 196 7.8 100 39periodori8×10 −3 3.98 Hardwood 15 0.8 250 53periodori3×10 −3 13.33 Aluminium alloy 69 2.7 400 39periodori1×10 −3 15.64 GFRP 40 1.8 1000 45periodori0×10 −3 45.0 Titanium alloy 120 4.5 10,000 37periodori5×10 −3 375 CFRP 200 1.5 20,000 7periodori5×10 −3 150 There is not much to choose between steel, hardwood, aluminium alloy, GFRP and titanium alloys. But CFRP is much stiffer, for a given weight, and would permit an immense weight saving — by a factor of at least 5.0 — over other materials. (b) The frame of minimum material cost is that for which the relative price M˜p (where ˜p is relative the price per tonne) is a minimum — i.e. that for which ˜pSLrho/E is least. The table shows that steel is by far the most attractive material: a CFRP frame will cost 38 times more. This is nothing if it permits you to win the Tour de France — so bicycles can be made of CFRP. 7.3. p b = 0periodori3E parenleftBig t r parenrightBig 2 periodori M b = 4SLpir 2 tSLrhoperiodori parenleftBig t r parenrightBig 2 = p b 0periodori3E periodori t r = parenleftbigg p b 0periodori3E parenrightbigg 1/2 periodori t = parenleftbigg p b 0periodori3E parenrightbigg 1/2 rperiodori M b = 4SLpir 3 SLrho parenleftbigg p b 0periodori3E parenrightbigg 1/2 = 22periodori9r 3 p b 1/2 parenleftBig SLrho E 1/2 parenrightBig periodori Merit index is parenleftbigg E 1/2 SLrho parenrightbigg periodori 14 Solutions Manual: Engineering Materials I 8.1. Nominal stress=Hardness/3 lparenori1+SLepsilon n rparenori where SLepsilon n is the nominal strain as given in the question plus 0.08. Data for nominal stress–nominal strain curve Nominal stress lparenoriMN m −2 rparenori 129 171 197 210 216 217 214 209 188 Nominal strain 0.09 0.18 0.28 0.38 0.48 0.58 0.68 0.78 1.08 8.2. (a) From graph, tensile strength is 217 MN m −2 (the stress maximum of the curve). (b) The strain at the stress maximum is 0.6 approximately. (c) From eqn. (8.4), A o l o =Al, and l o l = A A o . lparenoril −l o rparenori/l o = 0periodori6semicolonori A/A o = 1/1periodori6periodori 1− A A o = 0periodori38semicolonorilparenoriA o −Arparenori/A o = 0periodori38periodori Thus percentage reduction in area = 38%. (d) 109 MJ from graph. Solutions Manual: Engineering Materials I 15 8.3. During a tensile test, unstable necks develop when the maximum in the nominal stress–nominal strain curve is reached. Neck growth then leads rapidly to failure. Physically, necks become unstable when the material in the elongating neck work hardens insufficiently to make up for the decrease in load-bearing capacity at the neck. In rolling, the material is deforming mainly in compression, and the load-bearing area is always on the increase. Tensile instabilities cannot therefore form, and failure occurs at the much larger strains required to cause failure by cracking. 8.4. (a) Tensile strength = 101periodori5kN/160 mm 2 = 634 MN m −2 . Working stress = 160 MN m −2 . 16 Solutions Manual: Engineering Materials I (b) Load corresponding to 0.1% Proof strain = 35 kN. ∴ 0.1% Proof stress = 219 MN m −2 . Working stress = 131 MN m −2 . 8.5. The indentation hardness is defined by H = F SLpia 2 commaori where a is the radius of the circle of indentation. Simple geometry (see figure) gives lparenorir−hrparenori 2 +a 2 =r 2 commaori or 2rh−h 2 =a 2 commaori or h = a 2 2r if h lessmuchrperiodori Thus H = F 2SLpirh . The indenter penetrates a distance h which is proportional to the load. Solutions Manual: Engineering Materials I 17 8.6. (a) Conservation of energy gives mgh =U el ,som=U el /gh. Thus m max = lparenori1000N×15m+0periodori5×1500N×15mrparenori 9periodori81×30 = 89periodori2kgperiodori (b) Taking the maximum extension, in one cycle of loading/unloading, 500N×15m = 7500J is dissipated by hysteresis in the rope, compared to a U el of 26250J. Thus 29% of the energy input is lost. This results in the jumper rebounding to a position which is well below the bridge deck (this is obviously essential for a safe jump). 9.1. See Section 9.2. 9.2. The fractional volume change is UPDeltaV V =lparenori8periodori9323−8periodori9321rparenori/8periodori9323 = 2periodori24×10 −5 periodori (a) Define the dislocation density as SLrholparenorim/m 3 or m −2 rparenori. Then UPDeltaV V = 1 4 b 2 SLrhocommaori from which SLrho= 1periodori4×10 15 m −2 periodori (b) The energy is: U = 1 2 Gb 2 SLrho= 2 UPDeltaV V · 3 8 E = 2periodori1MJ m −3 periodori 9.3. See Section 9.3. 10.1. (a) See Section 10.5. (b) See Section 10.3. (c) See Section 10.4. 10.2. (a) Balance line tension against force on dislocation (Section 10.4 and Fig. 10.2(c)): SLtau y bL = 2Tperiodori So SLtau y = 2T bL ≈ Gb L periodori 18 Solutions Manual: Engineering Materials I 10.3. (a) For the alloy: L = 5×10 −8 mcommaori b = 2periodori86×10 −10 mcommaori G = 26×10 9 Nm −2 periodori So SLtau y = Gb L = 1periodori5×10 8 Nm −2 periodori But SLsigma y = 3SLtau y for polycrystals. Hence SLsigma y ≈ 450MN m −2 periodori (b) New L= 15×10 −8 m. Repeating the calculationcommaoriSLsigma y = 149MN m −2 periodori Drop in yield strength UPDeltaSLsigma y ≈ 300MN m −2 periodori 10.4. d = 4 × 10 −6 mcommaorid −1/2 = 10 3 /2 = 500m −1/2 periodorilparenori120 − 20rparenori MN m −2 = SLbeta500m −1/2 periodoriSLbeta= 100MN m −2 /500m −1/2 = 0periodori2MN m −3/2 . 11.1. Suppose first that the shaft yields. The stress in the shaft is F/SLpi·10 2 . If this is equal to the tensile yield stress SLsigma y , we have F = 100SLpiSLsigma y = 200SLpikcommaori since SLsigma y = 2k where k is the shear-yield strength. Now consider shearing-off of the head, as shown. At yield F = 2SLpirtk = 180SLpik Thus the head will shear off. 11.2. (a) Lubricated Anvils Work balance gives, from upper-bound theorem, Fu ≤ 8× wL 4 √ 2×k× u 2 √ 2 = 2wLkuperiodori F ≤ 2wLkperiodori Solutions Manual: Engineering Materials I 19 (b) Welded Interfaces Fu ≤ 2wLku+4× wL 2 ×k× u 2 = 2wLku+wLku= 3wLkuperiodori F ≤ 3wLkperiodori General formula gives, for w d = 2, F wL ≤ 2k parenleftbigg 1+ 1 2 parenrightbigg = 3kperiodori F ≤ 3wLkcommaori verification demonstratedperiodori 11.3. 