Engineering Materials 1 Solutions

SOLUTIONS MANUAL Engineering Materials I An Introduction to Properties, Applications and Design, 3rd edn

Solutions to Examples 2.1.

rA t 100 r and for commodity B Qt = CB exp B t 100 where CA and CB are the current rates of consumption t = t0 and Pt and Qt are the values at t = t . Equating and solving for t gives 100 CA t= ln rB − rA CB

(a) For commodity A

Pt = CA exp

(b) The doubling time, tD , is calculated by setting Ct = t = 2C0 , giving tD =

70 100 ln 2 ≈ r r

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

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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 r dC = C dt 100 where r is the fractional rate of growth in % per year. Integrating gives rt − t0 C = C0 exp 100 where C0 was the consumption rate at time t = t0 . (b) Set Q t1/2 = C dt 2 0 where C = C0 exp Then

rt 100

rt t1/2 100 Q C0 exp 2 r 100 0

which gives the desired result. 2.4.

See Chapter 2 for discussion with examples.

3.1.

(a) Poisson’s ratio, , can be defined as the negative of the ratio of the lateral strain to the tensile strain in a tensile test. =−

lateral strain tensile strain

(Note that the lateral strain, here, is a negative quantity so that is positive.) The dilatation, , is the change of volume over the whole volume.

Solutions Manual: Engineering Materials I

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(b) Volume change per unit volume (small strains) is 1 + 2 + 3 . But in a uniaxial extension 2 = −1 and 3 = −1 . Hence =

V = 1 − 2 V

where ≡ 1 is the tensile strain. Clearly is zero if = 05. (c) The dilatations are: Most metals ≈ 04 Cork = 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 relatively 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

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Solutions Manual: Engineering Materials I At the equilibrium distance, ro , the energy U is a minimum (i.e. F is zero, and U is the “dissociation energy” Uo ). mA nB dU = m+1 − n+1 = 0 dr r r m n−m or B = ro A n A 1 m Uo = − m + n · ron−m A ro ro n A m = − m 1− ro n Now, for ro = 03 nm Uo = −4 eV. 5 A = 4 · 032 = 045 eV nm2 4 = 72 × 10−20 J nm2 B = 15 038 × 045 = 059 × 10−5 eV nm10 = 94 × 10−25 J nm10 . Max force d2 U is at = 0. i.e. at value of r given by dr 2 1 mm + 1A nn + 1B B n n + 1 n−m + − = 0 which is r = r m+2 r n+2 A m m + 1 1 n−m 18 n+1 11 = ro = × 03 = 0352 nm m+1 3 mA ron−m dU = m+1 1 − n−m and Force = dr r r 2 × 045 038 = 1 − = 149 eV nm−1 03523 03528 =

149 × 1602 × 10−19 J m−1 10−9

= 239 × 10−9 N 4.2.

4.3.

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). The values of A˜ are shown below. The mean is 88. The calculated values of the moduli are


Material

Calculated from 88 kTM /

Measured

Ice Diamond

10 × 1010 N m−2 90 × 1011 N m−2

77 × 109 N m−2 10 × 1012 N m−2

5

The calculated values, for these extremes of elastic behaviour, are close to the measured values. The important point is that the moduli are roughly proportional 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 4 /6. Hence the density, , is given by =

4

√ = 0740 62 2

(b) Glassy nickel is less dense than crystalline nickel by the factor 0.636/0.740. The density is therefore 8900636/0740 = 765 Mg m−3 . 5.2. (a)

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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 2d a1 = √ 1 = 03524 nm 2


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(b) The weight of one atom is 5871 Atomic wt = = 9752 × 10−26 kg Avogadro’s number 6022 × 1026 There are 4 atoms per unit cell, so density =

49752 × 10−26 = 891 Mg m−3 03524 × 10−9 3

(c) If the atom diameter is d2 , then the lattice constant for the b.c.c. structure is 2d a2 = √ 2 = 02866 nm 3

(d) One atom weighs 5585 = 9274 × 10−26 kg 6022 × 1026 There are 2 atoms per unit cell. So density =

29274 × 10−26 = 788 Mg m−3 02866 × 10−9 3

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Solutions Manual: Engineering Materials I 5.4.

