Engineering Materials 1 Solutions

8/10/2019 Engineering Materials 1 Solutions

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SOLUTIONS MANUAL

Engineering Materials I

An Introduction to Properties, Applicationsand Design, 3rd edn

Solutions to Examples

2.1. (a) For commodity A Pt = CAexp rA100

t

and for commodity B Qt = CB exp rB100t

whereCA andCB are the current rates of consumption t= t0andPt andQtare the values at t= t. Equating and solving for tgives

t= 100rB rA

ln

CACB

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

tD=

100

r

ln 2

70

r

Substitution of the values given for rin the table into this equation gives thedoubling 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. Formany applications, substitutes are easily found at small penalty of cost. But in

certain specific uses, most elements are not easily replaced. Examples: tungstenin cutting tools and lighting (a fluorescent tube contains more tungsten, asa 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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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; useof 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 isCthen exponentialgrowth means that

dC

dt= r

100C

whereris the fractional rate of growth in % per year. Integrating gives

C= C0 exp

rt t0100

whereC0 was the consumption rate at time t= t0.(b) Set

Q

2=

t1/20

Cdt

where

C= C0 exp

rt100

Then

Q

2

C0

100

r exp

rt100

t1/20

which gives the desired result.

2.4. See Chapter 2 for discussion with examples.

3.1. (a) Poissons ratio, , can be defined as the negative of the ratio of the lateralstrain to the tensile strain in a tensile test.

= lateral straintensile strain

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


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Solutions Manual: Engineering Materials I 3

(b) Volume change per unit volume (small strains) is 1 + 2 + 3. But in auniaxial extension2 = 1 and3 = 1.

Hence = VV

= 1 2

where 1 is the tensile strain. Clearly is zero if = 05.(c) The dilatations are:

Most metals 04Cork=

Rubber= 0

3.2. The solid rubber sole is very resistant to being compressed, because it isrestrained against lateral Poissons ratio expansion by being glued to the rela-tively stiff sole. However, the moulded surface has a much lower resistance tobeing compressed, because the lateral Poissons ratio expansion of each separaterubber cube can occur without constraint (provided the gaps between adjacentcubes 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 zerolateral Poissons ratio expansion, so it does not become any harder to push thecork into the neck of the bottle. However, the axial force applied to the rubberbung results in a large lateral Poissons 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.

ForceFbetween atoms = dUdr


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At the equilibrium distance, ro, the energyU is a minimum (i.e. F is zero,andU is the dissociation energy Uo).

dUdr

= mArm+1

nBrn+1

= 0

or B= mn

rnmo A

Uo= A

rmo+ 1

rno m

nrnmo A

= Armo

1 m

n

Now, forro = 03 nm Uo = 4eV.

A = 4 54

032 = 045eV nm2

= 72 1020J nm2

B= 15

038 045 = 059 105 eV nm10 = 94 1025J nm10. Max forceis at

d2U

dr2= 0. i.e. at value ofrgiven by

mm + 1Arm+2

+ nn + 1Brn+2

= 0 which is r =

B

A

n

m

n + 1m + 1

1nm

= n

+1

m + 11

nm

ro = 11

3

18

03 = 0352nm

and Force = dUdr

= mArm+1

1 r

nmo

rnm

= 2 04503523

1 03

8

03528

= 149eV nm1

=149 1602 1019

109 J m1

= 239 109 N

4.2. The termA/rm is an attractive potential which depends on the type ofbonding. The B/rn term is a repulsive potential due to charge-cloud overlapand diminished screening of the nuclei (see Section 4.2).

4.3. The values ofA are shown below. The mean is 88. The calculated values ofthe moduli are


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(b)

(c)

(d)

5.3. (a) If the atom diameter isd, then the lattice constant for the f.c.c. structure is

a1 =2d1

2= 03524nm


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5.4. (a) Copper

Have 4 atoms per unit cell 8 18

from cube corners = 1 + 6 12

from

cube faces

=3

Atoms touch along cube-face diagonal:

This gives atom radius r =

2

4a.

