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Unformatted text preview: Problems  Chapter 5
l. FINB: Calculate the stress on a tensioned ﬁber. GIVEN: The ﬁber diameter is 25 micrometers. The elongationai load is 25 g.
ASSUMPTIONS: The engineering stress is requested. DATA: Acceleration due to gravity is 9.8 111/8602. A Newton is a kgm/secZ. A
Pascal is s N/m2. A We is 106 Pa.
SOLUTION: Stress is force per unit area. The crosssectional area is 1:112 = 1963.5 square
micrometers. The force is 25 g (kg/loongss m/sch) = 0.245 N. Thus, the stress is
0.24SN :O‘um =F/A= ,_.—.. ' 0.1253460
0 1960mm2x m ' a COMMENTS: You must team to do these sorts of problems, including the conversions. 2. GIVEN: FCC Cu with a, = 0.362nm
REQUIRED: A) Lowest energy Burgers vector, B) Length in terms of radius of Cu
atom, C) Family of planes
SOLUTION: We note that the Burgers vector is the shortest vector that connects
crystallographically eqmvalent positions. A diagram of the structure is shown below: FCC structure with (111) shown We note that atoms lying along face diagonals touch and are crystallographically
equivalent. Therefore, the shortest vector connecting equivalent positions is V2 face diagonal. For example, one such vector is 921 [710] as shown in (11 I). 2 2
A. The length ofthis vector is %+9‘1 =£i=m=a256nm B. By inspection, the size of the vector is 2 Cu atom radii. C. Siip occurs in the most densely packed plane which is ofthe type {Ill}. These are the 81 smoothest planes and contain the smallest Burgers vector, This means that the
dislocations move easily and the energy is low. 3. GIVEN: bl = 0.288nm‘in Ag
REQUIRED: Find lattice parameter SOLUTION: Recall the Ag is FCC. For FCC structures the Burgers vector is V: a face ﬂag: “9
2 J3 raft/3b b: = 2x0.288 a0 = 0. 4o 7nm diagonal as shown. We see that 4. A. FCC structure The (111) plane is shown in a unit cell with all atoms shown. Atoms touch along face
diagonals. The (1 l 1) plane is the most closely packed, and the vectors shown connect equivalent atomic position. Thus 11 = :—[ l1 0] etc. Then in general I) = g— < 110 > B. For NaCl We see that the shortest vector connecting equivalent positions is %[ l1 0] as shown. This direction lies in both the {100} and {110} planes and both are possible slip planes.
However {110} are the planes most frequently observed as the slip planes. This is
because repulsive interionic forces are minimized on these planes during dislocation
motion. Thus we expect l/2<110> Burgers vectors and {110} slip planes. GIVEN: Mo crystal
b= 0.272nm
a0 = 0.314nm For FCC, _ b _ = “3 (960 face diagonal) For BCC, _ b _ = ‘5‘!" (960 cube diagonal) 2
REQUIRED: Determine the crystal structure.
IfMo were FCC, then .. b _ = 0' 31 4 = 0.22211»: but lb = 0.272 2:. M0 is not FCC.
32
J} Assuming M0 is BCC, then _ b _ = —2— X 0.314 = 0.272117». Thus the Burgers vector is
consistent with Mo being BCC. 83 FIND: Is the fracture surface in ionic solids rough or smooth? SOLUTION: Cleavages surfaces of ionic materials are generally smooth. Once a crack is
started, it easily propagates in a straight line in a speciﬁc crystallographic direction on a
speciﬁc crystallographic plane. Ceramic fracture surfaces are rough when failure proceeds
through the noncrystalline boundaries between small crystals. 7. GIVEN: ECG or with b  0.25nm REQUIRED: Find lattice parameter a
ASSUME: b =§< 111 > for BCC structure 2 J36, 2 2
SOLUTION: _ b _ = %~+ 5:;— + 94— = T from the formula for the magnitude of a vector: 2 =ix0,25=—0—'_5_=0.289nm “=Tb .13 13 GIVEN: Normal stress of 123 MPa applied to BCC Fe in [110] direction
REQUIRED: Resolved shear in [101] on (010)
SOLUTION: Recall that the resolved shear stress is given by: 1: = 0 cost) cost (1)
where 9 = angle between slip direction and tensile axis; 4» = angle between normal to slip
plane and tensile axis 34 /\ /\ 9=[110][101],¢#[010] [110]
{1101[101]=_[110]__[101]__c050
=J§0J30050=1+0+0 l
cosB=—, 9: "
_ 2 30 [010]o[110]z _[010] ~_ q_ [110]_ cos¢
= 10 ﬁc059= I c0549 = Ioversqrt2 ,6 = 45° 2 4'2'
9. GIVEN: Stress in [123] direction of BCC crystal
REQUIRED: Find the stress needed to promote slip if “lick = 800 psi. The slip plane is (11
0) and slip direction is [111].
SOLUTION: Recall ‘t = o c050 cos¢ (l) Thusr=123[ l )[L]=43.5MPa e=[123] [111] =J17J3coso=1+2+3=6 2733x2 «503/3043 7J5 It; c056= =W~=W= [1231.[111]= l[123] l[111]cose
4,2 [123] [110] [1231 [1.1.01= l[123]l[110]005¢ 85 10. ll. l2. l3. =JﬁJ3cos¢=12+0=1 cos¢=.___e_1__
mJE
Tan 1/7
=—~—~——~=800 —— 14 2
a cosﬂcosd 1:“.fo—
my a =  800 =  4,572 51' com ressz‘on
J; P ( P ) Burgers vectors lie in the closest packed directions since the distance between equivalent
crystallographic positions is shortest in the closepacked directions. This means that the
energy associated with the dislocation will be minimum for such dislocations since the
energy is proportional to the square of the Burgers vector. Close packed planes are slip planes since these are the smoothest planes (on an atomic
level) and would then be expected to have the lowest critical resolved shear stress. GIVEN: Dislocation lies on (1_1_1) parallel to intersection of (111) and (1 11) with
Burgers vector parallel to {1_10]. Structure is FCC.