(a) Copper

1 1 Have 4 atoms per unit cell 8 × from cube corners = 1 + 6 × from 8 2 cube faces = 3 Atoms touch along cube-face diagonal: This gives atom radius r =

√

2 a. 4

4 √ 4 × r 3 × 100 16

2 2a 3 × 100 3 Required percentage = = · 3a 3 16 × 4 a3 √

2 × 100 = 74% = 6 Answer is same for magnesium — both are close-packed structures. (b) Copper 4×m Density = where m is atomic mass. a3 Atomic weight for Cu = 6354. m=

6354 kg = 1055 × 10−26 kg 6023 × 1026 4 × 1055 × 10−26 896 Mg m−3 = kg a3 a 3 = 471 × 10−29 m3 a = 361 × 10−10 m or 0361 nm


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Magnesium

1 1 from corners = 2 + 2 × Have 6 atoms per unit cell 12 × from 6 2 end faces = 1 + 3 inside √ 3 Volume of unit cell = 3 a 2 c 2 4m 2 × 6m Density = √ 2 = √ 2 3a c 3 3a c 2431 kg = 404 × 10−26 kg m= 6023 × 1025 4 × 404 × 10−26 kg 174 Mg m−3 = √ 2 3a c a 2 c = 536 × 10−29 m3 In face-centred structure, plane spacing √ 2a 3a where r = 3 4 √ 3 4r Spacing = √ 3 2 √

=

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Solutions Manual: Engineering Materials I √ 3 4r c = In close-packed-hexagonal structure, plane spacing = √ = 2 3 2 √ 3 4 a √ 3 22

c = 1633 a c = 1633a

Using value for a 2 c of 536 × 10−29 m3 we obtain: a = 320 × 10−10 m or 0320 nm c = 523 × 10−10 m or 0523 nm

5.5.

See Section 5.10.

5.6.

Let V be the volume fraction of the polyethylene which is crystalline. Then, (a) 1014V + 0841 − V = 092, giving V = 046, or 46%. (b) 1014V + 0841 − V = 097, giving V = 075, or 75%.

6.1.

The two sets of values for the moduli are calculated from the formulae Ecomposite = Vf Ef + 1 − Vf Em (upper values); Ecomposite =

1 Vf Ef

f + 1−V E

(lower values);

m

where Vf is the volume fraction of glass, Em the modulus of epoxy, and Ef 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 Vf = 05. 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 Vf = 05); it can be more for woods.


6.2.

Volume fraction of glass, Vf

Ecomposite GN m−2

Ecomposite (upper values) GN m−2

Ecomposite (lower values) GN m−2

0 0.05 0.10 0.15 0.20 0.25 0.30

5.0 5.5 6.4 7.8 9.5 11.5 14.0

5.0 8.8 12.5 16.3 20.0 23.8 27.5

5.0 5.2 5.5 5.8 6.2 6.5 7.0

Ec = Ef Vf + 1 − Vf Em . c = f Vf + 1 − Vf m . (a) c = 05 × 190 Mg m−3 + 05 × 115 Mg m−3 = 153 Mg m−3 . (b) c = 05 × 255 Mg m−3 + 05 × 115 Mg m−3 = 185 Mg m−3 . (c) c = 002 × 790 Mg m−3 + 098 × 240 Mg m−3 = 251 Mg m−3 . (a) Ec = 05 × 390 GN m−2 + 05 × 3 GN m−2 = 197 GN m−2 . (b) Ec = 05 × 72 GN m−2 + 05 × 3 GN m−2 = 375 GN m−2 . (c) Ec = 002 × 200 GN m−2 + 098 × 45 GN m−2 = 481 GN m−2 .

11

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E⊥ =

V1 1 − V1 + E1 E2

−1 see Section 6.4

E = V1 E1 + 1 − V1 E2 see Section 6.4 E V 1 − V1 = V1 E1 + 1 − V1 E2 1 + E⊥ E1 E2 E E = 1 − 2V1 + 2V12 + 1 − V1 V1 1 + 2 E2 E1 E d E E = 1 − 2V1 1 + 2 − 2 dV1 E⊥ E2 E1 For E1 = E2 E d = 0 when V1 = 05 (the only turning point). dV E⊥ 6.4.

Refer to Chapters 4, 5, 6 and the appropriate References.

7.1.

Following eqn (7.6), mass of beam M = wdc . Substituting d using the equation given in the example gives 6 2 1/3

c F w M= = K c /Ec1/3 1/3 4 Ec where K is a constant. Values of c /Ec1/3 taken from the answers to Example 6.2 are as follows. (a) Carbon fibre-epoxy resin = 026. (b) Glass fibre-polyester resin = 055. (c) Steel-concrete = 069. The lightest beam is (a) carbon fibre-epoxy resin.

