sol manual ashby jones eng mat-1 4e

8/13/2019 Sol Manual Ashby Jones Eng Mat-1 4e

1/52

Copyright 2012, Michael F. Ashby and David R.H. Jones. Published by E lsevier Ltd. All rights reserved

SOLUTIONS MANUAL

Engineering Materials I

An Introduction to Properties, Applications and Design, Fourth Edition

Solutions to Examples

2. 1. (a) For commodity A AA

( ) = exp ,100

rP t C t

and for commodity B BB

( ) = exp ,100

rQ t C t

where CAand CBare the current rates of consumption (t= t0) and P(t) and Q(t) are the

values at t= t. Equating and solving for tgives

A

B A B

100= ln .

Ct

r r C

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

D

100 70= ln 2 .t

r r

Substitution of the values given for rin 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 201years; 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.

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.


2/52


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

d= ,

d 100

C rC

t

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

00

( )= exp ,

100

r t tC C

where C0was the consumption rate at time t= t0.

(b) Set

1/2

0= d ,

2

tQC t

where

0= exp .

100

rtC C

Then

1/2

0

0

100exp ,

2 100

tQ rtC

r =

which gives the desired result.

2.4. See Chapter 2 for discussion with examples.

3.1. Refer to the results at the end of Chapter 3 for the elastic buckling of struts (pp 52 and

53), and second moments of area (pp 49 and 50). Appropriate situation is probably

Case 2 (left hand side drawing).

Fcr = 9.87 EI

l2

, I=

r4

4,

Fcr =9.87E

4

r

l

2

r2 =9.87 2 104 N mm2

4

8.5 mm

750 mm

2

8.52 mm2

=1439 N =148 kgf.

This gives a factor of safety of about 148/90 = 1.65, so he should be OK.

3.2. Refer to the results at the end of Chapter 3 for the mode 1 natural vibration

frequencies of beams (pp 50 and 51), and second moments of area (pp 49 and 50).


3/52


The appropriate situation is Case 2.

f=0.560 EI

Ml3, I=

bd3

12, M=lbd.

f=0.560 Ed2

12l4

1/ 2

=0.1617 dl2

E

1/ 2

,

E= f

0.1617

2l2

d

2

.

Because the frequency of natural vibration involves a force acting on a mass to give it

an acceleration, it is crucial when real numbers are put into the governing equation

that the basic SI units are used as follows

E= 4400.1617

2

0.085

2

0.00386

2

7.85 103=204 10

9 N m2 =204 GN m2.

3.3. This is a consequence of the equations of static equilibrium.

3.4. Principal planes have no components of shear stress acting on them. Principal

directions are normal to principal planes. Principal stresses are normal stresses acting

on principal planes. The shear stress components all vanish.

3.5. (a)

1 0 0

0 0 00 0 0

, (b)

1

2

0 0

0 00 0 0

, (c)

0 0

0 00 0

p

pp

, (d)

1

2

3

0 0

0 00 0

.

3.6.

0 0

0 0

0 0 0

. No.

3.7.

1

1

0 0

0 0.5 0

0 0 0

. There are no shear stress components normal to these axes.

3.8. Because the two shear strain terms on any given axis plane are defined so there is no

rotation.

3.9. Principal strains are axial strains. The shear strain components all vanish.

3.10. (a)

1

1

1

0 0

0 0

0 0

. (b)

0 0

0 0

0 0

.


4/52


5/52


.(1 2 )3 3(1 2 )

E EK p

p = =

This shows that K=Ewhen = 1/3.

3.16. The solid rubber sole is very resistant to being compressed, because it is restrained

against lateral Poissons 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 Poissons ratio expansion of each separate rubber cube can occur

without constraint (provided the gaps between adjacent cubes do not close-us

completely). So your colleague is correct.

3.17. The axial force applied to the cork to push it into the bottle results in a zero lateral

Poissons 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 Poissons ratio expansion, which can make it almost impossible to forcethe bung into the neck of the bottle.

4. 1. Refer to Fig. 4.11.

dForce between atoms = .

d

UF

r

At the equilibrium distance, ro, the energy Uis a minimum (i.e. Fis zero, and Uis the

dissociation energy Uo).

1 1

o

o o

o o

o

d

= = 0,d

or = .

1=

= 1 .

m n

n m

n m

m n

m

U mA nB

r r r

mB r A

n

A mU r A

r r n

A m

r n

+ +

+

Now, for ro= 0.3 nm, Uo= 4 eV.

2 2

20 2

5= 4 (0.3) = 0.45eV nm

4

= 7.2 10 J nm .

A

8 5 10 25 1015

= (0.3) 0.45 = 0.59 10 eV nm = 9.4 10 J nmB . Max force is at2

2

d= 0

d

U

r.

i.e. at value of rgiven by


6/52


1

1 18

2 2

o

( 1) ( 1) ( 1)= 0 which is =

( 1)

1 11= = 0.3 = 0.352nm.

1 3

n m

n m

m n

m m A n n B B n nr

r r A m m

nr

m

+ +

+ + + +

+

+ +

o

1

81

3 8

191

9

9

dand Force = = 1

d

2 0.45 (0.3)= 1 = 14.9eV nm

(0.352) (0.352)

14.9 1.602 10= J m

10

= 2.39 10 N.

n m

m n m

rU mA

r r r

+

4.2. The term A/rmis an attractive potential which depends on the type of bonding. The

B/rn 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 themoduli are

Material Calculated from 88 kTM/ MeasuredIce 1.0 1010N m2 7.7 109N m2

Diamond 9.0 1011N m2 1.0 1012N m2

The calculated values, for these extremes of elastic behaviour, are close to the

measured values. The important point is that the moduli are roughly propertional 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.

4.4. From eqn (4.3),2

2 1

0

d

d 4 nU q nB

r r r+

= .

Setting0

d

0d r r

U

r =

= gives

2

2 1

0 0 04 n

q nB

r r += .

Therefore,2 1

0

04

nq r

Bn

= as required.

From eqn (4.8),

0

2

0 2

d

dr r

US

r=

=

.

From eqn (4.3) and the result derived above forB,2 12

0i

0 0

1( )

4 4

n

n

q rqU r U

r n r

= + .


7/52


2 12

0

2 1

0 0

d

d 4 4

n

n

q rU q

r r r += .

2 12 2

0

2 3 2

0 0

( 1)d 2

d 4 4

n

n

q n rU q

r r r ++

= + .

Therefore, ( )2 2 2

0 3 3 3

0 0 0 0 0 0

2 ( 1) 2 1

4 4 4

q q n qS n

r r r

+= + = + + .

Finally,2 2

0 3 3

0 0 0 0

( 1)

4 4

n q qS

r r

= = .

4.5. See Section 4.2, last paragraph.

4.6. See Fig. 4.10.

4.7. See Section 4.5, paragraphs 1 to 3.

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= = 0.740.