Required percentage =4 4

3r3 100

a3 = 16

3a3

2 2a3 100

16 4

=

2

6 100

=74%

Answer is same formagnesium both are close-packed structures.(b) Copper

Density= 4 ma3

wherem is atomic mass.

Atomic weight for Cu = 6354.

m = 6354kg6023 1026= 1055 10

26 kg

896Mg m3 = 4 1055 1026

a3 kg

a3 = 471 1029 m3a= 361 1010 m or 0361nm


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Magnesium

Have 6 atoms per unit cell 12 16

from corners= 2 + 2 12

from

end faces = 1+ 3 inside

Volume of unit cell= 3

a2

3

2 c

Density= 2

6m

33a2c= 4m

3a2cm= 2431kg

6023 1025= 404 1026 kg

174 Mg m3 = 4 404 1026 kg

3a2c

a2c= 536 1029 m3

In face-centred structure, plane spacing

=

3a

3 where r =

2a

4

Spacing=

3

3

4r2


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In close-packed-hexagonal structure, plane spacing = c2=

3

3

4r2=

33

42

a2

c

a= 1633

c= 1633a

Using value for a2cof 536 1029 m3 we obtain:a= 320 1010 m or 0320nmc= 523 1010 m or 0523nm

5.5. See Section 5.10.

5.6. LetVbe the volume fraction of the polyethylene which is crystalline. Then,

(a) 1014V+ 0841 V = 092, givingV= 046, or 46%.(b) 1014V+ 0841 V = 097, givingV= 075, or 75%.

6.1. The two sets of values for the moduli are calculated from the formulae

Ecomposite = VfEf + 1 VfEm (upper values);Ecomposite =

1VfEf

+ 1VfEm

(lower values);

whereVfis the volume fraction of glass,Em the modulus of epoxy, andEf thatof glass.

Values are given in the table and plotted in the figure. The data lie near thelower level. This is because the approximation from which the lower values arederived (that the stress is equal in glass and epoxy) is nearer reality than theapproximation from which the upper values are derived (that the strains areequal 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 themaximum composite modulus. Both materials, tested at right angles to thefibres or grain lie near the lower modulus. They are, therefore, veryanisotropic:the ratio of the two moduli can be as much as a factor of 4.5 for fibreglass (seethe figure, atVf= 05); it can be more for woods.


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Ecomposite EcompositeVolume fraction Ecomposite (upper values) (lower values)

of glass,Vf GN m2 GN m2 GN m2

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. Ec = EfVf + 1 VfEm.c = fVf + 1 Vfm.

(a) c = 05 190Mg m3 +05 115Mg m3 = 153Mg m3.(b) c = 05 255Mg m3 +05 115Mg m3 = 185Mg m3.(c) c = 002 790Mg m3 +098 240Mg m3 = 251Mg m3.(a) Ec = 05 390 GN m2 +05 3 GN m2 = 197GN m2.(b) Ec = 05 72GN m2 + 05 3 GN m2 = 375 GN m2.(c) Ec = 002 200 GN m2 + 098 45GN m2 = 481 GN m2.


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6.3.

E= V1E1 +1

V1

E21

see Section 6.4

E= V1E1 + 1 V1E2 see Section 6.4EE

= V1E1 + 1 V1E2

V1E1

+ 1 V1E2

= 1 2V1 + 2V21+ 1 V1V1

E1E2

+ E2E1

d

dV1

EE

= 1 2V1

E1E2

+ E2E1

2

ForE1= E2d

dV

EE

= 0 whenV1 = 05 (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= wdc. Substituting d using theequation given in the example gives

M=

c

E1/3c

F6w2

4

1/3= Kc/E1/3c

whereK is a constant. Values ofc/E1/3c 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

F

= 3Er

3t

l3

The mass,M, is given by

M=

2rtl

Substituting fortin (1) gives

M= 2l3r2

F

E


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

sensible) materials.

Material E GN m2) Mg m3) p $ tonne1 E

p E

Mildsteel 196 7.8 100 398 103 3.98Hardwood 15 0.8 250 533 103 13.33Aluminium alloy 69 2.7 400 391 103 15.64GFRP 40 1.8 1000 450 103 45.0Titanium alloy 120 4.5 10,000 375 103 375CFRP 200 1.5 20,000 75 103 150

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 least5.0 over other materials.