7.2.

(a) Define the bending stiffness of the tube as F/ . Then 3 Er 3 t F = l3 The mass, M , is given by M = 2 rtl Substituting for t in (1) gives 2l M= 2 3r

F · E


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The lightest bicycle (for a given stiffness) is that made of the material for which /E is least. The table shows data for six possible (and quite sensible) materials.

Material

E GN m−2 )

Mg m−3 )

p˜ $ tonne−1

E

p˜ E

Mildsteel Hardwood Aluminium alloy GFRP Titanium alloy CFRP

196 15 69 40 120 200

7.8 0.8 2.7 1.8 4.5 1.5

100 250 400 1000 10,000 20,000

398 × 10−3 533 × 10−3 391 × 10−3 450 × 10−3 375 × 10−3 75 × 10−3

3.98 13.33 15.64 45.0 375 150

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 p˜ /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.

pb = 03 E

t 2

r Mb = 4 r 2 t t 2 pb = r 03 E 1/2 pb t = r 03 E 1/2 pb t = r 03 E 1/2 pb 3 Mb = 4 r = 229r 3 pb 1/2 03 E E 1/2 1/2 E Merit index is

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Nominal stress = Hardness/3 1 + n where n is the nominal strain as given in the question plus 0.08. Data for nominal stress–nominal strain curve Nominal stress MN m−2 Nominal strain

8.2.

129 0.09

171 0.18

197 0.28

210 0.38

216 0.48

217 0.58

214 0.68

209 0.78

(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), Ao lo = Al , and llo = AA . o

l − lo /lo = 06 A/Ao = 1/16 A = 038 Ao − A/Ao = 038 1− Ao Thus percentage reduction in area = 38%. (d) 109 MJ from graph.

188 1.08


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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 Working stress

= 1015 kN/160 mm2 = 634 MN m−2 . = 160 MN m−2 .

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Solutions Manual: Engineering Materials I (b) ∴

8.5.

Load corresponding to 0.1% Proof strain 0.1% Proof stress Working stress

= 35 kN. = 219 MN m−2 . = 131 MN m−2 .

The indentation hardness is defined by H=

F

a 2

where a is the radius of the circle of indentation. Simple geometry (see figure) gives r − h2 + a 2 = r 2 or 2rh − h 2 = a 2 or h= Thus H =

a2 if h r 2r

F . 2 rh

The indenter penetrates a distance h which is proportional to the load.


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(a) Conservation of energy gives mgh = U el , so m = U el /gh. Thus mmax =

1000 N × 15 m + 05 × 1500 N × 15 m = 892 kg 981 × 30

(b) Taking the maximum extension, in one cycle of loading/unloading, 500 N × 15 m = 7500 J is dissipated by hysteresis in the rope, compared to a U el of 26250 J. 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 V = 89323 − 89321/89323 = 224 × 10−5 V (a) Define the dislocation density as m/m3 or m−2 . Then 1 V = b 2 V 4 from which = 14 × 1015 m−2 (b) The energy is: V 3 1 U = Gb 2 = 2 · E = 21 MJ m−3 2 V 8

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)): y bL = 2T So y =

Gb 2T ≈ bL L

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(a) For the alloy: L = 5 × 10−8 m b = 286 × 10−10 m G = 26 × 109 N m−2 Gb So y = = 15 × 108 N m−2 L But y = 3y for polycrystals. Hence y ≈ 450 MN m−2 (b) New L = 15 × 10−8 m. Repeating the calculation y = 149 MN m−2 Drop in yield strength y ≈ 300 MN m−2

10.4.

d = 4 × 10−6 m d −1/2 = 103 /2 = 500 m−1/2 120 − 20 MN m−2 = 500 m−1/2 = 100 MN m−2 /500 m−1/2 = 02 MN m−3/2 .

11.1.

Suppose first that the shaft yields. The stress in the shaft is F/ · 102 . If this is equal to the tensile yield stress y , we have F = 100 y = 200 k since y = 2k where k is the shear-yield strength. Now consider shearing-off of the head, as shown. At yield F = 2 rtk = 180 k

Thus the head will shear off. 11.2.