4000MN m −2 = 350MN m −2 periodori parenleftBig 1+ w 4d parenrightBig . w 4d = 10periodori43periodori w d = 41periodori7periodori Take w d = 42 to produce an integral value of w 2d in a safe direction. d = w 42 = 20UPmum 42 = 0periodori48UPmumperiodori (This gives a volume fraction of cobalt of 7.2% — a typical value for a rock-drilling grade of WC-Co cement.) 20 Solutions Manual: Engineering Materials I 11.4. (a) See Section 8.4. (b) See Section 11.4. dSLsigma dSLepsilon =SLsigmaperiodori 11.5. SLsigma = 350SLepsilon 0periodori4 MN m −2 . At onset of necking dSLsigma dSLepsilon =SLsigma. 350×0periodori4SLepsilon 0periodori6 = 350SLepsilon 0periodori4 commaori or SLepsilon = 0periodori4periodori SLsigma = 350×lparenori0periodori4rparenori 0periodori4 MN m −2 = 242periodori6MN m −2 periodori Nominal stresscommaoriSLsigma n = SLsigma lparenori1+SLepsilon n rparenori commaori where SLepsilon n is the nominal strainperiodori SLepsilon = lnlparenori1+SLepsilon n rparenoriperiodori Tensile strengthcommaoriSLsigma TS = 242periodori6 antilnlparenori0periodori4rparenori MN m −2 = 242periodori6 1periodori492 MN m −2 = 163MN m −2 periodori Work = integraldisplay SLepsilon 0 SLsigmadSLepsilon per unit volume = integraldisplay 0periodori4 0 350SLepsilon 0periodori4 dSLepsilon MN m −2 = 350 bracketleftbigg SLepsilon 1periodori4 1periodori4 bracketrightbigg 0periodori4 0 MN m −2 = 350 lparenori0periodori4rparenori 1periodori4 1periodori4 MN m −2 = 69periodori3MJ per 1m 3 periodori Solutions Manual: Engineering Materials I 21 11.6. (a) Given SLsigma = ASLepsilon n . From eqn (11.4), dSLsigma dSLvarepsilon = SLsigma at onset of necking. Thus dSLsigma dSLvarepsilon = AnSLvarepsilon n−1 = ASLvarepsilon n commaori so SLvarepsilon = n and SLsigma = An n . From eqn (8.13), SLsigma TS = SLsigma lparenori1+SLvarepsilon n rparenori = An n lparenori1+SLvarepsilon n rparenori . From eqn (8.15), SLvarepsilon= lnlparenori1+SLvarepsilon n rparenori=n. Thus lparenori1+SLvarepsilon n rparenori=e n . Finally, SLsigma TS = An n e n . (b) Inserting A = 800MN m −2 and n = 0periodori2 gives SLsigma TS = 800lparenori0periodori2rparenori 0periodori2 e 0periodori2 = 475MN m −2 . SLsigma =SLsigma TS lparenori1+SLvarepsilon n rparenori=SLsigma TS e n = 580MN m −2 periodori 11.7. We have that SLsigma y at 8% plastic strain is H/3 or 200MN m −2 . Thus A= 200 lparenoriln1periodori08rparenori 0periodori2 = 334MN m −2 periodori Using the result of Example 11.6(a), SLsigma TS = An n e n = 198MN m −2 periodori 12.1. p f = 2SLsigma f parenleftBig t r parenrightBig periodori M f = 4SLpir 2 tSLrhoperiodori parenleftBig t r parenrightBig = p f 2SLsigma f periodori t = p f r 2SLsigma f periodori M f = 4SLpir 3 p f SLrho 2SLsigma f = 2SLpir 3 p f parenleftbigg SLrho SLsigma f parenrightbigg periodori Merit index is parenleftbigg SLsigma f SLrho parenrightbigg periodori 22 Solutions Manual: Engineering Materials I 12.2. M b = 22periodori9r 3 p b 1/2 parenleftBig SLrho E 1/2 parenrightBig periodori t b = M b 4SLpir 2 SLrho periodori M f = 2SLpir 3 p f parenleftbigg SLrho SLsigma f parenrightbigg periodori t f = M f 4SLpir 2 SLrho periodori Set r = 1m and p b =p f = 200MPa. Values for SLrhocommaori E and SLsigma f are to be taken from the table of data. Material M b (tonne) t b (mm) M f (tonne) t f (mm) Mechanism Al 2 O 3 2.02 41 0.98 20 Buckling Glass 3.18 97 1.63 50 Buckling Alloy steel 5.51 56 4.90 50 Buckling Ti alloy 4.39 74 4.92 83 Yielding Al alloy 3.30 97 6.79 200 Yielding Optimum material is Al 2 O 3 with a mass of 2.02 tonne. The wall thickness is 41mm and the limiting failure mechanism is external-pressure buckling. 12.3. When the bolts yield, the connection can be approximated as a mechanism which hinges at X. The cross-sectional area of one bolt is SLpilparenori1periodori25/2rparenori 2 in 2 = 1periodori23in 2 . The yield load of one bolt is 1periodori23in 2 ×11tsi=13periodori5tons. The moment at yield is given by M ≈ lparenori2×13periodori5×25rparenori+lparenori2×13periodori5×19rparenori+lparenori2×13periodori5×9rparenori+lparenori2×13periodori5×3rparenori = 1512ton in = 126ton ftperiodori The hinge must react the total load from the bolts, which is 108 tons. This means that in practice the hinge will extend over a finite area of contact. X will lie in from the outer edge of the flange by about 1 in to 2 in but the effect on the bending moment will be small. 12.4. The yield load of each link plate in tension is given approximately by the min- imum cross-sectional area multiplied by the yield strength. The total breaking load of the two links in parallel is double this figure and is given by T = 2lparenori1500N mm −2 ×1mm×4periodori5mmrparenori = 1periodori35×10 4 Nperiodori Solutions Manual: Engineering Materials I 23 From eqn (11.2), the shear yield strength k =SLsigma y /2. k for the pin is therefore 1500/2 = 750MN m −2 . The yield load of the pin in double shear is obtained by multiplying k by twice the cross-sectional area of the pin to give T = 2 braceleftBigg 750N mm −2 ×SLpi parenleftbigg 3periodori5 2 parenrightbigg 2 mm 2 bracerightBigg = 1periodori44×10 4 Nperiodori This load is 7% greater than the load needed to yield the links and the strength of the chain is therefore given by the lower figure of 1periodori35×10 4 N. To estimate the tension produced in the chain during use we take moments about the centre of the chain wheel to give 90kgf ×170mm ≈T × 190mm 2 commaori T ≈ 161kgf ≈ 1periodori58×10 3 Nperiodori The factor of safety is then given by 1periodori35×10 4 N 1periodori58×10 3 N = 8periodori5periodori Comments (a) The factor of safety is calculated assuming static loading conditions. The maximum loadings experienced in service might be twice as much due to dynamic effects. (b) The chain must also be designed against fatigue and this is probably why the factor of safety is apparently so large. 