(b) The frame of minimum material cost is that for which the relative priceMp (wherep is relative the price per tonne) is a minimum i.e. thatfor whichp/E is least. The table shows that steel is by far the mostattractive material: a CFRP frame will cost 38 times more. This is nothingif it permits you to win the Tour de France so bicycles can be made ofCFRP.

7.3.

pb=

03 E tr

2

Mb= 4r2t tr

2= pb

03 E

t

r=

pb03 E

1/2

t=

pb03 E

1/2r

Mb=

4r3 pb03 E

1/2

=229r3pb

1/2 E1/2

Merit index is

E1/2


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8.1. Nominal stress=Hardness/31 + nwhere n is the nominal strain as givenin the question plus 0.08.

Data for nominal stressnominal strain curve

Nominal stressMN m2 129 171 197 210 216 217 214 209 188Nominal 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 m2 (the stress maximum of thecurve).

(b) The strain at the stress maximum is 0.6 approximately.(c) From eqn. (8.4), Aolo =Al, and lol= AAo .

l lo/lo = 06 A/Ao = 1/16

1 AAo

= 038 Ao A/Ao = 038

Thus percentage reduction in area= 38%.(d) 109 MJ from graph.


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8.3. During a tensile test, unstable necks develop when the maximum in the nominalstressnominal strain curve is reached. Neck growth then leads rapidly to failure.

Physically, necks become unstable when the material in the elongating neckwork hardens insufficiently to make up for the decrease in load-bearing capacityat the neck. In rolling, the material is deforming mainly in compression, andthe load-bearing area is always on the increase. Tensile instabilities cannottherefore form, and failure occurs at the much larger strains required to causefailure by cracking.

8.4.

(a) Tensile strength = 1015 kN/160 mm2= 634 MN m2.

Working stress = 160 MN m2.


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(b) Load corresponding to 0.1%

Proof strain = 35 kN. 0.1% Proof stress = 219 MN m2.

Working stress = 131 MN m2.

8.5. The indentation hardness is defined by

H= Fa2

wherea is the radius of the circle of indentation. Simple geometry (see figure)gives

r h2

+ a2

= r2

or

2rh h2 = a2

or

h = a2

2r if h r

Thus H= F2rh

.

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


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8.6. (a) Conservation of energy givesmgh = Uel, som = Uel/gh. Thus

mmax = 1000N 15 m + 05 1500N 15 m981 30 = 892 kg

(b) Taking the maximum extension, in one cycle of loading/unloading,500N 15 m = 7500J is dissipated by hysteresis in the rope, compared toaUel of 26250 J. Thus 29% of the energy input is lost. This results in thejumper rebounding to a position which is well below the bridge deck (thisis obviously essential for a safe jump).


9.2. The fractional volume change is

V

V = 89323 89321/89323 = 224 105

(a) Define the dislocation density as m/m3 or m2. Then

V

V = 1

4b2

from which

= 14 1015 m2

(b) The energy is:

U= 12

Gb2 = 2 VV

38

E= 21 MJ m3


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 andFig. 10.2(c)):

ybL= 2T

Soy= 2T

bL Gb

L


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(b) Welded Interfaces

Fu 2wLku + 4 wL2 k u

2= 2wLku + wLku = 3wLku

F 3wLk

General formula gives, for w

d= 2,

F

wL 2k

1+ 1

2

= 3k

F

3wLk verification demonstrated

11.3. 4000 MN m2 = 350 MN m2

1+ w4d

.

w

4d= 1043

w

d= 417

Take w

d= 42 to produce an integral value of w

2din a safe direction.

d= w42

=20 m42

= 048 m

(This gives a volume fraction of cobalt of 7.2% a typical value for arock-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 m2.

At onset of necking

d

d= .350 0406 = 35004

or = 04= 350 0404 MN m2 = 2426MN m2

Nominal stress n=

1 + n where n is the nominal strain

= ln1 +n

Tensile strength TS=

2426

antiln04 MN m2

= 24261492

MN m2

= 163MN m2

Work=

0dper unit volume

= 04

035004dMN m2

=35014

14

04

0

MN m2

= 3500414

14 MN m2

= 693MJ per 1m3


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11.6. (a) Given = An. From eqn (11.4), dd

= at onset of necking. Thusd

d= Ann1 =An so = n and =Ann. From eqn (8.13), TS=

1 + n= An

n

1 + n.