(a) Lubricated Anvils Work balance gives, from upper-bound theorem, wL √ u√ 2×k × 2 = 2wLku 4 2 F ≤ 2wLk

Fu ≤ 8 ×


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(b) Welded Interfaces wL u ×k × 2 2 = 2wLku + wLku = 3wLku

Fu ≤ 2wLku + 4 × F ≤ 3wLk

w = 2, d F 1 ≤ 2k 1 + = 3k wL 2 F ≤ 3wLk verification demonstrated

General formula gives, for

11.3.

w . 4000 MN m−2 = 350 MN m−2 1 + 4d w = 1043 4d w = 417 d Take

w w = 42 to produce an integral value of in a safe direction. d 2d d =

w 20 m = = 048 m 42 42

(This gives a volume fraction of cobalt of 7.2% — a typical value for a rock-drilling grade of WC-Co cement.)

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11.4.

(a) See Section 8.4. (b) See Section 11.4. d = d

11.5.

= 35004 MN m−2 . d = . At onset of necking d 350 × 0406 = 35004 = 04

or = 350 × 04 MN m 04

Nominal stress n =

−2

= 2426 MN m−2

where n is the nominal strain 1 + n

= ln1 + n 2426 MN m−2 antiln04 2426 = MN m−2 1492 = 163 MN m−2

Tensile strength TS =

Work = =

d per unit volume

0 04 0

35004 d MN m−2

14 = 350 14

04 MN m−2 0

0414 = 350 MN m−2 14 = 693 MJ per 1m3


11.6.

d (a) Given = An . From eqn (11.4), = at onset of necking. Thus d d = Ann−1 = An so = n and = Ann . From eqn (8.13), TS = d Ann = . 1 + n 1 + n From eqn (8.15), = ln1 + n = n. Ann Thus 1 + n = e n . Finally, TS = n . e 8000202 −2 = (b) Inserting A = 800 MN m and n = 02 gives TS = e 02 −2 475 MN m . = TS 1 + n = TS e n = 580 MN m−2

11.7.

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We have that y at 8% plastic strain is H/3 or 200 MN m−2 . Thus A=

200 = 334 MN m−2 ln 10802

Using the result of Example 11.6(a), TS =

Ann = 198 MN m−2 n e

12.1.

pf = 2f

t

r 2 Mf = 4 r t t p = f r 2f pr t = f 2f

4 r 3 pf Mf = = 2 r 3 pf 2f Merit index is

f

f

22


Mb = 229r 3 pb 1/2

E 1/2

Mb 4 r 2 Mf = 2 r 3 pf f Mf tf = 4 r 2 tb =

Set r = 1 m and pb = pf = 200 MPa. Values for E and f are to be taken from the table of data. Material

Mb (tonne)

tb (mm)

Mf (tonne)

tf (mm)

Mechanism

Al2 O3 Glass Alloy steel Ti alloy Al alloy

2.02 3.18 5.51 4.39 3.30

41 97 56 74 97

0.98 1.63 4.90 4.92 6.79

20 50 50 83 200

Buckling Buckling Buckling Yielding Yielding

Optimum material is Al2 O3 with a mass of 2.02 tonne. The wall thickness is 41 mm 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 125/22 in2 = 123 in2 . The yield load of one bolt is 123 in2 ×11 tsi = 135 tons. The moment at yield is given by M ≈ 2 × 135 × 25 + 2 × 135 × 19 + 2 × 135 × 9 + 2 × 135 × 3 = 1512 ton in = 126 ton ft 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 minimum 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 = 21500 N mm−2 × 1 mm × 45 mm = 135 × 104 N


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From eqn (11.2), the shear yield strength k = y /2. k for the pin is therefore 1500/2 = 750 MN 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 2 35 T = 2 750 N mm−2 ×

mm2 = 144 × 104 N 2 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 135 × 104 N. To estimate the tension produced in the chain during use we take moments about the centre of the chain wheel to give 190 mm 2 T ≈ 161 kgf ≈ 158 × 103 N

90 kgf × 170 mm ≈ T ×

The factor of safety is then given by 135 × 104 N = 85 158 × 103 N

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 = 22 − 122 mm2 = 80 mm2 The force needed to shear this area is fs = kA = 750 N mm−2 × 80 mm2 = 6000 N Finally, the failure torque is = 2 fs × 10 mm = 2 × 6000 N × 10 mm = 12 × 105 N mm = 120 N m ≈ 12 kgf m

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The maximum tensile stress at the surface of a beam loaded in three-point bending (see eqn (12.3)) is 3Fl = 2bt 2 at the mid-span of the beam. Fracture occurs when √ a = Kc i.e. when

3Fl √

a = Kc 2bt 2 Hence, the maximum load which can be sustained by the adhesive joint is F=

2bt 2 Kc √ 3l a

For the joint shown 2 · 013 · 05 √ 32 · 0001 = 297 kN

F =

13.2.