12.5. The cross-sectional area of the pin is A=SLpilparenori2 2 −1periodori2 2 rparenori mm 2 = 8periodori0mm 2 periodori The force needed to shear this area is f s =kA= 750N mm −2 ×8periodori0mm 2 = 6000Nperiodori Finally, the failure torque is UPGamma = 2lparenorif s ×10mmrparenori = 2×6000N×10mm = 1periodori2×10 5 Nmm = 120N m ≈ 12kgf mperiodori 24 Solutions Manual: Engineering Materials I 13.1. The maximum tensile stress at the surface of a beam loaded in three-point bending (see eqn (12.3)) is SLsigma = 3Fl 2bt 2 at the mid-span of the beam. Fracture occurs when SLsigma √ SLpia =K c commaori i.e. when 3Fl 2bt 2 √ SLpia =K c periodori Hence, the maximum load which can be sustained by the adhesive joint is F = 2bt 2 K c 3l √ SLpia periodori For the joint shown F = 2·lparenori0periodori1rparenori 3 ·0periodori5 3periodori2 √ SLpi·0periodori001 = 2periodori97kNperiodori 13.2. Calculate the stress for failure by (a) general yield and (b) fast fracture. (a) SLsigma = 500MN m −2 for general yield. (b) SLsigma = K c √ SLpia = 40 √ SLpiperiodori0periodori005 = 319MN m −2 commaori for fast fracture, assuming that a crack on the limit of detection is present. The plate will fail by fast fracture before it fails by general yield. 13.3. SLsigma = pr t = 0periodori06× 7000 2 × 1 3 = 70MN m −2 periodoriK c =YSLsigma √ SLpiacommaori Y = 1. a = 1 SLpi parenleftbigg K c a parenrightbigg 2 = 1 SLpi parenleftbigg 100 70 parenrightbigg 2 = 0periodori65mperiodori 14.1. K c = YSLsigma √ SLpiacommaori Y = 1commaori K c = SLsigma √ SLpiacommaori a = 1 SLpi parenleftbigg K c SLsigma parenrightbigg 2 = 1 SLpi parenleftbigg 30MPa √ m 60MPa parenrightbigg 2 = 0periodori080m = 80mmperiodori 2a = 160mmperiodori Solutions Manual: Engineering Materials I 25 14.2. lparenori3 × 1periodori0 + 4rparenori = 7periodori0MN m −2 periodoriK= SLsigma √ SLpia = 7periodori0 √ SLpi×0periodori010 = 1periodori24MN m −3/2 . This is only 5% less than the value of K c obtained from tests. Experimental scatter in the test data, dynamic loads and errors in the stress analysis are more than enough to account for this small discrepancy. 14.3. (a) See Sections 14.2 and 14.3. (b) See Section 14.4. 14.4. Classic features of tensile failure by microvoid coalescence. See Fig. 14.2. 14.5. See Fig. 14.3 and Section 14.3, paragraph 2. Atomically flat cleavage planes can be seen. Many fracture facets have “river markings”, produced by fracture on multiple parallel cleavage planes. 15.1. From Section 13.3 (“A note on the stress intensity, K”), K =YSLsigma √ SLpia. (a) From Fig. 15.3, a/W = 5/10 = 0periodori5commaoriY= 3periodoriK= 3×100 √ SLpi×0periodori005 = 37periodori6MN m −3/2 . (b) From Section 13.3 (“A note on the stress intensity, K”), Y = 1. K = 1×100 √ SLpi×0periodori020 = 25periodori1MN m −3/2 periodori 15.2. See Section 15.3 (“Failure analysis” and “Conclusions”). PMMA is a poor choice of material because it has a very low fracture toughness. Under a tensile hoop stress, the connector is liable to suffer catastrophic fast fracture from a small defect. 15.3. Reinforce the foam with polymer fibres. These will bridge any incipient cracks, and prevent crack propagation. Layers of fibre mesh can be incorporated into the foam as it is sprayed on. 15.4. Fix each end of the top rail directly to the brick wall, using a steel bracket bolted to both the top rail and the brick wall. 15.5. The low fracture toughness of wood along the grain allows wood to be split very easily along the grain. This permits easy splitting of logs into kindling and wood for fires, production of wood shingles for roofing material, finishing/sizing by planing, shaping by routing, turning and chiselling, and even pencil sharpening. 16.1. See Section 16.1 Cracks propagate in an unstable way in tension, but in a stable way in compression. (a) From eqn (16.1) SLsigma TS = K c √ SLpia = 3MPa m 1/2 √ SLpi×30×10 −6 m = 309MPaperiodori (b) From eqn (16.3) SLsigma C ≈ 15SLsigma TS = 4635MPaperiodori 26 Solutions Manual: Engineering Materials I 16.2. From eqn (16.2) SLsigma r = 6 M r bd 2 = 6× F 2 × l 2 × 1 bd 2 = 6× 330N 2 × 50mm 2 × 1 5 3 mm 3 = 198 N mm 2 = 198MPaperiodori 16.3. V V o = SLpilparenori11/2rparenori 2 mm 2 ×50mm SLpilparenori5/2rparenori 2 mm 2 ×25mm = 9periodori7. For the test specimens, eqn (16.7) gives 0periodori5 = exp braceleftbigg − V o V o parenleftbigg SLsigma SLsigma 0 parenrightbigg m bracerightbigg periodori For the components, eqn (16.7) gives 0periodori99 = exp braceleftbigg − V V o parenleftbigg SLsigma prime SLsigma 0 parenrightbigg m bracerightbigg periodori Thus lnlparenori0periodori5rparenori lnlparenori0periodori99rparenori = braceleftbigg − V o V o parenleftbigg SLsigma SLsigma 0 parenrightbigg m bracerightbiggbraceleftbigg − V o V parenleftBig SLsigma 0 SLsigma prime parenrightBig m bracerightbigg periodori 69periodori0 = 1 9periodori7 parenleftBig SLsigma SLsigma prime parenrightBig m periodori lparenoriSLsigma prime rparenori m = 1 669 lparenoriSLsigmarparenori m periodori SLsigma prime = 0periodori272×120MPa = 32periodori6MPaperiodori 16.4. See Answer to this Example. 16.5. Specimen measuring 100mm×10mm×10mm will have median SLsigma TS of 300 MPa/1.73. Specimen volume V = 10 4 mm 3 . Eqn (16.7) then gives 0periodori5 = exp braceleftbigg − 10 4 mm 3 V o SLsigma o 10 SLsigma TS 10 bracerightbigg periodori Taking natural logs gives −0periodori69 =− 10 4 mm 3 V o SLsigma o 10 SLsigma TS 10 periodori Solutions Manual: Engineering Materials I 27 Component volume = 1periodori25×10 3 mm 3 periodoriP f = 10 −6 gives P s = 1−10 −6 . Thus 1−10 −6 = exp braceleftbigg − 1periodori25×10 3 mm 3 V o SLsigma o 10 SLsigma 10 bracerightbigg periodori Taking logs, and assuming that lnlparenori1−xrparenori=−x for small x, gives −10 −6 =− 1periodori25×10 3 mm 3 V o SLsigma o 10 SLsigma 10 periodori Thus 0periodori69 10 −6 = 10 4 mm 3 V o SLsigma o 10 SLsigma TS 10 × V o SLsigma 10 o 1periodori25×10 3 mm 3 1 SLsigma 10 = 10 4 1periodori25×10 3 parenleftBig SLsigma TS SLsigma parenrightBig 10 periodori SLsigma 10 = 10 4 1periodori25×10 3 × 10 −6 0periodori69 SLsigma TS 10 periodori SLsigma = 3periodori21×10 −1 SLsigma TS = 3periodori21×10 −1 × 300MPa 1periodori73 = 55periodori7MPa 16.6. (a) Weight of material below section at x is SLrhog× SLpi 3 lparenoriSLalphaxrparenori 2 xperiodori Cross-sectional area =SLpilparenoriSLalphaxrparenori 2 . Stress = force/area = 1 3 SLrhogx. (b) Integrate over the volume, using disc of thickness dx with volume dV = SLpilparenoriSLalphaxrparenori 2 dx. Eqn (16.8) then gives P s lparenoriLrparenori= exp ⎧ ⎨ ⎩ integraldisplay V − parenleftbigg SLsigma SLsigma 0 parenrightbigg m dV V 0 ⎫ ⎬ ⎭ = exp ⎧ ⎨ ⎩ L integraldisplay 0 − parenleftbigg SLrhogx 3SLsigma 0 parenrightbigg m SLpilparenoriSLalphaxrparenori 2 dx V 0 ⎫ ⎬ ⎭ periodori Integrating gives P s lparenoriLrparenori= exp braceleftbigg − parenleftbigg SLrhog 3SLsigma 0 parenrightbigg m SLpiSLalpha 2 L m+3 lparenorim+3rparenoriV 0 bracerightbigg periodori The probability of survival falls with increasing SLalpha because, although the stresses are the same, the amount of material which is stressed increases with SLalpha, and hence the chances of meeting a critical flaw increase. 