From eqn (8.15), = ln1 +n = n.Thus1 + n = en. Finally, TS =

Ann

en .

(b) Inserting A= 800MN m2 and n= 02 gives TS= 8000202

e02 =

475 MN m2.

= TS1 + n = TSen = 580 MN m2

11.7. We have thatyat 8% plastic strain is H/3 or 200MN m2. Thus

A = 200ln 10802

= 334 MN m2

Using the result of Example 11.6(a),

TS =Ann

en = 198 MN m2

12.1.

pf= 2f t

r

Mf= 4r2t tr

= pf

2f

t= pfr2f

Mf

=

4r3pf

2f =2r3pf

f

Merit index is

f


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12.2.

Mb= 229r3

pb1/2

E1/2

tb= Mb4r2

Mf= 2r3pf

f

tf= Mf4r2

Setr= 1m and pb =pf= 200MPa. Values for E andfare to be takenfrom the table of data.

Material Mb (tonne) tb (mm) Mf (tonne) tf (mm) Mechanism

Al2O3 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 Al2O3 with a mass of 2.02 tonne. The wall thickness is41 mm and the limiting failure mechanism is external-pressure buckling.

12.3. When the bolts yield, the connection can be approximated as a mechanismwhich hinges at X. The cross-sectional area of one bolt is 125/22 in2

=123in2. The yield load of one bolt is 123in2 11tsi = 135 tons. The momentat yield is given by

M 2 135 25 + 2 135 19 + 2 135 9 + 2 135 3= 1512ton in = 126ton ft

The hinge must react the total load from the bolts, which is 108 tons. Thismeans that in practice the hinge will extend over a finite area of contact. X willlie in from the outer edge of the flange by about 1 in to 2 in but the effect onthe 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 breakingload of the two links in parallel is double this figure and is given by

T= 21500N mm2 1 mm 45 mm= 135 104 N


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From eqn (11.2), the shear yield strength k= y/2.kfor the pin is therefore1500/2 = 750MN m2.

The yield load of the pin in double shear is obtained by multiplyingkbytwice the cross-sectional area of the pin to give

T= 2

750 N mm2

35

2

2mm2

= 144 104 N

This load is 7% greater than the load needed to yield the links and the strengthof 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 momentsabout the centre of the chain wheel to give

90kgf

170mm

T

190mm

2

T 161kgf 158 103 N

The factor of safety is then given by

135 104 N158 103 N= 85

Comments

(a) The factor of safety is calculated assuming static loading conditions. Themaximum loadings experienced in service might be twice as much due todynamic effects.

(b) The chain must also be designed against fatigue and this is probably whythe factor of safety is apparently so large.

12.5. The cross-sectional area of the pin is

A = 22 122mm2 = 80 mm2

The force needed to shear this area is

fs = kA = 750 N mm2 80 mm2 = 6000N

Finally, the failure torque is

= 2 fs 10mm= 2 6000N 10mm = 12 105 N mm= 120N m 12kgf m


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13.1. The maximum tensile stress at the surface of a beam loaded in three-pointbending (see eqn (12.3)) is

= 3Fl2bt2

at the mid-span of the beam. Fracture occurs when

a= Kci.e. when

3Fl

2bt2

a= KcHence, the maximum load which can be sustained by the adhesive joint is

F= 2bt2Kc

3l

a

For the joint shown

F= 2 013 05

32

0001= 297kN

13.2. Calculate the stress for failure by (a) general yield and (b) fast fracture.

(a) = 500 MN m2 for general yield.(b) = Kc

a= 40

0005= 319 MN m2

for fast fracture, assuming that a crack on the limit of detection is present. Theplate will fail by fast fracture before it fails by general yield.

13.3. = prt = 006 7000

2 1

3= 70MN m2 Kc = Y

a Y= 1.

a= 1

Kca

2= 1

100

70

2= 065 m

14.1.