Calculate the stress for failure by (a) general yield and (b) fast fracture. (a) = 500 MN m−2 for general yield. K 40 (b) = √ c = √ = 319 MN m−2

a

0005

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. √ 7000 1 pr 13.3. = = 006 × × = 70 MN m−2 Kc = Y a Y = 1. t 2 3 2 2 1 Kc 1 100 a= = = 065 m

a

70 14.1.

√ Kc = Y a Y = 1 √ Kc = a √ 2 2 1 Kc 1 30 MPa m a= = = 0080 m = 80 mm

60 MPa 2a = 160 mm


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√ √ 3 × 10 + 4 = 70 MN m−2 K = a = 70 × 0010 = 124 MN m−3/2 . This is only 5% less than the value of Kc 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.

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 = Y a. √ (a) From Fig. 15.3, a/W = 5/10 = 05 Y = 3 K = 3 × 100 × 0005 = 376 MN m−3/2 . (b) From Section 13.3 (“A note on the stress intensity, K ”), Y = 1. √ K = 1 × 100 × 0020 = 251 MN m−3/2 14.5.

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) K 3 MPa m1/2 = 309 MPa TS = √ c = √

a

× 30 × 10−6 m (b) From eqn (16.3) C ≈ 15TS = 4635 MPa

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From eqn (16.2) 6 Mr 1 F l = 6× × × 2 2 bd 2 2 bd 330 N 50 mm 1 × = 6× × 3 2 2 5 mm3 N = 198 = 198 MPa mm2

r =

16.3.

V

11/22 mm2 × 50 mm = = 97. Vo

5/22 mm2 × 25 mm For the test specimens, eqn (16.7) gives m Vo 05 = exp − Vo 0 For the components, eqn (16.7) gives m V

099 = exp − Vo 0 Thus

m ln05 V V 0 m = − o − o ln099 Vo 0 V

1 m 690 = 97

1 m = m 669 = 0272 × 120 MPa = 326 MPa

16.4.

See Answer to this Example.

16.5.

Specimen measuring 100 mm × 10 mm × 10 mm will have median TS of 300 MPa/1.73. Specimen volume V = 104 mm3 . Eqn (16.7) then gives 104 mm3 10 05 = exp − Vo o 10 TS Taking natural logs gives −069 = −

104 mm3 10 Vo o 10 TS


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Component volume = 125 × 103 mm3 Pf = 10−6 gives Ps = 1 − 10−6 . Thus 125 × 103 mm3 10 −6 1 − 10 = exp − Vo o 10 Taking logs, and assuming that ln1 − x = −x for small x, gives −10−6 = −

125 × 103 mm3 10 Vo o 10

Thus 069 104 mm3 1 Vo o10 10 = × TS 10 −6 3 3 10 V o o 125 × 10 mm 10 10 104 TS = 125 × 103 104 10−6 10 = 10 × 125 × 103 069 TS 300 MPa = 321 × 10−1 TS = 321 × 10−1 × 173 = 557 MPa 16.6.

(a) Weight of material below section at x is g ×

x2 x 3

Cross-sectional area = x2 . Stress = force/area = 13 gx. (b) Integrate over the volume, using disc of thickness dx with volume dV =

x2 dx. Eqn (16.8) then gives ⎫ ⎧ ⎫ ⎧ m ⎨ L gx m x2 dx ⎬ ⎨ dV ⎬ Ps L = exp = exp − − ⎩ ⎭ ⎩ 0 V0 ⎭ 30 V0 V

0

Integrating gives m

2 Lm+3 g Ps L = exp − 30 m + 3V0 The probability of survival falls with increasing because, although the stresses are the same, the amount of material which is stressed increases with , and hence the chances of meeting a critical flaw increase.