28 Solutions Manual: Engineering Materials I 16.7. (a) For the specimen in uniform tension, eqn (16.7) gives P st = 0periodori5 = exp braceleftbigg − SLpir 2 lscript V 0 parenleftbigg SLsigma t SLsigma 0 parenrightbigg m bracerightbigg periodori Setting P s lparenoriLrparenori=P st gives exp braceleftbigg − parenleftbigg SLrhog 3SLsigma 0 parenrightbigg m SLpiSLalpha 2 L m+3 lparenorim+3rparenoriV 0 bracerightbigg = exp braceleftbigg − SLpir 2 lscript V 0 parenleftbigg SLsigma t SLsigma 0 parenrightbigg m bracerightbigg periodori Thus L= braceleftbigg lscriptr 2 lparenorim+3rparenori SLalpha 2 parenleftbigg 3SLsigma t SLrhog parenrightbigg m bracerightbigg 1 m+3 . (b) Flaws induced during sample preparation, maintaining the correct environ- mental conditions, gripping without causing failure in the grips. 17.1. Catastrophic failure will occur when UPDeltaSLsigma √ SLpia = 54MN m −3/2 commaori or when a = 0periodori029mperiodori Now da dN = 4×10 −13 parenleftbigg MN m 2 parenrightbigg −4 m −1 lparenoriUPDeltaK rparenori 4 = 4×10 −13 parenleftbigg MN m 2 parenrightbigg −4 m −1 lparenoriUPDeltaSLsigmarparenori 4 SLpi 2 a 2 = 4periodori14×10 −3 a 2 m −1 periodori Integrating from a = 10 −4 matN = 0 to a = 2periodori9×10 −2 matN =N f gives N f = 10 3 4periodori14 bracketleftbigg 1 10 −4 − 1 2periodori9×10 −2 bracketrightbigg = 2periodori4×10 6 cycles to failureperiodori Solutions Manual: Engineering Materials I 29 17.2. UPDeltaSLsigmalparenoriN f rparenori a =C. 280lparenori10 5 rparenori a = 200lparenori10 7 rparenori a periodori 280 200 = 1periodori4 = parenleftbigg 10 7 10 5 parenrightbigg a =lparenori10 2 rparenori a periodori log 1periodori4 = a2 = 0periodori146periodori a = 0periodori073 parenleftbigg or 1 13periodori7 parenrightbigg periodori C = 280lparenori10 5 rparenori 0periodori073 MN m −2 periodori C = 649 MN m −2 periodori At 150MN m −2 commaoriN f = C 1/a UPDeltaSLsigma 1/a = parenleftbigg 649 150 parenrightbigg 13periodori7 = 5periodori2×10 8 cycles. 17.3. The total strain range is: UPDeltaSLepsilon = SLalphaUPDeltaT = 2periodori4×10 −3 periodori The plastic strain range is: UPDeltaSLepsilon pl =UPDeltaSLepsilon−UPDeltaSLepsilon el = 2periodori0×10 −3 periodori The cycles to failure are: N f = braceleftbigg 0periodori2 2×10 −3 bracerightbigg 2 = 10 4 periodori 17.4. (a) SLsigma working = pr t = 5periodori1×7periodori5 0periodori04×2 MN m −2 = 478MN m −2 . K c =SLsigma working √ SLpia at fractureperiodori a = 1 SLpi braceleftbigg 200MN m −3/2 478MN m −2 bracerightbigg 2 = 5periodori6×10 −2 mperiodori This critical depth for fast fracture is greater than the wall thickness of 40mm. The vessel will fail by leaking before the crack length becomes critical and it fails by fast fracture. (In practice we should allow for the complicated geometry of the crack, by looking up the geometry calibration factor Y in a stress intensity factor handbook. This will be particularly important as the crack approaches the outside of the wall). 30 Solutions Manual: Engineering Materials I (b) Rearranging the crack growth equation da dN = 2periodori44×10 −14 braceleftbigg MN m 2 bracerightbigg −4 m −1 lparenoriUPDeltaK rparenori 4 = 2periodori44×10 −14 braceleftbigg MN m 2 bracerightbigg −4 m −1 lparenoriUPDeltaSLsigmarparenori 4 SLpi 2 a 2 periodori We can then integrate this, from the initial condition after the proof test (assuming that a crack of length a 0 is present), to the required end point where after 3000 cycles the crack has grown out to the wall, where failure would occur by leakage. 2periodori44×10 −14 braceleftbigg MN m 2 bracerightbigg −4 m −1 lparenori478rparenori 4 braceleftbigg MN m 2 bracerightbigg 4 SLpi 2 integraldisplay N f 0 dN = integraldisplay 4periodori0×10 −2 m a 0 da a 2 1periodori257×10 −2 m −1 N f = bracketleftbigg − 1 a bracketrightbigg 4periodori0×10 −2 m a 0 a 0 = 0periodori016mperiodori This is the initial flaw size that will penetrate the wall after 3000 loading cycles. The proof stress P proof must be sufficient to cause flaws of this size to propagate by fast fracture. K = SLsigma proof √ SLpia 0 ≥K c Where SLsigma proof = P proof r t Hence P proof ≥ tK c r √ SLpia 0 = 0periodori04×200×10 6 lparenori7periodori5/2rparenori √ SLpi0periodori016 = 9periodori5MN m −2 periodori 17.5. Each time the iron was moved backwards and forwards the flex would have expe- rienced a cycle of bending where it emerged from the polymer sheath. The sheath is intended to be fairly flexible to avoid concentrating the bending in one place. Possibly the sheath was not sufficiently flexible and the flex suffered a significant bending stress at the location of failure. The number of cycles of bending is well into the range for high-cycle fatigue and fatigue is the likely cause. The scenario is that the individual strands in the live conductor broke one by one until the current became too much for the remaining strands to carry. At this stage the last strands would have acted as a fuse and melted, causing the fire. If 23 strands are ratedtocarry13A,thenasinglestrandshouldcarryabout0.57Asafely.Theiron draws 4.8 A, which is 8.4 times the safe capacity of one strand. It is therefore not surprising that, when only a few wires were left intact, the flex was no longer able to take the current without overheating. Failures of this sort have also occurred with appliances such as vacuum cleaners. However, these tend to have a smaller current rating and failure does not always result in a fire. Solutions Manual: Engineering Materials I 31 17.6. (a) LHS = 0730hrs position. RHS = 0600hrs position. (b) LHS = 1300hrs position. RHS = 1200hrs position. 