Kc= Y

a

Y

=1

Kc= a

a= 1

Kc

2= 1

30MPa

m

60MPa

2= 0080m = 80mm

2a= 160mm


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14.2. 3 10+ 4 = 70 MN m2 K = a = 70

0010 =124MN m3/2.

This is only 5% less than the value ofKc obtained from tests. Experimentalscatter in the test data, dynamic loads and errors in the stress analysis are morethan 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 planescan be seen. Many fracture facets have river markings, produced by fractureon multiple parallel cleavage planes.

15.1. From Section 13.3 (A note on the stress intensity,K),K= Ya.(a) From Fig. 15.3,a/W

=5/10

=05 Y

=3 K

=3

100

0005

=376 MN m3/2.(b) From Section 13.3 (A note on the stress intensity, K),Y= 1.

K= 1 100

0020 = 251MN m3/2

15.2. See Section 15.3 (Failure analysis and Conclusions). PMMA is a poorchoice of material because it has a very low fracture toughness. Under a tensilehoop stress, the connector is liable to suffer catastrophic fast fracture from asmall 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 intothe foam as it is sprayed on.

15.4. Fix each end of the top rail directly to the brick wall, using a steel bracketbolted 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 veryeasily along the grain. This permits easy splitting of logs into kindling and woodfor fires, production of wood shingles for roofing material, finishing/sizing byplaning, 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 astable way in compression.

(a) From eqn (16.1)

TS = Kca

= 3 MPa m1/2

30 106 m

= 309MPa

(b) From eqn (16.3)

C 15TS = 4635MPa


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16.2. From eqn (16.2)

r= 6Mr

bd2 = 6 F

2l

2 1

bd2

= 6 330N2

50mm2

153 mm3

= 198 Nmm2

= 198MPa

16.3.V

Vo= 11/2

2 mm2 50mm5/22 mm2 25mm = 97.

For the test specimens, eqn (16.7) gives

05 = expVoVo

0m

For the components, eqn (16.7) gives

099 = exp V

Vo

0

m

Thus

ln05

ln099=

Vo

Vo

0

mVo

V

0

m

690

=

1

97

m

m = 1669

m

= 0272 120MPa= 326MPa

16.4. See Answer to this Example.

16.5. Specimen measuring 100 mm 10mm 10mm will have median TS of300 MPa/1.73.

Specimen volumeV= 104 mm3. Eqn (16.7) then gives

05

=exp

104 mm3

Vo o10

TS10

Taking natural logs gives

069 = 104 mm3

Vo o10

TS10


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17.2. Nfa = C.

280105a

=200107a

280

200= 14=

107

105

a= 102a

log 14= a2 = 0146

a= 0073

or 1

137

C= 2801050073 MN m2C= 649 MN m2

At 150 MN m2 Nf=

C1/a

1/a= 649

150

137

=52

108 cycles.

17.3. The total strain range is:

= T= 24 103

The plastic strain range is:

pl = el = 20 103The cycles to failure are:

Nf= 022 1032

= 104

17.4. (a) working =pr

t = 51 75

004 2 MN m2 = 478 MN m2.

Kc = working

aat fracture

a= 1

200 MN m3/2

478 MN m2

2= 56 102m

This critical depth for fast fracture is greater than the wall thickness of40 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 factorhandbook. This will be particularly important as the crack approaches theoutside of the wall).


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(b) Rearranging the crack growth equation

da

dN= 244 1014 MN

m24

m1

K 4

= 244 1014

MN

m2

4m142a2

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 pointwhere after 3000 cycles the crack has grown out to the wall, where failurewould occur by leakage.

244 1014

MN

m2

4m14784

MN

m2

42

Nf0

dN= 40102 m

a0

da

a2

1257 102 m1Nf=1

a

4010

2

m

a0

a0 = 0016mThis is the initial flaw size that will penetrate the wall after 3000 loadingcycles.

The proof stress Pproof must be sufficient to cause flaws of this size topropagate by fast fracture.