28


(a) For the specimen in uniform tension, eqn (16.7) gives m

r 2 t Pst = 05 = exp − V0 0 Setting Ps L = Pst gives m m

2 Lm+3

r 2 t g = exp − exp − 30 m + 3V0 V0 0 1 m m+3 2 r m + 3 3t . Thus L = 2 g (b) Flaws induced during sample preparation, maintaining the correct environmental conditions, gripping without causing failure in the grips.

17.1.

Catastrophic failure will occur when √ a = 54 MN m−3/2 or when a = 0029 m Now −4 MN da −13 = 4 × 10 m−1 K 4 dN m2 −4 MN −13 = 4 × 10 m−1 4 2 a 2 m2 = 414 × 10−3 a 2 m−1 Integrating from a = 10−4 m at N = 0 to a = 29 × 10−2 m at N = Nf gives

103 1 1 Nf = − 414 10−4 29 × 10−2 = 24 × 106 cycles to failure


29

Nf a = C . 280105 a = 200107 a 7 a 10 280 = 14 = = 102 a 200 105 log 14 = a2 = 0146 1 a = 0073 or 137 C = 280105 0073 MN m−2

C = 649 MN m−2 137 C 1/a 649 −2 = = 52 × 108 cycles. At 150 MN m Nf = 1/a 150 17.3. The total strain range is: = T = 24 × 10−3 The plastic strain range is: pl = − el = 20 × 10−3 The cycles to failure are:

02 Nf = 2 × 10−3 17.4.

(a)

2 = 104

51 × 75 pr = MN m−2 = 478 MN m−2 . t 004 × 2 √ Kc = working a at fracture 2 1 200 MN m−3/2 a= = 56 × 10−2 m

478 MN m−2

working =

This critical depth for fast fracture is greater than the wall thickness of 40 mm. 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 −4 da −14 MN = 244 × 10 m−1 K 4 dN m2 −4 −14 MN = 244 × 10 m−1 4 2 a 2 m2 We can then integrate this, from the initial condition after the proof test (assuming that a crack of length a0 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. −4 4 N 40×10−2 m da f −14 MN −1 4 MN 2 244 × 10 m 478

dN = m2 m2 a2 a0 0 40×10−2 m 1 −2 −1 1257 × 10 m Nf = − a a0 a0 = 0016 m This is the initial flaw size that will penetrate the wall after 3000 loading cycles. The proof stress Pproof must be sufficient to cause flaws of this size to propagate by fast fracture. √ K = proof a0 ≥ Kc Where proof = Hence Pproof 17.5.

Pproof r

t tK 004 × 200 × 106 ≥ √ c = = 95 MN m−2 √ r a0 75/2 0016

Each time the iron was moved backwards and forwards the flex would have experienced 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 rated to carry 13 A, then a single strand should carry about 0.57 A safely. The iron 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.


31

17.6.

(a) LHS = 0730 hrs position. RHS = 0600 hrs position. (b) LHS = 1300 hrs position. RHS = 1200 hrs position.

17.7.

The 50 m marker at the top of the photograph measures 33 mm in length. The most distinct striations are located approximately 70 mm up from the bottom of the photograph and 50 mm in from the left-hand side of then photograph. 5 spacings = 10 mm in this region, giving a striation spacing of 2 mm on the photograph. The spacing is thus 2/33 × 50 = 3 m on the fracture surface itself. N 4 × 108 For 4 × 108 cycles, = = 077. Nf 52 × 108

18.1.

N N1 + 1 = 1. Nf Nf

Miner’s rule gives: ∴

N1 = 1 − 077 = 023 Nf1 N1 4 × 108 = = 174 × 108 cycles 023 023 C 649 MN m−2 = 1 0073 = 174 × 108 0073 Nf Nf1 =

For this

Decrease = 13 MN m

−2

= 137 MN 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 and reduce the fatigue life. (c) A mean tensile stress will decrease Nf 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 07 N mm−2 × 45 mm2 = 4453 N The stress in the connecting rod next to the joint is 4453 N/28 mm × 11 mm = 145 N mm−2 Since the locomotive is double-acting the stress range is twice this value, or 29 N mm−2 . The number of revolutions that the driving wheel is likely to make in 20 years is 20 × 6000 × 1000 m/ × 0235 m = 16 × 108 .