6.2 2

(b) Glassy nickel is less dense than crystalline nickel by the factor 0.636/0.740. The

density is therefore 38.90(0.636/0.740) = 7.65 Mg m .

5.2. (a)


8/52


(b)

(c)

(d)

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

11 2= = 0.3524 nm.

2da

(b) The weight of one atom is


9/52


26

26

Atomic wt. 58.71= = 9.752 10 kg.

Avogadros number 6.022 10

There are 4 atoms per unit cell,

263

9 3

4(9.752 10 )so density = = 8.91Mg m .

(0.3524 10 )

(c) If the atom diameter is d2, then the lattice constant for the b.c.c. structure is

22

2= = 0.2866nm

3

da .

(d) One atom weighs

26

26

55.85= 9.274 10 kg.

6.022 10

There are 2 atoms per unit cell.

263

9 3

2(9.274 10 )So density = = 7.88 Mg m .

(0.2866 10 )

5.4. (a) Copper

Have 4 atoms per unit cell1

(88

from cube corners1

= 1 62

+ from cube faces = 3)


10/52


Atoms touch along cube-face diagonal:

This gives atom radius2

4=r a .3

3

3 3

44 100

16 2 2 1003Required percentage = =3 16 4

2= 100 74%.

6

ra

a a

=

Answer is same for magnesiumbothare close-packed srructures.

(b) Copper

Density 3

4

=

m

a

where mis atomic mass.Atomic weight for Cu = 63.54.

26

26

263

3

3 29 3

10

63.54kg= 10.55 10 kg.

6.023 l0

4 10.55 108.96 Mg m = kg.

= 4.71 10 m .

= 3.61 10 m, or 0.361 nm.

m

a

a

a

=

Magnesium

Have 6 atoms per unit cell1

(126

from corners1

= 2 22

+ from end faces = 1 + 3

inside)


11/52


2

2 2

26

25

263

2

2 29 3

3Volume of unit cell = 3 .

2

2 6 4Density = = .

3 3 3

24.31kg= = 4.04 10 kg.

6.023 l0

4 4.04 10 kg1.74 Mg m = .

3

= 5.36 10 m .

a c

m m

a c a c

m

a c

a c

In face-centred structure, plane spacing

3 2

= , where = .3 4

3 4Spacing .

3 2

a a

r

r=

In close-packed-hexagonal structure, plane spacing. 3 4 3 4

2 3 3 22 2

c r a= = = .

= 1.633,

1.633 .

c

ac a=

Using value for a2cof 5.36 10

29m3we obtain:

10

10

= 3.20 10 m or 0.320 nm;

= 5.23 10 m or 0.523 nm.

a

c

5.5. See Section 5.10.

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

(a) 1.014 V+ 0.84(1 V) = 0.92, giving V= 0.46, or 46%.

(b) 1.014 V+ 0.84(1 V) = 0.97, giving V= 0.75, or 75%.

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

f f

f m

composite f f f m

composite 1

= (1 ) (upper values);

1= (lower values);

V V

E E

E V E V E

E

+

+


12/52


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 fromwhich the upper values are derived (that the strainsare equal in the two components).

Note that the sets of values are widely separated near Vf = 0.5. 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= 0.5); it can be

more for woods.

Volume

fraction of

glass, Vf

Ecomposite

(GN m2

)

Ecomposite

(upper values)

(GNm2

)

Ecomposite

(lower values)

(GNm2

)

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 Vf)Em.

c=fVf+ (1 Vf)m.

(a) c= 0.5 1.90 Mg m3+ 0.5 1.15 Mg m3= 1.53 Mg m3

(b) c= 0.5 2.55 Mg m3+ 0.5 1.15 Mg m3=

.

1.85 Mg m3

(c) c= 0.02 7.90 Mg m3+ 0.98 2.40 Mg m3=

.

2.51 Mg m3.

(a) Ec= 0.5 390 GN m2+ 0.5 3 GN m2= 197 GN m2

(b) Ec= 0.5 72 GN m2+ 0.5 3 GN m2=

.

37.5 GN m2

(c) Ec= 0.02 200 GN m2+ 0.98 45 GN m2=

.

48.1 GN m2

.


13/52


6.3.

( )4.6Sectionsee;

V1

E

V1

1

1

1

+=

2EE

( ) 4.6Sectionsee;1 2111|| EVEVE += .

( ){ } ( )

++= 2

1

1

12111

\\

E

V

E

VEVEV

E

E 11

( ) .1221 2

+++=1

2

2

11111

E

E

E

EVVVV

( ) .221d

d

+=

1

2

2

11

\\

1

E

E

E

EV

E

E

V

,For 21 EE

( ).intpoturningonlythe5.0when0d

d==

1\\ V

E

E

V

6.4. to 6.8 Refer to Chapters 4, 5, 6 and the appropriate References.

7.1. By inspection from the results given on pp 48 and 49, the values of Care respectively:

Case 1 = 1/3; Case 2 = 1/8; Case 3 = 1/48; Case 4 = 5/384; Case 5 = 1/192; Case 6 =

1/384.

7.2. Refer to the results at the end of Chapter 3 for the elastic bending of beams (pp 48 and


14/52


49), and second moments of area (pp 49 and 50).

(a)3 4

3 3

2 4 2 2 2

2

3 2 2

1/2 1/21/2 2 21/2 5/2

1 1, ,

12

, / ,

1 1,

12

(12 ) .

Fl F EI E aC

EI C l C l

M a l a M l

F E M

C l l

FM C l

E E

= = =

= =

=

=

(b)3

3

3 3 3 3 3

3

3 3 3 3

1/3 1/31/3 3 31/ 3 2 2 /3

1

,12

, / ,

1,

12

(12 ) .

F E bd

C l

M bdl d M b l

F E b M

C l b l

FM C l b

E E

== =

=

=

(c)4

3

2 4 2 2 2 2

2

3 2 2 2

1/2 1/21/2 2 21/2 5/2

1 ,4

, / ,

1,

4

(4 ) .

F E rC l

M r l r M l

F E M

C l l

FM C l

E E

=

= =

=

=

(d)

3

3

3

3

4

2

1 ,

2 , / 2 ,

1,

2

2 .

F E r tC l

M rtl t M rl

F E Mr

C l rl

F lM C

r E E

=

= =

=

=

7.3. By inspection from the results given on pp 52 and 53, the values of Care respectively:

Case 1 = 2.47; Case 2 = 9.87; Case 3 = 39.5.


15/52


7.4. Refer to the results at the end of Chapter 3 for the elastic buckling of struts (pp 52 and

53), and second moments of area (pp 49 and 50).

(a)4

cr cr 2 2

2 4 2 2 2

2

cr 2 2 2

1/2 1/21/2 2 21/2 2

cr

, ,12

, / ,

1,

12

12( ) .