17.7. The 50UPmum marker at the top of the photograph measures 33mm in length. The most distinct striations are located approximately 70mm up from the bottom of the photograph and 50mm in from the left-hand side of then photograph. 5 spacings=10mm in this region, giving a striation spacing of 2mm on the photograph. The spacing is thus lparenori2/33rparenori×50 = 3UPmum on the fracture surface itself. 18.1. For 4×10 8 cycles, N N f = 4×10 8 5periodori2×10 8 = 0periodori77. Miner’s rule gives: N N f + N 1 N 1 f = 1. ∴ N 1 N 1 f = 1−0periodori77 = 0periodori23periodori N 1 f = N 1 0periodori23 = 4×10 8 0periodori23 = 17periodori4×10 8 cyclesperiodori For this UPDeltaSLsigma = C N 10periodori073 f = 649MN m −2 lparenori17periodori4×10 8 rparenori 0periodori073 = 137MN m −2 periodori Decrease = 13MN m −2 18.2. (a) A good surface finish will increase the fatigue life by increasing the time required for fatigue-crack initiation. (b) A rivet hole will cause a local stress concentration which will increase UPDeltaSLsigma and reduce the fatigue life. (c) A mean tensile stress will decrease N f as in Goodman’s rule (see eqn (17.3)). (d) Corrosion may reduce the fatigue life by corrosion fatigue, or by creating pits in the surface from which fatigue cracks can initiate more easily (see Section 26.5). 18.3. The maximum pressure in the cylinder occurs at the point of admission. The maximum force acting on the piston is therefore given by 0periodori7Nmm −2 ×SLpilparenori45mmrparenori 2 = 4453Nperiodori The stress in the connecting rod next to the joint is 4453N/lparenori28mm×11mmrparenori= 14periodori5N mm −2 periodori Since the locomotive is double-acting the stress range is twice this value, or 29N mm −2 . The number of revolutions that the driving wheel is likely to make in 20 years is lparenori20×6000×1000mrparenori/lparenoriSLpi×0periodori235mrparenori= 1periodori6×10 8 . 32 Solutions Manual: Engineering Materials I Data for the fatigue strengths of welded joints are given in Fig. 18.4. The type of weld specified is a Class C. The design curve shows that this should be safe up to a stress range of 33N mm −2 , which is slightly more than the figure calculated above. Of course, our calculations have ignored dynamic effects due to the reciprocating masses of the piston and connecting rod and these should be investigated as well before taking the design modification any further. 18.4. Data for the fatigue strengths of welds are given in Fig. 18.4. The weld is a surface detail on the stressed plate and the weld classification is Class F2. We extrapolate the curve following the dashed line for Class F2 until we hit the stress range of 8N mm −2 . The mean-line fatigue curve gives the data for a 50% chance of cracking. For the stress range of 8N mm −2 the cycles to failure are 3×10 9 . The time to failure is lparenori3×10 9 rparenori/lparenori20×60×60×12×6×52rparenori= 11 years. The design curve gives the data for a 2.3% chance of cracking. The number of cycles to failure is 10 9 and the time to failure is therefore 4 years. 19.1. (a) Because the end of the crack was subjected to a large tensile stress every time the cyclist pushed the pedal down. There were clearly enough cycles of stress ofsufficientamplitudetomakefatiguecracksinitiateandgrowtofinalfailure. (b) There are two fatigue cracks, one on either side of the hole. They are smooth and dark in appearance, and are located in the lower half of each fracture surface. The dark appearance is caused by a compact layer of iron oxide produced by slow long-term corrosion, indicating that the cracks had been present for a long time. In addition, the smooth appearance of the crack surfaces is consistent with high-cycle fatigue. (c) The final fracture surfaces are grainy and bright in appearance, and are located in the upper half of each fracture surface. They consist of fresh, un-corroded metal, indicating that the final fracture was not exposed to a corrosive environment after failure. In addition, the grainy appearance of the crack surfaces is consistent with a single overload failure. (d) The RHS crack probably formed first, because it is larger than the LHS crack. It probably initiated at the surface of the hole, because the local stress would have been larger than the average stress over the whole cross-section. The LHS crack appears to have initiated at the 0630hrs position, since the crack appears to radiate from this position. (e) Moderate, since the fatigue cracks had spread across 50% of the cross- section on average before they reached the critical size for fast fracture. 19.2. (a) Inthereducedsectionofthepivotpin,atthelowerendofthereducedsection. (b) The horizontal force from the top of the door pulls the bottom of the pivot pin to the right. As a result, the pin is made to rotate around the point at which it touches the bottom of the frame housing. This rotation pushes the top of the pin to the left, against the left-hand wall of the hole in the frame housing. The reaction to this force at the top of the pin generates a bending stress in the reduced section of the pin. The maximum value Solutions Manual: Engineering Materials I 33 of this bending stress occurs at the lower end of the reduced section. In addition, the sharp change of section at this location introduces a large SCF eff (see Sections 18.3 and 18.4). (c) The bending stress at the failure location cycled from tension to compression every time the door swung from one extreme position to the other (e.g. from fully open inwards to fully open outwards). (d) When the fracture took place at the lower end of the reduced section, the length of pivot pin below the fracture fell down into the hole in the door housing, and came clear of the bottom of the frame housing. 19.3. Do away with the lifting eye altogether, and extend the top of the pulley block to provide a horizontal hole to take the pin of the shackle. Because the rotational degree of freedom provided by the lifting eye assembly has now been lost, it will be necessary to insert an in-line swivel-link (a standard item) between the shackle and the crane boom. 19.4. See Fig. 17.7, Fig. 17.9 and Section 17.4, paragraph 1. 