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 33 N 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 8 N mm−2 . The mean-line fatigue curve gives the data for a 50% chance of cracking. For the stress range of 8 N mm−2 the cycles to failure are 3 × 109 . The time to failure is 3 × 109 /20 × 60 × 60 × 12 × 6 × 52 = 11 years. The design curve gives the data for a 2.3% chance of cracking. The number of cycles to failure is 109 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 of sufficient amplitude to make fatigue cracks initiate and grow to final failure. (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 0630 hrs position, since the crack appears to radiate from this position. (e) Moderate, since the fatigue cracks had spread across 50% of the crosssection on average before they reached the critical size for fast fracture.

19.2.

(a) In the reduced section of the pivot pin, at the lower end of the reduced section. (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


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 SCFeff (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 C

T K

1/T K −1

˙ s−1

ln ˙

618 640 660 683 707

891 913 933 956 980

0.00112 0.00110 0.00107 0.00105 0.00102

10 × 10−7 17 × 10−7 43 × 10−7 77 × 10−7 20 × 10−6

−1612 −1559 −1466 −1408 −1312

510

783

0.00128

83 × 10−10 From graph

−2090

34

Solutions Manual: Engineering Materials I The hoop stress in the tube is =

6 MN m−2 × 20 mm pr = t 2 mm = 60 MN m−2 ˙ = 83 × 10−10 s−1 at = 200 MN m−2 5 60 ˙ = 83 × 10−10 s−1 . 200

at 510 C, Under 60 MN m−2 ,

= 20 × 10−12 s−1 . = 20 × 10−12 × 60 × 60 × 24 × 365 × 9

Strain in 9 years

= 57 × 10−4 or 000057 Design safe 20.2.

At 25 MN m−2 and 620 C ˙ = 31 × 10−12 s−1 . At 30 MN m

−2

and 620 C ˙ =

30 25

5 × 31 × 10−12 s−1

= 771 × 10−12 s−1 ˙ = A 5 e−Q/RT Q 1 1 ln ˙ 1 − ln ˙ 2 = − − R T1 T2 Q 1 1 ln ˙ 2 = + ln ˙ 1 − R T1 T2 160 × 103 1 1 = − 8313 893 923 + ln771 × 10−12 s−1 = 0700 − 2558 = −2488 ˙ 2 = 155 × 10−12 s−1 at 30 MN m−2 and 650 C ˙ = 31 × 10−12 s−1 × 30 100 s−1

×10−12 s−1 × = 682 × 10−12

70 + 155 100


35

20.3.

From Table 20.1, the softening temperature of soda glass is in the range 700–900 K. The operating temperature of window glass is rarely more than 293 K, so T/TS is at most 0.42. Ceramics only begin to creep when T/TM > 04 (see Section 20.1), and then only under a large stress (far greater than the selfweight 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.

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. M 21.1. Measure the rate by the mass injected per second, . Then, if this rate follows t an Arhennius Law, we have M Q = A exp − t RT 20.6.

Or, combining the constants M and A: 1 Q = B exp − t RT Then, converting temperatures to kelvin: 1 Q s−1 = B exp − 30 R450 1 Q s−1 = B exp − 815 R430 Solving for Q and B, using R = 831 J mol−1 K −1 , we have Q = 804 × 104 J mol−1 B = 725 × 107 s−1 We may now calculate the time required to inject the mass M of polymer at 227 C (500 K): 804 × 104 1 7 = 725 × 10 exp − s−1 t 8313 × 500 giving t = 35 seconds.

36


See Sections 21.2 and 21.4 (“Fast diffusion paths: grain boundary and dislocation 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 concentration 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 boundaries. 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 , so t = x 2 /D.

D = Do e = t = = = =

−Q/RT

−1

= 95 mm s 2

−159 × 103 J mol−1 exp 8313 J K −1 mol−1 1023 K

95 mm2 s−1 = 720 × 10−8 mm2 s−1 132 × 108 x2 10−2 mm2 = D 72 × 10−8 mm2 s−1 108 10−4 mm2 × s 72 mm2 104 s = 01389 × 104 s 72 1389 s or 23 min

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).


37

22.5.

The rate of bulk vacancy diffusion decreases rapidly with decreasing temperature 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 dislocation 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 shortcircuit 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.

dc . dx m = AJ t where A = constant. c −c dc m = −A D t = AD 1 2 t . dx b Fick’s first law J = −D

38

Solutions Manual: Engineering Materials I b m = ADc1 − c2 t . b = Bm where B = constant. Bm m = ADc1 − c2 t . B m2 = ADc1 − c2 t , 2 i.e., m2 = CDt , where C = constant. At constant temperature m2 = kP t . 24.2.