E E a

F C I F C l l

M a l a M l

E MF C

l l

M F lC E E

= == =

=

=

(b)3

cr 2

3 3 3 3 3

3

cr 2 3 3 3

1/3 1/31/3 3 31/3 5/3 2/3

cr

,12

, / ,

,12

12( ) .

E bdF C

l

M bdl d M b l

E b MF C

l b l

M F l bC E E

=

= =

=

=

(c)4

cr 2

2 4 2 2 2 2

2

cr 2 2 2 2

1/2 1/21/2 2 21/2 2

cr

,4

, / ,

,4

4( ) .

E rF C

l

M r l r M l

E MF C

l l

M F lC E E

=

= =

=

=

(d)

3

cr 2

3

cr 2

3

cr 2

,

2 , / 2 ,

,2

2( ) .

EF C r t

l

M rtl t M rl

E MF C r

l rl

lM F

C r E E

=

= =

=

=


16/52


17/52


proportional to t. The minimum value ofMcorresponds to the value of twhich gives

a cross sectional area just sufficient to carry the torque which the prop shaft is

required to transmit.

8.1. Nominal stress = Hardness/3 (1 + n) where n is the nominal strain as given in the

question plus 0.08.

Data for nominal stressnominal strain curve

Nominal stress (MN m2) 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 m2

(b) The strain at the stress maximum is

(the stress maximum of the curve).

0.6

(c) From eqn. (8.4),Aolo=Al, and

approximately.

o=

l Al A .

o o o

o o

o

( )/ = 0.6; / = 1/1.6.

1 = 0.38; ( )/ = 0.38.

l l l A A

AA A A

A

Thus percentage reduction in area = 38% .

(d) 109 MJ

from graph.


18/52


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 atthe much larger strains required to cause failure by cracking.

8.4.

(a) Tensile strength = 101.5 kN/160 mm2

= 634 MN m2.2mWorking stress = 160 MN .

(b) Load corresponding to 0.1%

Proof strain = 35 kN.

0.1% Proof stress = 219 MN m2.2mWorking stress = 131 MN .


19/52


8.5. The indentation hardness is defined by

2= ,

FH

a

where ais the radius of the circle of indentation. Simple geometry (see figure) gives

2 2 2( ) = ,r h a r +

or

2 22 = ,rh h a

or

rhr2

ah

2


20/52


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).


9.2. The fractional volume change is

5= (8.9323 8.9321) / 8.9323 = 2.24 10 .V

V

(a) Define the dislocation density as(m/m3or m2). Then21= ,

4

Vb

V

from which15 2=1.4 10 m .

(b) The energy is:2 31 3= = 2 = 2.1 MJ m .

2 8

VU Gb E

V

9.3. to 9.7. Refer to Chapter 9.

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

y

= 2 .

2So = .

bL T

T Gb

bL L

10.3. (a) For the alloy:8

10

9 2

8 2

y

y y

2

y

= 5 10 m,

= 2.86 10 m,

= 26 10 Nm .

So = = 1.5 10 N m .

But = 3 for polycrystals. Hence

450 MN m .

L

b

G

Gb

L


21/52


22/52


(b) Welded Interfaces

2 42 2

= 2 = 3 .

3 .

wL uFu wLku k

wLku wLku wLku

F wLk

+

+

General formula gives, for = 2w

d,

12 1 = 3 .

2

3 , verification demonstrated.

Fk k

wL

F wLk

+

11.3. 4000 MN m2= 350 MN m2. 14

w

d

+

.

= 10.43.4

= 41.7.

w

d

w

d

Take = 42w

dto produce an integral value of 2

w

din a safe direction.

20m= = = 0.48m.

42 42

wd

(This gives a volume fraction of cobalt of 7.2% a typical value for a rock-drilling

grade of WC-Co cement.)


23/52


11.4. (a) See Section 8.4.

(b) See Section 11.4.

d= .

d

11.5. = 3500.4MN m2.

At onset of necking =d

d

.

0.6 0.4

0.4 2 2

350 0.4 350 ,

or 0.4.

350 (0.4) MN m 242.6 MN m .

=

=

= =

n n

n

n

Nominal stress, = , where is the nominal strain.

(1 )= ln(1 ).

++

2

TS

2

2

242.6Tensile strength, = MN m

antiln (0.4)

242.6= MN m

1.492

= 163 MN m .

0

0.40.4 2

0

0.41.4

2

0

1.42

3

Work = d per unit volume

= 350 d MN m

= 350 MN m1.4

(0.4)= 350 MN m

1.4

= 69.3MJ per1m .


24/52


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

d=

d

at onset of necking. Thus

1d = =d

n nAn A

, so = n and = Ann. From eqn (8.13),

TS

n n

= = .(1 ) (1 )

n

An + +

From eqn (8. 15), = ln (1 + n) = n.

Thus (1 + n) = en. Finally,

TS=

n

n

An

e .

(b) InsertingA= 800 MN m2and n= 0.2 gives

0.22

TS 0.2

800(0.2)= = 475 MN m

e .

2

TS n TS= (1 ) = = 580 MN m .ne +

11.7. We have that yat 8% plastic strain isH/3 or 200 MN m2. Thus

2

0.2

200= = 334 MN m .

(ln1.08)A

Using the result of Example 11.6(a),

2

TS= = 198 MN m .

e

n

n

An

12.1.

f f

2

f

f

f

f

f

33f

f f

f f

= 2 .

= 4 .

= .2

= .2

4= = 2 .

2

tp

r

M r t

pt

r

p rt

r pM r p

Merit index is f .

12.2.


25/52


3 1/2

b 1/2

b

b 2

3

f f

f

ff 2

= 22.9 .

= .4

= 2 .

= .4

bM r pE

Mt

r

M r p

Mt

r

Set r= 1 m and pb=pf= 200 MPa. Values for ,Eand fare to be taken from thetable of data.

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

Al2O3 2.02 41 0.98 20 BucklingGlass 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 Al2O3with 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 (1.25/2)2 in2 = 1.23 in2. The

yield load of one bolt is 1.23 in2 11 tsi = 13.5 tons. The moment at yield is given by

(2 13.5 25) (2 13.5 19) (2 13.5 9) (2 13.5 3)

= 1512 ton in 126 ton ft.

M + + +

=

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 minimumcross-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

2

4

2(1500 N mm 1 mm 4.5 mm)

1.35 10 N.

T =

=

From eqn (11.2), the shear yield strength k= y/2. k for the pin is therefore 1500/2 =750 MN m2.

The yield load of the pin in double shear is obtained by multiplying k by twice thecross-sectional area of the pin to give


26/52


2

2 2 43.5= 2 750N mm mm = 1.44 10 N.2

T

This load is 7% greater than the load needed to yield the links and the strength of thechain is therefore given by the lower figure of 1.34 104N.

To estimate the tension produced in the chain during use we take moments about the

centre of the chain wheel to give

3

190 mm90kgf 170 mm ,

2

161 kgf 1.58 10 N.