K= proof

a0 Kc

Whereproof=Pproofr

tHencePproof

tKcr

a0= 004 200 10

6

75/2

0016= 95MN m2

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 sheathis 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 significantbending stress at the location of failure. The number of cycles of bending is wellinto the range for high-cycle fatigue and fatigue is the likely cause. The scenariois that the individual strands in the live conductor broke one by one until thecurrent 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 arerated to carry 13 A, then a single strand should carry about 0.57 A safely. The irondraws 4.8 A, which is 8.4 times the safe capacity of one strand. It is therefore notsurprising that, when only a few wires were left intact, the flex was no longer ableto take the current without overheating. Failures of this sort have also occurredwith appliances such as vacuum cleaners. However, these tend to have a smallercurrent rating and failure does not always result in a fire.


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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 thebottom of the photograph and 50 mm in from the left-hand side of thenphotograph. 5 spacings = 10 mm in this region, giving a striation spacing of2 mm on the photograph. The spacing is thus 2/33 50 = 3 m on thefracture surface itself.

18.1. For 4 108 cycles, NNf

= 4 108

52 108= 077.

Miners rule gives: N

Nf+ N

1

N1f= 1.

N1

N1f = 1 077 = 023

N1f = N1

023=4 10

8

023 = 174 108 cycles

For this = CN1 0073f

= 649MN m2

174 1080073= 137MN m2

Decrease = 13MN m218.2. (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 decreaseNfas in Goodmans 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 (seeSection 26.5).

18.3. The maximum pressure in the cylinder occurs at the point of admission. Themaximum force acting on the piston is therefore given by

07 N mm2 45mm2 = 4453N

The stress in the connecting rod next to the joint is

4453N/28mm 11mm = 145 N mm2

Since the locomotive is double-acting the stress range is twice this value, or29N mm2.

The number of revolutions that the driving wheel is likely to make in20 years is 20 6000 1000m/ 0235m = 16 108.


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Data for the fatigue strengths of welded joints are given in Fig. 18.4. Thetype 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 figurecalculated above. Of course, our calculations have ignored dynamic effects dueto the reciprocating masses of the piston and connecting rod and these shouldbe 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 asurface detail on the stressed plate and the weld classification is Class F2. Weextrapolate the curve following the dashed line for Class F2 until we hit thestress range of 8N mm2.

The mean-line fatigue curve gives the data for a 50% chance of cracking.For the stress range of 8N mm2 the cycles to failure are 3 109. The timeto failure is3 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 stressof sufficient amplitude to make fatigue cracks initiate andgrow to final failure.

(b) There are two fatigue cracks, one on either side of the hole. They aresmooth and dark in appearance, and are located in the lower half of eachfracture surface. The dark appearance is caused by a compact layer of ironoxide produced by slow long-term corrosion, indicating that the cracks hadbeen present for a long time. In addition, the smooth appearance of thecrack surfaces is consistent with high-cycle fatigue.

(c) The final fracture surfaces are grainy and bright in appearance, and arelocated in the upper half of each fracture surface. They consist of fresh,

un-corroded metal, indicating that the final fracture was not exposed to acorrosive environment after failure. In addition, the grainy appearance ofthe crack surfaces is consistent with a single overload failure.

(d) The RHS crack probably formed first, because it is larger than the LHScrack. It probably initiated at the surface of the hole, because the local stresswould 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 thecrack 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) 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 pointat which it touches the bottom of the frame housing. This rotation pushesthe top of the pin to the left, against the left-hand wall of the hole in theframe housing. The reaction to this force at the top of the pin generatesa bending stress in the reduced section of the pin. The maximum value


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of this bending stress occurs at the lower end of the reduced section. Inaddition, 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 compressionevery 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, thelength of pivot pin below the fracture fell down into the hole in the doorhousing, 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 pulleyblock to provide a horizontal hole to take the pin of the shackle. Because therotational degree of freedom provided by the lifting eye assembly has nowbeen 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.

TemperatureC T K 1/T K1 s1 ln

618 891 0.00112 10 107 1612640 913 0.00110 17 107 1559660 933 0.00107 43 107 1466683 956 0.00105 77 107 1408707 980 0.00102 20 106 1312

510 783 0.00128 83 1010 2090From graph


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The hoop stress in the tube is

=pr

t =6MN m2

20mm

2 mm

= 60MN m2at 510C, = 83 1010 s1 at = 200 MN m2

Under 60 MN m2, = 83 1010

60

200

5s1.