Ohm’s Law V = IR. m = PI t where P = constant PV t m = PI t = R R m = PV t R = Qb where Q = constant = Sm where S = constant Sm m = PV t S m2 = PVt ie m2 = kP t 2

24.3.

138 × 103 J mol−1 AT 500 C kP = 37 exp − 8313 J K −1 mol−1 773 K

kg 2 m−4 s−1

= 174 × 10−8 kg 2 m−4 s−1 m2 = kP t = 174 × 10−8 kg 2 m−4 s−1 × 3600 × 24 × 365 s = 050 kg 2 m−4 m = 074 kg m−2 i.e., each square metre of metal surface absorbs 0.74 kg of oxygen from the atmosphere in the form of FeO. Number of oxygen atoms absorbed = 074 kg m−2 , where NA is Avogadro’s number. 16/NA 074 kg m−2 ∴ Number of iron atoms removed from metal as FeO = . 16/NA


39

074 kg m−2 Weight of iron removed from metal = × 559/NA = 16/NA −2 259 kg m . 259 kg m−2 m3 7870 kg = 033 mm

Thickness of metal lost =

at 600 C kP = 204 × 10−7 kg 2 m

−4

s−1 −4

m2 = 204 × 10−7 × 3600 × 24 × 365 kg 2 m −4

= 643 kg 2 m m = 254 kg m−2 giving

loss = 113 mm

24.4.

As shown in Table 24.1, gold is the only metal which requires energy to make it react with oxygen 80 kJ mol−1 of O2 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 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 Cr2 O3 , produced by the chromium content of the stainless steel.

25.3.

See Sections 25.2 and 25.3. Examples are Cr in stainless steel (Cr2 O3 film), Cr in nickel alloys (Cr2 O3 film), Al in aluminium bronzes (Al2 O3 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 pretinned copper wire, the pre-tinning solder melts and this protects the copper surfaces from oxidation.

40


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, SiO2 Al2 O3 (see Table 24.1). Others, e.g. SiC and Si3 N4 form protective oxide layers of SiO2 when exposed to oxygen at high temperature.

26.1.

da ∝ a at constant 4 MN m−2 . dt da ∝ 2 at constant a025 mm dt da ∴ = A 2 a = AK 2 n = 2 dt 1 03 mm year−1 da =

A = dt 2 a 16MN m−2 2 025 mm 0075 −2 = A = 00239 m4 MN year−1 −2 2 MN m year

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 C (summer time, heating off) to 70 C (winter time, heating on). As shown in Fig. 26.7, the corrosion rate at 70 C will be approximately twice that at 20 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 surface 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, windscreen 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.


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++ + 2e Number of electrons to give a current of 6 × 10−3 A m−2 for 5 years = = ∴ Mass zinc = Thickness = =

27.5.

6 × 10−3 × 5 × 315 × 107 = 589 × 1024 electrons 16 × 10−19 589 × 1024 atoms Zn 2 589 × 1024 × 654 = 0320 kg m−2 2 × 602 × 1026 0320 m = 45 × 10−5 m 7130 0045 mm

Following 44, if steel were lost uniformly over a square metre at 2×10−3 A m−2 , thickness lost by reaction: Fe → Fe++ + 2e would be 0.0116 mm on each side of the plate. If this loss is concentrated over 0.5% of the surface, the loss there will be 00116 ×

100 995 × mm = 231 mm 05 100

The sheet will thus rust through.

42


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 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 pw r 2 . This force is balanced by the frictional force between the olive and the pipe so we can write pw r 2 = P2 rl Combining the two equations to eliminate P gives t l pw = 2y r r Using the data given we get

065 pw = 2 × 015 × 120 MPa 75

75 75

= 31 MPa

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 machinery, 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.


43

(b) Typical examples are as follows. Bearings and sliding surfaces in machinery, 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: s = tan 24 = 045 If the slope of the roof is greater than 24 the static frictional force is exceeded and the snow will slide off. On a 2 slope, with a ski already moving, it is the sliding friction which counts: k = tan 2 = 0035

29.3.

The work done = 2100 gk = 68 J. The melts a volume of water V =

69 = 21 × 10−7 m3 330 × 106

If spread uniformly over the undersurface of two skis this would give a film of thickness 21 × 10−7 m = 05 m 2 × 2 × 01 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.

Engineering Materials 1 Solutions

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