T

T

The factor of safety is then given by

4

3

1.35 10 N= 8.5.

1.58 10 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

2 2 2 2= (2 1.2 ) mm 8.0 mm .A =

The force needed to shear this area is

2 2s= = 750N mm 8.0 mm = 6000 N.f kA

Finally, the failure torque is

5

s= 2( 10 mm) = 2 6000 N 10 mm 1.2 10 N mm

= 120 N m 12 kgf m.

f =

12.6. From Example 3.7, 26.303 71.76

944 MN m0.479

pr

t

= = = . y

12441.32

944S

= = = .

12.7.


27/52


2 2 2

y y

3

y

,

2 , ,2

2 ,

2 2 1.32 2 .

/ 7.8 / 1244 6.3 10

pr

t

MM rtl t

rl

rlprM

M pr l S pr l pr l

=

= =

=

= = =

= =

12.8. Because the applied stress is not unidirectional it is equi-biaxial (see Eqn 12.10,

and Fig. 12.5). When the applied stress is unidirectional i.e. when the only stress

acting is the hoop stress, and there is no axial stress. This stress state is satisfied when

the cylinder has no end caps the opposite of the conditions shown in Example 3.7.A good example is a long pressurised water main, with the axial load removed by

fixing the pipe to reinforced concrete thrust blocks. In such an application, CFRP is

clearly the winner, resulting in a pressure vessel approximately three times lighter

than one made from alloy steel or aluminium alloy. However, cost is still a major

issue, and may well rule out CFRP for other than special applications (e.g. aerospace).

12.9. Refer to the results at the end of Chapter 3 for the elastic bending of beams (p 47),

and second moments of area (pp 49 and 50). Refer to the results at the end of Chapter

11 for plastic moments (p 169).

(a)

y y

y y 3 2

2

y

y

2

y

p

p

y

612,

2

.6

.4

61.5.

4

M c MdM

I bd bdbd

M

bdM

M

M

= = =

=

=

= =

(b)


28/52


y y

y y 4 3

3

y

y

3y

p

p

y

44,

.4

4.

3

4 41.70.

3

M c MM r

I r r

rM

rM

M

M

= = =

=

=

= =

(c)

y y

y y 3 2

2

y y

2

p y

p

y

1,

.

4 .

41.27.

M c MM r

I r t r t

M r t

M r t

M

M

= = =

=

=

= =

12.10. See the section Plastic buckling on p 170. Once the metal yields, the slope of the

stress/strain curve decreases markedly. Because the buckling load is determined by

the slope of the stress/strain curve (Youngs modulus below yield, and tangent

modulus above) it decreases markedly as well.

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

2

3=

2

Fl

bt

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

c= ,Y a K

i.e. when

c2

3= .

2

Y Fla K

bt

Hence, the maximum load which can be sustained by the adhesive joint is

2

c2

= .3

bt KF

Y l a

For the joint shown


29/52


32 (0.1) 0.54.65 kN.

0.64 3 2 0.001F

= =

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

(a) = 500 MN m2for general yield.

(b) 2c 40

285 MN m ,1.12 0.005

K

Y a

= = =

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. 27000 1

= = 0.06 = 70 MN m2 3

pr

t .

c= , = 1K Y a Y .

2 2

c1 1 100= = = 0.65 m.70

Ka

a

13.4.

c

c

22

c

= ,

= 1,

= ,

1 1 30MPa m

= = = 0.080 m = 80 mm.60MPa

2 = 160 mm.

K Y a

Y

K a

K

a

a

13.5. (3 1.0 + 4) = 7.0 MN m2. K=Y a =1.2 7.0 0.010 =1.49 MN m3/2.

This is 15% more than the value of Kcobtained from tests. Experimental scatter in the

test data, and errors in the stress analysis are more than enough to account for this

small discrepancy.

13.6. The total stress on the crack is 285 + 100 = 385 MN m2

. Fast fracture occurs whenthe crack length ac is given by Kc=Y ac . For a semicircular surface crack in a

thick plate (a= 6 mm


30/52


14.3. 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.

14.4. The mode of failure is shown in the diagram and the photograph.

The collapse of the railing was caused by the propagation of pre-existing cracks in the

end posts, which are safety-critical components. The collapse was caused by a lack of

awareness by both architect and maintenance personnel of the poor along-grainfracture toughness of wood, the potential sources of crack-opening forces in the

design, the propensity of externally exposed timber to develop cracks and the

significance of cracks for the integrity of the structure.

You could recommend fixing 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. This designs-out the end

posts and their problems entirely.

14.5. 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.

15.1.From eqn (15.1)

1/2

cTS

6

3MP a m= = = 276 MPa.

1.12 1.12 30 10 m

K

a

15.2. From eqn (15.6)

A horizontal outward force applied to

the to rail tends to s lit the end ost.

The crack path in a failed end post


31/52


rr 2 2

3 3

2

6 1= = 6

2 2

330 N 50 mm 1= 6

2 2 5 mm

N= 198 = 198 MPa.mm

M F l

bd bd

15.32 2

2 2

o

(11/ 2) mm 50mm= = 9.7

(5 / 2) mm 25mm

V

V

.

For the test specimens, eqn (15.4) gives

o

o 0

0.5 = exp .

m

V

V

For the components, eqn (15.4) gives

o 0

0.99 = exp .

m

V

V

Thus

o o 0

o 0

ln(0.5) = .ln(0.99)

169.0 = .

9.7

1( ) = ( ) .

669

= 0.272 120 MPa 32.6 MPa.

m m

m

m m

V VV V

=

15.4. See Answer to this Example.

15.5. Specimen measuring 100 mm 10 mm 10 mm will have median TS of 300Mpa/1.73.

Specimen volume V= 104mm3. Eqn (15.4) then gives

4 310

TS10

o o

10 mm0.5 = exp .

V

Taking natural logs gives


32/52


4 310

TS10

o o

10 mm0.69 = .

V

Component volume = 1.25 103mm3. Pf= 106gives Ps= 1 10

6. Thus

3 36 10

10

o o

1.25 10 mm1 10 = exp .

V

Taking logs, and assuming that ln (1 x) = xfor smallx, gives

3 36 10

10

o o

1.25 10 mm10 = .

V

Thus

104 310 o o

TS6 10 3 3 10

o o

104

TS

3

4 610 10

TS3

1 1

TS

0.69 10 mm 1=

10 1.25 10 mm

10= .

1.25 10

10 10= .

1.25 10 0.69

300 MPa= 3.21 10 = 3.21 10

1.7355.7 MPa

V

V

=

15.6.(a) Weight of material below section atxis

2( ) .3

g x x

Cross-sectional area = (ax)2. 1

3Stress force/area gx= = .

(b) Integrate over the volume, using disc of thickness dxwith volume dV= (x)2

dx.Eqn (15.5) then gives

m m2

s

0 0 0 00

d ( ) d ( ) = exp = exp .