= 20 1012 s1.Strain in 9 years = 20 1012 60 60 24 365 9

= 57 104 or 000057Design safe

20.2. At 25MN m2 and 620C = 31 1012 s1.

At 30 MN m2 and 620C=

30

25

5 31 1012 s1

= 771 1012 s1= A5eQ/RT

ln 1 ln 2= Q

R 1

T1 1

T2

ln 2= Q

R

1

T1 1

T2

+ ln 1

= 160 103

8313

1

893 1

923

+ ln771 1012 s1

= 07002558 = 24882= 155 1012 s1 at 30MN m2

and 650C

= 31 1012

s1

70

100 + 1551012 s1 30

100

= 682 1012 s1


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20.3. From Table 20.1, the softening temperature of soda glass is in the range700900 K. The operating temperature of window glass is rarely more than

293 K, soT/TSis at most 0.42. Ceramics only begin to creep whenT/TM> 04(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 usedto manufacture panes of glass in the past.


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 atwhich it would creep under the applied loads, and the subsequent deformationcould trigger the collapse of the building. This is why the World Trade Centertowers collapsed on 9/11.

21.1. Measure the rate by the mass injected per second,Mt

. Then, if this rate follows

an Arhennius Law, we have

M

t =A exp

Q

RT

Or, combining the constants M andA:

1

t= Bexp

Q

RT

Then, converting temperatures to kelvin:

130

= Bexp Q

R450

s1

1

815= Bexp

Q

R430

s1

Solving forQ andB, usingR = 831J mol1K1, we haveQ= 804 104 J mol1B= 725 107 s1

We may now calculate the time required to inject the massMof polymer at

227C (500 K):1

t= 725 107 exp

804 10

4

8313 500

s1

givingt= 35 seconds.


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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 remainingunfilled by carbon. The probability of a carbon atom being next to analternative interstitial site is therefore high, and hence the probability of thecarbon atom moving into a different position is also high, i.e. it will diffuserelatively rapidly. Chromium forms a random substitutional solid solutionin iron at room temperature (that it should not form aninterstitialsolutionis evident from its large atomic radius, comparable to that of iron). Theprobability of a vacancy appearing next to a chromium atom is thereforesmall, and the diffusion of chromium is correspondingly slow.

(b) Diffusion in grain boundaries is generally more rapid than in the grainsthemselves because of the geometrically more open structure of grain bound-aries. A small grain size produces a larger contribution from grain boundarydiffusion than does a large grain size, and thus increases the overall diffusioncoefficient of the polycrystal.

21.4. See Section 21.4 (A useful approximation).

x=

Dt, sot=x2/D.

D= DoeQ/RT = 95 mm2 s1 exp 159 103J mol1

8313 J K1 mol11023K

= 95 mm2

s1

132 108 = 720 108 mm2 s1

t= x2

D= 10

2 mm2

72 108 mm2 s1

= 104 mm2

72 10

8

mm2 s

= 104

72 s = 01389 104 s

= 1389s 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).


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22.5. The rate of bulk vacancy diffusion decreases rapidly with decreasing tempera-ture because the activation energyQis 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. Becausethe activation energyQ is small for short-circuit diffusion, the rate of short-circuit diffusion decreases only slowly with decreasing temperature, so creepcontinues to occur at temperatures well below the transition from bulk toshort-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 creepand Designing metals and ceramics to resist diffusional flow).


23.1. See Section 23.1.23.2. See Section 23.3, paragraph 4, and Figs 23.4 and 23.5.



24.1.

Ficks first lawJ= Ddcdx

.

m =AJtwhereA = constant.m = A

D

dc

dx

t=AD

c1 c2b

t.


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bm =ADc1 c2t.b = BmwhereB= constant.Bmm =ADc1 c2t.B

2m2 =ADc1 c2t,

i.e.,m2 = CDt, whereC= constant.At constant temperaturem2 = kPt.