20.1. Temperature lparenori compB Crparenori T lparenoriKrparenori 1/T lparenoriK −1 rparenori ˙SLepsilonlparenoris −1 rparenori ln˙SLepsilon 618 891 0.00112 1periodori0×10 −7 −16periodori12 640 913 0.00110 1periodori7×10 −7 −15periodori59 660 933 0.00107 4periodori3×10 −7 −14periodori66 683 956 0.00105 7periodori7×10 −7 −14periodori08 707 980 0.00102 2periodori0×10 −6 −13periodori12 510 783 0.00128 8periodori3×10 −10 −20periodori90 From graph 34 Solutions Manual: Engineering Materials I The hoop stress in the tube is SLsigma = pr t = 6MN m −2 ×20mm 2mm = 60MN m −2 periodori at 510 compB C, ˙SLepsilon= 8periodori3×10 −10 s −1 at SLsigma = 200MN m −2 Under 60 MN m −2 , ˙SLepsilon= 8periodori3×10 −10 braceleftbigg 60 200 bracerightbigg 5 s −1 . = 2periodori0×10 −12 s −1 . Strain in 9 years = 2periodori0×10 −12 ×60×60×24×365×9 = 5periodori7×10 −4 or 0periodori00057periodori Design safe 20.2. At 25MN m −2 and 620 compB Ccommaori ˙SLepsilon= 3periodori1×10 −12 s −1 . At 30 MN m −2 and 620 compB Ccommaori ˙SLepsilon = braceleftbigg 30 25 bracerightbigg 5 ×3periodori1×10 −12 s −1 = 7periodori71×10 −12 s −1 periodori ˙SLepsilon = ASLsigma 5 e −Q/RT periodori ln ˙SLepsilon 1 −ln ˙SLepsilon 2 =− Q R braceleftbigg 1 T 1 − 1 T 2 bracerightbigg periodori ln ˙SLepsilon 2 = Q R braceleftbigg 1 T 1 − 1 T 2 bracerightbigg +ln ˙SLepsilon 1 commaori = 160×10 3 8periodori313 braceleftbigg 1 893 − 1 923 bracerightbigg +lnlparenori7periodori71×10 −12 s −1 rparenori = 0periodori700−25periodori58 =−24periodori88periodori ˙SLepsilon 2 = 15periodori5×10 −12 s −1 at 30MN m −2 and 650 compB Cperiodori ˙SLepsilon = 3periodori1×10 −12 s −1 × 70 100 +15periodori5 ×10 −12 s −1 × 30 100 = 6periodori82×10 −12 s −1 periodori Solutions Manual: Engineering Materials I 35 20.3. From Table 20.1, the softening temperature of soda glass is in the range 700–900K. The operating temperature of window glass is rarely more than 293 K, soT/T S is at most 0.42. Ceramics only begin to creep whenT/T M >0periodori4 (see Section 20.1), and then only under a large stress (far greater than the self- weight stress). Thus the flow marks cannot possibly be due to creep. In fact, the flow marks come from the rather crude high-temperature processes used to manufacture panes of glass in the past. 20.4. See Section 20.3. 20.5. See Section 20.4 and Fig. 20.9. 20.6. A major fire would increase the temperature of the steelwork to the point at which it would creep under the applied loads, and the subsequent deformation could trigger the collapse of the building. This is why the World Trade Center towers collapsed on 9/11. 21.1. Measure the rate by the mass injected per second, M t . Then, if this rate follows an Arhennius Law, we have M t =Aexp braceleftbigg − Q RT bracerightbigg periodori Or, combining the constants M and A: 1 t =Bexp braceleftbigg − Q RT bracerightbigg periodori Then, converting temperatures to kelvin: 1 30 = Bexp braceleftbigg − Q R450 bracerightbigg s −1 commaori 1 81periodori5 = Bexp braceleftbigg − Q R430 bracerightbigg s −1 periodori Solving for Q and B, using R = 8periodori31J mol −1 K −1 , we have Q = 8periodori04×10 4 J mol −1 commaori B = 7periodori25×10 7 s −1 periodori We may now calculate the time required to inject the mass M of polymer at 227 compB C (500K): 1 t = 7periodori25×10 7 exp braceleftbigg − 8periodori04×10 4 8periodori313×500 bracerightbigg s −1 commaori giving t = 3periodori5 seconds. 36 Solutions Manual: Engineering Materials I 21.2. See Sections 21.2 and 21.4 (“Fast diffusion paths: grain boundary and dislo- cation core diffusion”). 21.3. See Section 21.4. More specifically, (a) Carbon forms an interstitial solid solution in iron at room temperature containing up to 0.007% by weight carbon. Even at this maximum concen- tration there is a large proportion of alternative interstitial sites remaining unfilled by carbon. The probability of a carbon atom being next to an alternative interstitial site is therefore high, and hence the probability of the carbon atom moving into a different position is also high, i.e. it will diffuse relatively rapidly. Chromium forms a random substitutional solid solution in iron at room temperature (that it should not form an interstitial solution is evident from its large atomic radius, comparable to that of iron). The probability of a vacancy appearing next to a chromium atom is therefore small, and the diffusion of chromium is correspondingly slow. (b) Diffusion in grain boundaries is generally more rapid than in the grains themselves because of the geometrically more open structure of grain bound- aries. A small grain size produces a larger contribution from grain boundary diffusion than does a large grain size, and thus increases the overall diffusion coefficient of the polycrystal. 21.4. See Section 21.4 (“A useful approximation”). x = √ Dt,sot =x 2 /D. D = D o e −Q/RT = 9periodori5mm 2 s −1 exp braceleftbigg −159×10 3 J mol −1 8periodori313J K −1 mol −1 1023K bracerightbigg = 9periodori5mm 2 s −1 1periodori32×10 8 = 7periodori20×10 −8 mm 2 s −1 periodori t = x 2 D = lparenori10 −2 mmrparenori 2 7periodori2×10 −8 mm 2 s −1 = 10 −4 mm 2 7periodori2 × 10 8 mm 2 s = 10 4 7periodori2 s = 0periodori1389×10 4 s = 1389scommaori or 23 minperiodori 22.1. See Fig. 22.2. 22.2. See Fig. 22.3. 22.3. See eqn (22.1). 22.4. See Fig. 22.4 and eqn (22.2). Solutions Manual: Engineering Materials I 37 22.5. The rate of bulk vacancy diffusion decreases rapidly with decreasing tempera- ture because the activation energy Q is large for bulk diffusion (see Table 21.1). At a sufficiently low temperature, short-circuit diffusion (in this case along dis- location cores, see Fig. 21.9) takes over from bulk vacancy diffusion. Because the activation energy Q is small for short-circuit diffusion, the rate of short- circuit diffusion decreases only slowly with decreasing temperature, so creep continues to occur at temperatures well below the transition from bulk to short-circuit diffusion. 