3

L

V

V gx x xP L

V V

Integrating gives

m2 m 3

s

0 0

( ) = exp .

3 (m 3)

g LP L

V

+

+


33/52


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.

16.1.From Section 13.3, =K Y a .

(a) From Fig. 16.3, 3/2/ = 5/10 = 0.5, = 3. = 3 100 0 005 = 37.6 MN ma W Y K

(b) From Section 13.3, Y= 1.

(c)3/2=1 100 0.020 = 25.1MN m .K

16.2.See Section 16.3 (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.

16.3. 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.

16.4.From Section 16.3 (p 235), the maximum (hoop) stress occurs at the bore of the

cylinder, and is given by

2 2 2 22 2

2 2 2 2

279 1271.52 1.52 83 MN m 126 MN m .

279 127

A Bp p p

A B

+ += = = = =

Fracture will commence at the bore of the cylinder when the hoop stress reaches the

tensile strength of the material, i.e. the tensile strength = 126 MN m2.

Grey cast iron is too brittle a material to use for making a pressure vessel. The fracture

toughness is typically 10 MN m3/2(see Table 13.1 on p 195). The critical defect size is

given by the equation Kc a (see p 235), i.e.2 2

c1 1 10 0.002 m 2.0 mm126

Ka

= = = =

. This is far too small a critical defect

size a low-tech casting process cannot achieve this degree of control over castingdefects.

17.1.Answers obtained by reading directly off the graph in the diagram.

17.2.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


34/52


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 tohave a smaller current rating and failure does not always result in a fire.

17.3. Fast fracture occurs when the (final) crack length a2 is given by c max 2K Y a = ,

where maxis the maximum tensile stress in the loading cycle. For a long surface crack

in a thick plate, Y = 1.12 (see Case 3 on p 199). Therefore,

2 254 1.12 180 , 0.0228 ma a= = .

The initial crack length a1 is 2 mm, which we need to write as 0.002 m in the crack

growth equation in order to get the units right.

We integrate the crack growth equation over the range of crack lengths from 0.002 m to

0.0228 m to find the number of cycles. This is done as follows

2

1

2

1

3 3 3 3/2

3

3/2

3

1 2

12 3

5 5

d( ) ( ) ( ) ,

d

d( ) d ,

1 1 1

2 2 ( ) ,

1 12 31.48 6 10 (1.12 180 ) ,

0.002 0.0228

1.15 10 , rounded up to 1.2 10 .

a

a

a

a

aA K A Y a A Y a

N

aA Y N

a

A Y Na a a

N

N

= = =

=

= =

= =

=

17.4.(a) Circular = 0730 hrs position. Rectangular = 0600 hrs position.

(b) Circular = 1300 hrs position. Rectangular = 1200 hrs position.

The maximus stress is greater in the rectangular tool than in the circular tool.

17.5. (Nf)a= C.


35/52


5 7

72

5

5 0.073 2

2

13.71/2 8

f 1/

280(10 ) = 200(10 ) .

280 10= 1.4 = = (10 ) .

200 10

log 1.4 = 2 = 0.146.

1= 0.073 or .

13.7

= 280(10 ) MN m .

= 649 MN m .

649At 150 MN m , = = = 5.2 10 cycles.

150

a a

a

a

a

a

a

a

C

C

CN

.

17.6.The 1 m marker at the bottom LHS of the photograph measures 3.0 mm in length.

Take a length of 20 mm, and count the number of striation spacings over this length onvarious parts of the photograph. The number of spacings is typically about 14. One

spacing therefore corresponds to about 20 mm divided by 14, i.e. about 1.43 mm. This

corresponds to a real length of about 1.43/3.0 m, i.e. about 0.48 m, which should be

rounded up to 0.5 m.

17.7. The maximum thermal shock that could be experienced by the boiler plate next to the

feed water inlet pipe is 180 10 = 170C. This produces a thermal strain in the platesurface of T = 12 106 170 = 2.04 103. The strain required to give yield is

2y 3

y 2

250 MN m1.25 10

200 GN mE

= = = . This is only 61% of the maximum thermal strain.

Since the thermal shock is cyclic (it occurs every time water is put into the boiler) it is

likely that the cracks formed and grew by low-cycle fatigue, with 104cycles required

for failure (see Fig. 17.4). Over a three-year period, we would only have to put water

into the boiler ten times a day to achieve this number of cycles. In practice the number

of cycles could have been much more. The baffle would have made the cold water

sweep sideways over the inner surface of the boiler plate, ensuring efficient cooling.

The problem would have been much less serious if a baffle had not been fitted.

18.1.(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 andreduce the fatigue life.

(c) A mean tensile stress will decreaseNf(see eqns (17.3) and (17.4)).

(d) Corrosion may reduce the fatigue life, by creating pits in the surface from which

fatigue cracks can initiate more easily.

18.2.The maximum pressure in the cylinder occurs at the point of admission. The maximum

force acting on the piston is therefore given by

2 20.7 N mm (45 mm) 4453 N. =


36/52


The stress in the connecting rod next to the joint is

24453 N / (28 mm 11 mm) = 14.5 N mm .

Since the locomotive is double-acting the stress rangeis twice this value, or 29 N mm2.

The number of revolutions that the driving wheel is likely to make in 20 years is (20

6000 1000m) / ( 0.235m) = 1.6 108.

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 mm2, 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.

18.3.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 mm2.

The mean-line fatigue curve gives the data for a 50% chance of cracking. For the stress

range of 8 N mm2the 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 109and the time to failure is therefore 4 years

.

18.4.

(a) Because the shoulder combined with the zero fillet radius produced a stress

concentration (see the bottom diagram in Fig. 18.1).

(b) At the top of the 4 mm diameter section the position of maximum tensile

bending stress.

(c) Eliminate the shoulder, by making the arm from rod having a uniform diameter of

2 mm, and fixing it into the roller with an adhesive like Loctite (this has the

counter-intuitive consequence that removing material actually increases the

fatigue strength).

(d) Titanium alloy (see Table 18.1). This has a maximum fatigue strength of 600 MN

m2

compared to a maximum fatigue strength for aluminium alloy of 170 MN m2

,yet is only about 50% heavier (see Table 5.1).

=4Fl

c3=

4 9.81157

23=90 N mm2 =90 MN m2.

The stress amplitude is therefore 45 MN m2 and the mean stress is 45 MN m2

(tensile).

Refer to Eqn (17.4),


37/52


m2

=(f

' m )

2f'

=

21

mf

'

,

m2

=

21

45

250

=0.82

2,

2=

1

0.82m

2=

1

0.8245 =55 MN m2.

18.5.

plate =F

Lt, throat =

F

2gL.

For plate = throat,F

Lt=

F

2gL, and t=2g=

2a

2= 2a,

t

a

= 2.