24.2. Ohms LawV= IR.m= PItwhere P= constant

m= PIt=PVtR

Rm= PVtR= QbwhereQ= constant

= SmwhereS= constantSmm= PVt

S

2m2 = PVt ie m2 = kPt

24.3.

AT 500C kP= 37 exp 138 103J mol1

8313J K1 mol1773K

kg2

m4

s1

= 174 108 kg2 m4 s1m2 = kPt

= 174 108 kg2 m4 s1 3600 24 365s= 050kg2 m4

m= 074kg m2

i.e., each square metre of metal surface absorbs 0.74 kg of oxygen from theatmosphere in the form of FeO. Number of oxygen atoms absorbed

=074kg m216/NA

, whereNA is Avogadros number.

Number of iron atoms removed frommetalas FeO = 074kg m2

16/NA.


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Weight of iron removed from metal= 074kg m2

16/NA 559/NA =

259kg m2

.

Thickness of metal lost= 259kg m2 m3

7870kg

= 033mmat 600C kP= 204 107 kg2 m4 s1

m2 = 204 107 3600 24 365kg2 m4= 643kg2 m4

m

=254kg m2

giving loss= 113mm

24.4. As shown in Table 24.1, gold is the only metal which requires energy tomake it react with oxygen 80kJ mol1 of O2 in fact). It therefore remains asun-reacted metal.

24.5. If there is any contact resistance across a pair of silver contacts, the surfacesof the contacts will be heated up by the current passing through the contactresistance. If the temperature goes above 230C, any oxide will decompose toleave pure metal-to-metal contact. Gold contacts would not form an oxide filmat 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 Cr2O3,produced by the chromium content of the stainless steel.

25.3. See Sections 25.2 and 25.3. Examples are Cr in stainless steel (Cr2O3 film),Cr in nickel alloys (Cr2O3 film), Al in aluminium bronzes (Al2O3 film).

25.4. See Table 24.2. The refractory metals oxidize very rapidly in air at hightemperature. Lamp filaments are surrounded by a glass bulb, which is eitherevacuated 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 coppersurfaces from oxidation.


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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.


28.2. IfPis the radial pressure that the olive exerts on the outside of the pipe thenwe can write

y =Pr

t

provided we neglect the strengthening effect of the sections of pipe that lieoutside the olive. If we assume that the end of the pipe far away from thefitting has an end cap (or a bend that functions as an end cap) then the forcetrying to push the pipe out of the fitting is pwr

2. This force is balanced bythe frictional force between the olive and the pipe so we can write

pwr2 = P2rl

Combining the two equations to eliminate Pgives

pw = 2y t

r

lr

Using the data given we get

pw = 2 015 120MPa

065

75

75

75

= 31MPa

The hydrostatic head of water in a seven-floor building is about 2 bar, so thejoint could actually cope with pumping water to the top of a seventy-floorskyscraper and still have a factor of safety of 1.5. However, the pressure inwater systems frequently exceeds the static head, often substantially, becauseof water hammer. This is the dynamic overpressure that arises when taps aresuddenly shut off.

28.3. (a) Typical examples are as follows. Car tyres/road surfaces, brake pads/brakediscs, clutches, shoe soles/walking surfaces, climbing shoes/rock faces,knots in ropes, V-belt drives, interference fits, compression joints (seeExample 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 watchmechanisms.

28.4. (a) Typical examples are as follows. Metal finishing by linishing, grinding andpolishing. Wood finishing by sanding. Removal of surface scale by gritblasting. Bedding-in of brake pads, clutch linings and plain bearings.


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(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:

s = tan24 = 045

If the slope of the roof is greater than 24 the static frictional force is exceededand the snow will slide off. On a 2 slope, with a ski already moving, it is thesliding friction which counts:

k = tan2 = 0035

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

V= 69330 106= 21 10

7 m3

If spread uniformly over the undersurface of two skis this would give a film ofthickness

21 1072 2 01 m = 05 m


29.5. Secondary roads are often covered in ice, because it is uneconomical to treatthem with rock salt. As shown in Example 29.3, as soon as the tyre starts to slideon the ice, it will melt the surface of the ice, reducing the friction coefficientto zero. The hard studs bite into the ice, so the (non-zero) friction coefficientis 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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