22.6. For reasons analogous to those in Example 22.5. However, in the present case, the short-circuit diffusion takes place along grain boundaries (see Fig. 21.8). 22.7. See Section 22.2 (“Designing metals and ceramics to resist power-law creep” and “Designing metals and ceramics to resist diffusional flow”). 22.8. See Section 22.3. 23.1. See Section 23.1. 23.2. See Section 23.3, paragraph 4, and Figs 23.4 and 23.5. 23.3. See Section 23.6. 23.4. See Section 23.4. 24.1. Fick’s first law J =−D dc dx . SLdeltalparenoriUPDeltamrparenori=AJSLdeltat where A= constant. SLdeltalparenoriUPDeltamrparenori=−A parenleftbigg D dc dx parenrightbigg SLdeltat =AD parenleftBig c 1 −c 2 b parenrightBig SLdeltat. 38 Solutions Manual: Engineering Materials I bSLdeltalparenoriUPDeltamrparenori=ADlparenoric 1 −c 2 rparenoriSLdeltat. b =BUPDeltam where B = constant. BUPDeltamSLdeltalparenoriUPDeltamrparenori=ADlparenoric 1 −c 2 rparenoriSLdeltat. B 2 lparenoriUPDeltamrparenori 2 =ADlparenoric 1 −c 2 rparenorit, i.e., lparenoriUPDeltamrparenori 2 =CDt, where C = constant. At constant temperature lparenoriUPDeltamrparenori 2 =k P t. 24.2. Ohm’s Law V =IR. SLdeltalparenoriUPDeltamrparenori = PISLdeltat where P = constantperiodori SLdeltalparenoriUPDeltamrparenori = PISLdeltat = PVSLdeltat R periodori RSLdeltalparenoriUPDeltamrparenori = PVSLdeltatperiodori R = Qb where Q = constant = SUPDeltam where S = constantperiodori SUPDeltamSLdeltalparenoriUPDeltamrparenori = PVSLdeltatperiodori S 2 lparenoriUPDeltamrparenori 2 = PVtcommaori iperiodorieperiodoricommaori lparenoriUPDeltamrparenori 2 =k P tperiodori 24.3. AT 500 compB Ccommaorik P = 37 exp braceleftbigg − 138×10 3 J mol −1 8periodori313J K −1 mol −1 773K bracerightbigg kg 2 m −4 s −1 = 1periodori74×10 −8 kg 2 m −4 s −1 periodori lparenoriUPDeltamrparenori 2 = k P t = 1periodori74×10 −8 kg 2 m −4 s −1 ×3600×24×365s = 0periodori50kg 2 m −4 periodori UPDeltam = 0periodori74kg m −2 periodori i.e., each square metre of metal surface absorbs 0.74kg of oxygen from the atmosphere in the form of FeO. Number of oxygen atoms absorbed = 0periodori74kg m −2 16/N A , where N A is Avogadro’s number. ∴ Number of iron atoms removed from metal as FeO = 0periodori74kg m −2 16/N A . Solutions Manual: Engineering Materials I 39 Weight of iron removed from metal = 0periodori74kg m −2 16/N A × 55periodori9/N A = 2periodori59kg m −2 . Thickness of metal lost = 2periodori59kg m −2 m 3 7870kg = 0periodori33mmperiodori at 600 compB Ccommaorik P = 2periodori04×10 −7 kg 2 m −4 s −1 periodori lparenoriUPDeltamrparenori 2 = 2periodori04×10 −7 ×3600×24×365kg 2 m −4 commaori = 6periodori43kg 2 m −4 periodori UPDeltam = 2periodori54kg m −2 commaori giving loss = 1periodori13mmperiodori 24.4. As shown in Table 24.1, gold is the only metal which requires energy to make it react with oxygen lparenori80kJ mol −1 of O 2 in fact). It therefore remains as un-reacted metal. 24.5. If there is any contact resistance across a pair of silver contacts, the surfaces of the contacts will be heated up by the current passing through the contact resistance. If the temperature goes above 230 compB C, any oxide will decompose to leave pure metal-to-metal contact. Gold contacts would not form an oxide film at any temperature, but silver is used because it is much cheaper than gold. 24.6. See Section 24.3, paragraphs 1 to 3. 24.7. See Section 24.3, paragraph 4, and Fig. 24.2. 25.1. See Section 25.3, next-to-last paragraph, and Fig. 25.2. 25.2. See Section 25.2, paragraphs 1 and 2. The protective oxide film is Cr 2 O 3 , produced by the chromium content of the stainless steel. 25.3. See Sections 25.2 and 25.3. Examples are Cr in stainless steel (Cr 2 O 3 film), Cr in nickel alloys (Cr 2 O 3 film), Al in aluminium bronzes (Al 2 O 3 film). 25.4. See Table 24.2. The refractory metals oxidize very rapidly in air at high temperature. Lamp filaments are surrounded by a glass bulb, which is either evacuated or filled with an inert gas to remove any oxygen which would attack the filament. 25.5. The oxide layer prevents the molten solder or brazing alloy from wetting the surfaces to be joined. When the copper connection tabs are soldered to pre- tinned copper wire, the pre-tinning solder melts and this protects the copper surfaces from oxidation. 40 Solutions Manual: Engineering Materials I 25.6. The chromium and nickel form a protective oxide layer on the wire, which resists oxidation of the alloy at the high running temperature. Mild steel would oxidize far too rapidly at the running temperature, and would burn out in a matter of days (see Table 24.2). 25.7. Many are oxides already, e.g. MgO, SiO 2 commaori Al 2 O 3 (see Table 24.1). Others, e.g. SiC and Si 3 N 4 form protective oxide layers of SiO 2 when exposed to oxygen at high temperature. 26.1. da dt ∝a at constant SLsigmalparenori4MN m −2 rparenori. da dt ∝ SLsigma 2 at constant alparenori0periodori25mmrparenoriperiodori ∴ da dt = ASLpiSLsigma 2 a =AK 2 semicolonorin= 2periodori SLpiA = parenleftbigg da dt parenrightbigg 1 SLsigma 2 a = 0periodori3mm year −1 16lparenoriMN m −2 rparenori 2 0periodori25mm = 0periodori075 lparenoriMN m −2 rparenori 2 year semicolonoriA= 0periodori0239m 4 MN −2 year −1 periodori 26.2 Since water and air were in contact with the surface of the pipe, the cathodic oxygen-reduction reaction would have taken place easily. The temperature of the pipe would have varied from approximately 20 compB C (summer time, heating off) to 70 compB C (winter time, heating on). As shown in Fig. 26.7, the corrosion rate at 70 compB C will be approximately twice that at 20 compB C. Thus putting the heating on will double the rate of external corrosion. The pipe did not rust from the inside because there is little or no oxygen inside the heating water circuit. 26.3. Pitting attack. See Section 26.5 (“Pitting”) and Fig. 26.12. 26.4. Stress corrosion cracking can lead to complete fracture even though the sur- face of the component appears free from corrosion. See Section 26.5 (“Stress corrosion cracking”) and Fig. 26.9. Even if the stress corrosion cracks do not travel right across the component, they can still propagate to failure by fatigue or fast fracture. 