Refer to the weld fatigue curves of Fig. 18.4 (50% survival). At 107cycles, for classW = 32 MN m2, and for class F2 = 49 MN m2. For the same fatigue lives in theT joint, we therefore need

platethroat

=49

32=

F

Lt

2gL

F=

2g

t=

2a

t,

t

a= 2

32

49=0.92.

Note that the same result would be obtained for any given number of cycles, because

the fatigue curves are plotted on log-log scales, and the lines for the two weld classes

are parallel to one another, i.e. the ratios of the fatigue strengths are independent of the

number of cycles.

18.6.Because the hole produced a stress concentration. At the side of the hole (see the top

diagram in Fig. 18.1). In uniaxial loading, a hole produces a stress concentration factor

of 3 at the sides of the hole (see the top diagram in Fig. 18.1). Therefore the margin of

safety would have been approximately 3.

18.7. See Section 18.7 for how to design bolted connections so the bolts see only a smallfraction of the applied loading cycle. The bolts should be made from high tensile steel,

and should be waisted. This increases their give under tension, and also eliminates

unnecessary threads. For maximum fatigue resistance, the threads should be rolled, and

the fillet between the shank and the head should be roller peened (see Section 18.6,

paragraphs 1 and 2). Contact between the bolt head and the flange should only take

place on the flat part of the head, and not on the under-head fillet. This may require the

hole in the flange to be chamfered back to clear the fillet.

In spite of the above measures, the load shielding of the bolts is limited by their short

length, which is defined by the combined thickness of the flanges. The design

modification involves making the bolts much longer, so they have more give, and seeless of the applied loading cycle. This can be done by inserting a length of thick-walled


38/52


39/52


pin. The maximum value of this bending stress occurs at the lower end of the

reduced section. In addition, the sharp change of section at this location inrroduces

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 fullyopen 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.See Fig. 17.6, Fig. 17.8 and Section 17.4, paragraph 1.

19.4.The cracks were located in the welds between the trunnion shaft and the gussets on the

driving side. This is because the gussets are narrower on the driving side than the idling

side (100 mm compared to 180 mm), so the stress in the welds is greater on the drivingside than the idling side. The exact geometry of the cracking is given in Example 18.5,

where it is shown as either a class W or a class F2. In practice, the cracking is most

likely to be a class W, since T joints rarely have a leg length sufficiently large for

class F2 cracking to occur.

19.5.The effective fatigue stress amplitude at each fillet is given by the actual stress

amplitude multiplied by the effective stress concentration factor for fatigue loading (see

Section 18.4). Since failure did not occur in the 90 mm section after 10 million cycles

(and will not occur even if the number of cycles goes above 10 million), we can

determine the fillet radius required to avoid failure at the 81 mm section by equating the

effective fatigue stress amplitudes for the two fillets as follows

90 81eff,90 eff,81

SCF SCF2 2

= .

Then, 90eff,81 eff,90

81

SCF SCF

=

.

From the equation given for the maximum bending stress,

3

90

81

81= 0.729

90

=

.

Eqn (18.1) gives SCFeff= S (SCF 1) + 1.

The look-up table can be used to find values of S, and SCF corresponding to the twosteps in diameter, for a range of fillet radii. For the 2.6 mm fillet at the 180/90 step

(diameter ratio = 2), S= 0.81, and SCF = 2.64 (linear interpolation). This gives SCFeff,90

= 2.33. Then, SCFeff,81= 0.729 2.33 = 1.70.

We now need to use the look-up table to reverse engineer the fillet radius at the 90/81

step (diameter ratio = 1.11) which gives a SCFeffof 1.70. As a first trial, a fillet radius

of 4.5 mm gives S = 0.85 and SCF = 1.83. SCFeff is then found to be 1.71, so the

required radius is indeed 4.5 mm.

If the actual fillet radius has to be limited to 2.5 mm, the situation is not as bad as it

might appear, because the 2.6 mm fillet had obvious lathe tool markings. As long as thesurface at the 2.5 mm fillet does not show lathe tool markings, it will have a better


40/52


fatigue strength, other things being equal, than the 2.6 mm fillet. The relative fatigue

strength at the 2.5 mm fillet can be further improved by polishing the machined surface

(see Section 18.2, paragraph 1) and/or subjecting it to roller peening in order to

introduce a compressive residual stress (see Section 18.6, paragraph 2).

20.1.Temperature (C) T(K) 1/T(K

1) 1(s ) ln

618 891 0.00112 1.0 107 16.12

640 913 0.00110 1.7 107 15.59

660 933 0.00107 4.3 107 14.66

683 956 0.00105 7.7 107 14.08

707 980 0.00102 2.0 106 13.12

510 783 0.00128 8.3 1010From graph 20.90

The hoop stress in the tube is

2

2

10 1 2

5

2 10 1

12 1

12

4

6 MN m 2 0mm= =

2 mm

= 60 MN m .

at 510 C, 8.3 10 s at 200 MN m

60Under 60 MN m , = 8.3 10 s .

200

= 2.0 10 s .

Strain in 9 years = 2.0 10 60 60 24 365 9

= 5.7 10 or 0.0057.

pr

t

= =

.

Design safe


41/52


20.2.At 25 MN m2and 620C, 12 1= 3.1 10 s .5

2 o 12 1

12 1

30At 30 MN m and 620 C, = 3.1 10 s

25

= 7 71 10 s .

5 /= e .Q RTA

1 2

1 2

2 1

1 2

3

12 1

1 1ln ln = .

1 1ln = ln ,

160 10 1 1=

8.313 893 923

ln(7 71 10 s )

= 0.700 25.58 = 24.88.

Q

R T T

Q

R T T

+

+

12 1 2

2= 15.5 10 s at 30 MN m

and 650 C.

12 1

12 1

12 1

70= 3.1 10 s 15.5

100

3010 s

100

= 6.82 10 s .

+

20.3.From Table 20.1, the softening temperature of soda glass is in the range 700900 K.

The operating temperature of window glass is rarely more than 293 K, so T/Tsis at most

0.42. Ceramics only begin to creep when T/TM> 0.4. (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. 20.10.

20.6.A major fire would increase the temperature of the steelwork to the point at which itwould 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.

20.7.The linear elastic stress analysis ignores the fact that the material will deform by plastic

deformation and creep, which will keep the actual stress well below the tensile strength.

For this reason, the correct failure criterion is the one based on strain, not stress.

21. 1. Measure the rate by the mass injected per second, .M

tThen, if this rate follows an

Arhennius Law, we have


42/52


= exp .M Q

At RT

Or, combining the constantsMandA:

1= exp .

QB

t RT

Then, converting temperatures to kelvin:

11 = exp s ,30 450

QB

R

11 = exp s .81.5 430

QB

R

Solving for QandB, using = 8.31JR mo 1 11 K , we have4 1

7 1

= 8.04 10 J mo1 ,

= 7.25 10 s .

Q

B

We may now calculate the time required to inject the mass M of polymer ato227 C(500K) :

47 11 8.04 10= 7.25 10 exp s ,

8.313 500t

giving t= 3.5 seconds.

21.2.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 interstitialsolid 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

substitutionalsolid 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 2= /t x D .3 1

/ 2 1

o 1 1

159 10 J mo1= = 9.5mm s exp

8.313J K mo1 1023 K

Q RTD D e

2 1

8 2 1

89.5mm s= = 7.20 10 mm s .1.32 10


43/52


44/52


24. 1.

Ficks first lawd

=d

cJ D

x .

( ) =m AJ t whereA= constant.

1 2d

( ) = = .d

c ccm A D t AD t

x b

1 2

( ) = ( ) .b m AD c c t

=b B m whereB= constant.

1 2( ) = ( ) .B m m AD c c t

2

1 2( ) = ( ) ,

2

Bm AD c c t

i.e., 2( ) =m CDt , where C= constant.

At constant temperature 2P

( ) = .m k t

24.2. Ohms Law =V IR .

( ) =m PI t where P= constant.

( ) = = .PV t

m PI t R

( ) = .R m PV t

=R Qb where Q= constant

=S m where S= constant.( ) = .S m m PV t

2( ) = ,2

Sm PVt i.e., 2

P( ) = .m k t


45/52


24.3.3 1

o 2 4 1

P 1 1

8 2 4 1

2

p

8 2 4 1

2 4

2

138 10 J mo1AT 500 C, = 37 exp kg m s

8.313J K mo1 773K

=1.74 10 kg m s( ) =

=1.74 10 kg m s 3600 24 365 s

= 0.50 kg m .

= 0.74 kg m

k

m k t

m

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 =20.74 kg m

,16 /

AN

whereNAis Avogadros number.

Number of iron atoms removed from metalas20.74 kg m

FeO = .16 /

AN

Weight of iron removed from metal20.74 kg m

= 55.9 / =16 /

A

A

NN

2.59kg m2.

Thickness of merallost2 32.59 kg m m

=7870 kg

0.33 mm.=

at

o 7 2 4 1

P600 C, = 2.04 10 kg m s .k

2 7 2 4( ) = 2.04 10 3600 24 365kg m ,m 2 4= 6.43kg m .

= 2.54m kg 2m ,giving loss= 1.13 mm.

24.4. As shown in Table 24.1, gold is the only metal which requiresenergy to make it react

with oxygen (80 kJ mol1of O2in 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 230C, 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 is2 3

Cr O , produced by


46/52


the chromium content of the stainless steel.

25.3.See Sections 25.2 and 25.3. Examples are Cr in stainless steel (2 3

Cr O film), Cr in nickel

alloys (2 3

Cr O film), Al in aluminium bronzes (2 3

A1 O 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.

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 Table24.2).

25.7.Many are oxides already, e.g.2 2 3

MgO,SiO ,A1 O (see Table 24.1). Others, e.g. SiC and

Si3N4 form protective oxide layers of 2SiO when exposed to oxygen at high

temperature.

26.1.d

d

aa

t at constant (4 MN m2).

2d

d

a

t at constant a(0.25 mm).

2 2d = = ; = 2.d

aA a AK n

t

1

2 2 2

d 1 0.3mm year = =

d 16(MN m ) 0.25mm

aA

t a

.)mmMPa(yearmm0239.0 21 =A

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 o20 C (summer time, heating off) to o70 C (winter

time, heating on). The corrosion rate at o70 C will be approximately twice that at o20 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 26.7 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 26.7 and Fig. 26.12. Even if the

stress corrosion cracks do not travel right across the component, they can still propagateto failure by fatigue or fast fracture.


47/52


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.

(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 26.7 and Fig. 26.12. 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 alarge tensile stress to drive the initiation and growth of SCC cracks.

26.7.

(a) See Section 26.6 and Table 26.1. The standard electrode potentials for these three

metals are all greater than the SEP of the oxygen reduction reaction.

(b) Because of the protective effect of some oxide films See Section 26.4, last

paragraph.

(c) See Section 26.5, next to last paragraph, and Fig. 26.10. The surface oxide film is

only stable between these two pH values.

(d) See Section 26.5, paragraph 3, and Fig. 26.7. The surface oxide film is only stable

between these two pH values.

26.8.(a) See Section 26.4, first paragraph the three bullet points taken in that order define

the terms (i), (ii) and (iii).

(b) See Section 26.5, paragraph 3.

(c) See Section 26.5, last two paragraphs, and Fig. 26.11.

(d) Because it eliminates the most common cathodic (electron accepting) reaction the

oxygen reduction reaction.

27. 1. See Section 26.2 and Fig. 26.2. Because there was no oxygen in the system, theoxygen-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 27.1

27.3.See Section 27.1 and Fig. 27.4.

27.4.Corrosion of zinc:

Zn Zn 2e.++ +

Number of electrons to give a current of 36 10 A m2for 5 years3 7

24

19

6 10 5 3.15 10= = 5.89 10

1.6 10

elecrrons


48/52


245.89= 102

atoms Zn.

Mass zinc24

26

5.89 10 65.4= = 0.320

2 6.02 10

kg 2m

Thickness 50.320= m = 4.5 10 m7130

= 0.045 mm.

27.5. Following 27.4, if steel were lost uniformly over a square metre at 32 10 A 2m ,thickness lost by reaction:

Fe Fe 2e++ + would be 0.0116 mm on each side of the plate.

If this loss is concenrrated over 0.5% of the surface, the loss there will be

100 99.50.0116 mm 2.31 mm.

0.5 100 =

The sheet will thus rust through

.

27.6.See Section 27.4, and Fig. 27.12.

27.7.The following table gives the mass of iron per unit area of corrosion product for each

flake.

Sample number Mass of iron (mg mm2)

1

2

3

4

5

67

8

1.195

0.632

0.629

0.761

0.540

0.9830.670

0.864

The average value is 0.784 mg mm2.

The mill scale that covers an area of 1 mm2has a volume of 1 mm2x 0.10 mm = 0.1

mm

3

. The mass of Fe2O3in this volume is 5.26 g cm

3

x 0.1 mm

3

= 0.526 mg. Giventhat the atomic weight of iron is 55.85, the mass of iron in the volume of scale is

2 55.850.526 mg=0.368 mg.

2 55.85 3 16

+

The mass of iron taken up by the corrosion process is therefore 0.784 0.368 =

0.416 mg. The mass of iron is then divided by the density of iron (7.78 g cm3) to give

the volume of iron removed by corrosion = 0.053 mm3. The surface area over which

the volume of iron is lost is 1 mm2, so the thickness of the steel lost by corrosion is

0.053 mm, which is rounded down to 0.05 mm.