26.5. Typical applications are as follows. (a) Plastic water pipes in plumbing and drainage systems. (b) Plastic gas pipes for underground use. (c) Rubber hoses in automobile cooling systems, flexible brake hoses, wind- screen washer hoses. (d) Plastic containers for storing water, acids, alkalis, etc. (e) Plastics for exterior architectural use, e.g. window frames, roofing sheets, rainwater gutters and downpipes. Solutions Manual: Engineering Materials I 41 (f) Plastics for marine use, e.g. boat hulls (matrix of GFRP composite), mooring buoys, small fittings. (g) Plastics and rubbers in domestic appliances, e.g. water pumps, washing machine drums, flexible hoses, seals. 26.6. Stress corrosion cracking. See Section 26.5 (“Stress corrosion cracking”) and Fig. 26.9. Austenitic stainless steels are prone to SCC in hot chloride solutions. The solution in the present case was very hot and contained chloride ions from the zinc chloride corrosion inhibitor. There was also a large tensile stress to drive the initiation and growth of SCC cracks. 27.1. See Section 26.2 and Fig. 26.2. Because there was no oxygen in the system, the oxygen-reduction reaction could not take place. Therefore the anodic reaction could not take place and the steel was protected from corrosion. 27.2. See Section 27.2 (“Sacrificial protection”, paragraph 2) and Fig. 27.2. 27.3. See Section 27.2 (“Sacrificial protection”, paragraphs 1 and 2) and Fig. 27.1. 27.4. Corrosion of zinc: Zn → Zn ++ +2eperiodori Number of electrons to give a current of 6×10 −3 Am −2 for 5 years = 6×10 −3 ×5×3periodori15×10 7 1periodori6×10 −19 = 5periodori89×10 24 electrons = 5periodori89 2 ×10 24 atoms Znperiodori ∴ Mass zinc = 5periodori89×10 24 ×65periodori4 2×6periodori02×10 26 = 0periodori320kg m −2 periodori Thickness = 0periodori320 7130 m = 4periodori5×10 −5 m = 0periodori045mmperiodori 27.5. Following 44, if steel were lost uniformly over a square metre at 2×10 −3 Am −2 , thickness lost by reaction: Fe → Fe ++ +2e would be 0.0116mm on each side of the plate. If this loss is concentrated over 0.5% of the surface, the loss there will be 0periodori0116× 100 0periodori5 × 99periodori5 100 mm = 2periodori31mmperiodori The sheet will thus rust through. 42 Solutions Manual: Engineering Materials I 27.6. See Section 27.3, last paragraph, and Fig. 27.6. 27.7. See Section 27.4, last two paragraphs, and Fig. 27.7. 28.1. See Section 28.2. 28.2. If P is the radial pressure that the olive exerts on the outside of the pipe then we can write SLsigma y = Pr t provided we neglect the strengthening effect of the sections of pipe that lie outside the olive. If we assume that the end of the pipe far away from the fitting has an end cap (or a bend that functions as an end cap) then the force trying to push the pipe out of the fitting is p w SLpir 2 . This force is balanced by the frictional force between the olive and the pipe so we can write p w SLpir 2 =SLmuP2SLpirlperiodori Combining the two equations to eliminate P gives p w = 2SLmuSLsigma y parenleftBig t r parenrightBig parenleftbigg l r parenrightbigg periodori Using the data given we get p w = 2×0periodori15×120MPa parenleftbigg 0periodori65 7periodori5 parenrightbiggparenleftbigg 7periodori5 7periodori5 parenrightbigg = 3periodori1MPaperiodori The hydrostatic head of water in a seven-floor building is about 2 bar, so the joint could actually cope with pumping water to the top of a seventy-floor skyscraper and still have a factor of safety of 1.5. However, the pressure in water systems frequently exceeds the static head, often substantially, because of “water hammer”. This is the dynamic overpressure that arises when taps are suddenly shut off. 28.3. (a) Typical examples are as follows. Car tyres/road surfaces, brake pads/brake discs, clutches, shoe soles/walking surfaces, climbing shoes/rock faces, knots in ropes, V-belt drives, interference fits, compression joints (see Example 28.2). (b) Typical examples are as follows. Bearings and sliding surfaces in machin- ery, sledges and skis on snow, actuating mechanisms (e.g. car window mechanisms), door latches, ceramic discs in water taps, clock and watch mechanisms. 28.4. (a) Typical examples are as follows. Metal finishing by linishing, grinding and polishing. Wood finishing by sanding. Removal of surface scale by grit blasting. Bedding-in of brake pads, clutch linings and plain bearings. Solutions Manual: Engineering Materials I 43 (b) Typical examples are as follows. Bearings and sliding surfaces in machin- ery, car tyres, brake pads and discs, clutch linings, shoe soles, tools for metalworking and woodworking, grinding wheels and abrasive belts/discs. 29.1. See Sections 28.4 and 29.2, paragraph 1. 29.2. From the slope of the roof, the coefficient of static friction is: SLmu s = tan24 compB = 0periodori45periodori If the slope of the roof is greater than 24 compB the static frictional force is exceeded and the snow will slide off. On a 2 compB slope, with a ski already moving, it is the sliding friction which counts: SLmu k = tan2 compB = 0periodori035periodori 29.3. The work done = 2lparenori100gSLmu k rparenori= 68J. The melts a volume of water V = 69 330×10 6 = 2periodori1×10 −7 m 3 periodori If spread uniformly over the undersurface of two skis this would give a film of thickness 2periodori1×10 −7 2×2×0periodori1 m = 0periodori5UPmumperiodori 29.4. See Section 29.4. 29.5. Secondary roads are often covered in ice, because it is uneconomical to treat them with rock salt. As shown in Example 29.3, as soon as the tyre starts to slide on the ice, it will melt the surface of the ice, reducing the friction coefficient to zero. The hard studs bite into the ice, so the (non-zero) friction coefficient is determined by the resistance of the ice to ploughing (see Fig. 28.8). 29.6. See Section 29.2, paragraphs 1 and 2. 29.7. See Section 29.2, paragraphs 2, 5 and 6. Sriram Subramanaya 4711 2000 Nov 13 10:37:35 f:\pagination\elsevier uk\em2\latex-0750663812\0em1-sol.manual.dvi TeX output 2006.01.03:1617

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