49/52


27.8.At temperatures above 70C, austenitic stainless steel is susceptible to stress corrosioncracking (SCC) in the presence of tensile stress and chloride ions. Failures have been

reported even with low levels of stress and chloride ion concentration. In the present

case, the tensile stress was provided by the residual stresses from the welding process.

In addition, large concentrations of chloride ion were found. The chloride ions

originated from the sea water and were carried under the cladding by leaking rain water.There is little doubt that chloride-induced SCC must have been the cause of the failure.

See Section 26.7, last paragraph, and Fig. 26.12. See also Example 26.6.

The nitric acid, because it is a powerful oxidising agent, actually protectsthe stainless

steel by artificially thickening the protective oxide film. See Section 27.2, paragraph 4.


28.2.If P is the radial pressure that the olive exerts on the outside of the pipe then we can

write

=YPr

t

provided we neglect the srrengthening 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 2w

.p r This force is balanced by the frictional force between the olive and the

pipe so we can write2

w = 2 .p r P rl

Combining the two equations to eliminate Pgives

w = 2 .y t lpr r

Using the data given we get

.MPa1.35.7

5.7

5.7

65.0MPa12015.02w =

=p

The hydrostatic head of water in a seventy-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.

(b) Typical examples are as follows. Bearings and sliding surfaces in machinery, car


50/52


tyres, brake pads and discs, clutch linings, shoe soles, tools for metalworking and

woodworking, grinding wheels and abrasive belts/discs.

28.5. Let be the depth of metal lost from the surface of each contacting/loadbearing part

(pin or hole). Then it can easily be shown (draw a sketch of one pin, plus the two

links it passes through) that the elongation or stretch per 12.5 mm repeat unit of thechain is equal to 4.

The total stretch is 4x 110 = 440= 25 mm.

Therefore, = 25 mm/440 = 0.057, or 0.06 mm rounded up.

28.6. The seal is only perfectly gas-tight or liquid-tight when there is atom-to-atom contact

between the surfaces over the whole of the nominal area of contact. In this case, eqn

(28.3) is modified to Py = Ay , whereA is the nominal area of contact, and Py the

normal force is also the yield load.

The yield load of the bolt is 7.52 mm2 280 N mm2 = 49,480 N .The yield load of the washer is (112 7.52)mm2 50 N mm2 =10,150 N .

The required ratio is therefore 10,150/49,480 = 0.21.

When the gasket is compressed, it work hardens. The next time it is used, it will not

yield when the bolt is torqued up. It is heated to dull red heat to annealit - to remove

the work hardening, and restore the yield strength to its original value.

28.7. Also see Example 24.5. In contrast to the situation for silver, with copper, brass andin fact all metals other than silver and gold the oxide film is stable when the

contacts overheat. In fact, if the contacts overheat, the oxide films will grow, and the

contact resistance will increase. This can lead to a run-away situation, ultimately

causing connection burn-out. The answer is to do the screws up tightly, to increase

the actual area of contact abetween the surfaces, and make sure that the contacts can

transmit the current without heating up to a temperature at which the oxide films can

grow significantly.

28.8. Adhesive wear between the bore of the wheel and the surface of the axle. Because the

interference of fit was reduced by the additional pressing-on and pressing-off

operations.

28.9. Use a shrink-fit assembly process. The machined interference is 0.25 mm at a

workshop temperature of, say, 20C. Heat the compressor wheel up to, say, 140C (atemperature rise of 120C). Then the expansion of the bore of the wheel is110 mm 22.5 106 120 =0.297 mm. This is greater than the machinedinterference by approximately 0.05 mm (2 thou), so the bore of the wheel will clear

the shaft. Position the shaft vertically, with the compressor wheel seat uppermost. It

should then be possible to drop the compressor wheel over the shaft (it will stop when

it comes up against the step at the end of the seat). It is important that the wheel does

not lose any heat to the shaft during this process, otherwise it will contract and grabthe shaft during assembly - not a good idea! Hence the need to position the wheel


51/52


above the shaft so it can be allowed to fall down under gravity to the correct final

position.

28.10. The cycle tyres slip over the galvanised steel bars when they are wet because the

contact areas between the rubber and the zinc are partially separated by a thin film of

water. The contact areas are effectively lubricated by boundary lubrication (seeSection 28.4, paragraph 3). This should have been foreseen. Ironically, if the steel

had not been galvanised, but had been allowed to rust, the roughness of the rust layer

might have helped to prevent sliding (but note that heavy traffic might have worn

away the rust layer, and polished the surface of the steel).

The solution which was adopted is shown in the photograph.

The steel bars were all cut away from the frame, and new bars having a thread

machined along their length were welded into the frame (the threaded bars were also

galvanised to stop them rusting). The crests of the threads indent the rubber of thetyres (elastically), and stop them sliding sideways. More information on why this

works is given in Chapter 29, Section 29.4.

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:o

s= tan 24 = 0.45.

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:o

k= tan2 = 0.035.

29.3. The work done

k= 2(100g ) = 68J .

This melts a volume of water

7 3

6

69= = 2.1 10 m .

330 10V

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

thickness72.1 10

m = 0.5 m.2 2 0.1



52/52

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. 28.9).

29.6. See Section 29.2, paragraphs 1 and 2.

29.7. See Section 29.2, paragraphs 4 to 9.

29.8. When only one interface has sticking friction,

Fu2wLku+ 2 wL

2k

u

2=2wLku+

wLku

2=2.5wLk ,

F 2.5wLk.

This result was derived for the case of w/d= 2 (see Example 11.2). Inserting this

value of w/dinto the general formula,

F 2wLk1+ w

8d

, gives F2.5wL as

required.

For d= 0.5 mm and w= 40 mm, w/8d= 10. Then

F 2wLk(1+10) =22wL .

The yield load of the unconstrained sample is obtained by setting w/8d= 0, giving

F2wLk. The ratio of the constrained to unconstrained yield loads is then22wLk/2wLk= 11.

29.9. The main features of the damage are that large thin pancakes of white metal have

come away from the bearing. Extensive cracks have formed parallel to the bearing

surface, at a relatively small distance below the surface. Other cracks have initiated at

the bearing surface and propagated into the white metal at right angles to the surface.

These cracks have allowed the detachment of the white metal pancakes.

The likely mechanism of cracking is fatigue (see Section 29.2, last paragraph). This

could have been caused by misalignment of the propeller shaft (the vessel had been

involved in a grounding incident, which had bent the blades of the propeller).

29.10. When the rails are damp, the contact areas between the wheels and the rail arepartially separated by a thin film of water (which probably also contains oily residues

dropped from locomotives which have passed over the track before). The contact

areas are effectively lubricated by boundary lubrication (see Section 28.4, paragraph

3).

Sand particles are small and hard, and tend to bite through the surface film. Any

slipping between wheels and rail will be resisted by the ploughing action of the

sand over the metal surfaces (see Section 28.4, paragraphs 8 to 10, and Fig. 28.9).

sol manual ashby jones eng mat-1 4e

Documents