REQUIRED: A) Burgers vector of dislocation and, B) Character of dislocation. SOLUTION: A) Since the structure is FCC, the Burgers vector is parallel to <110> and
has magnitude 5—3. For a Burgers vector parallel to [_1_10} the scalar multiplier must be a12. Thus 13 = 2112 m0]. B) We must determine the line direction of the dislocation. From the diagram we see that the BV and line direction are at 60“ which means the
dislocation is mixed. [T ‘1‘ O] (1 T 1) plane GIVEN: Dislocation reaction below: REQUIRED: Show it is vectorially correct and energetically proper.
SOLUTION: % [111] +9717] = a [100} 86 The sum of the x, y & 2 components on the LHS must be equal to the corresponding
component on the right hand side. _?
a a _
3°(I)+3'(1)"ay95 x component (LHS) = x component (RIIS) .7
%(1)+%o(1)=oyes y component (LHS) = y component (RIIS) .7
%(1)+g~(I)=o yes 2 component (LHS) = 2 component (RHS)
Energy: The reaction is energetically favorable if  in l 2 +  b2] 2 >  b3} 3 2 2 2 2
‘ — U4 4 4 v 4 2 . . . 3 3
Thus the reaction 1s favorable since 7 a2 + — a2 > a2 4
l4. GIVEN: Dislocation in FCC 87 15. 16. 17. a
a“ 101
b 2[ J Parallel to [101] i.e. t_ = [101]
REQUIRED: Character and slip plane
SOLUTION: Character is found by angle between t_ and b_ . Note t_ o b_ = 1 + O + 1 = 0. Thus t___b¢. Sincct__b__ the dislocation is wedge. To ﬁnd the slip plane we note that the cross produce of b__ & 1_ gives a vector that is normal to the plane in which b_ & I_ lie. This vector so formed has the same indices as the
plane since we have a ﬁindamentally cubic structure. i j k
v=[iol]x[i01]= 1 o 1
1 o 1 = im) —j(2) +kk0) v = [050] We see from the diagram that these vectors lie on (010). Thus, we have the plane (010) which is the same as the (010) plane. This does not move
by glide since planes of the kind {100} are not slip planes for the FCC structure. FCC metals are more ductile than BCC or HCP because: 1) there is no easy mechanism
for nucleation of microcracks in FCC as there is for BCC and HCP; 2) the stresses for
plastic defamation are lower in FCC due to the (generally) smoother planes. This means that the microcracks that form in BCC & HCP will have high stresses tending to make
them propagate. For a simple cubic system, the lowest energy Burgers vectors are of the type <001> since
this is the shortest distance connecting equivalent atomic positions. This means that the
energy is lowest since the strain energy is proportional to the square of the Burgers vector. GIVEN: At. wt. 0 = 16
At. wt. Mg = 24.32 Same structure as NaCl
p = 3.65 g/cm3
REQUIRED: Find length of Burgers Vector in MgO
SOLUTION: The structure of MgO is shown schematically below along with the
shortest Burgers vector. To solve the problem we ﬁrst note that we require the lattice 88 We can calculate the total mass of this cube and the volume andcalculate the density. Since the mass is known and the density is known, the volume may be calculated from which a. may be extracted.
: Wo= + WM3++ We note that there are 40' ions and 4Mg++ ions located at the corners.
{ as T However, anion at a corner is shared by 8 such cubes. Thus, we have 1/2
T
\ " I =1.At.wtOH1 l6 : 8 “04ng : —.._x_....._...._..._.._____
W" 2 N” 2 6.02x10” 6.02 = I.33x10"”gm I At.wlMg 1 24.32
»=—x—————=—x—~m=2.02x '25
WM“ 2 NW 2 6.02x10” 10 gm O= and '/z Mg++ ions in our cube. 1. + _ as . _,,
““165:me
avD / 8 an 89 a’ = 8 x 3.35 1510'” Check 3.65
$ 2 r 2+ + r J
ao=0.419x10“’cm 2 M“ "
r. 0' 419 mu m; = 2(0. 078nm + 0. 132nm)
= 0.42nm b= “° =0.296nm f5 GIVEN: Critical resolved shear stress (0.34MPa), slip system (111)ﬂ10], and tensile axis
[101] REQUIRED: Applied stress at which crystal begins to deform and crystal structure.
SOLUTION: (A)The situation is shown below I“. = 6 cost)  cosd)
9 = angle between tensile axis and slip direction
9 = angle between tensile axis and normal to slip plane [101] T: [111] [101] e = [101] [101}
[1111[101]= [111]! l[101]lcos¢ [1011.{1101= [101] l[110]1cose 90 19. _[111]_ = J} . _[101]_ = J3 _[101]__ = __[11o]__ = J3 2 y _\/3 1 _ 3_1
leis—”5"" 3 7573'“ "3 TM, =o.34= U[%J[%]=% _0= 6 x0.34=0.83MPa (B): To have a {1 l l}<110> slip system, the material must have an FCC structure. GIVEN: Tu. = 55.2 We, (111)1101] slip system, [112] tensile axis REQUIRED: Find the highest normal stress that can be applied before dislocation
motion in the [10 1] direction. SOLUTION: The situation is shown below. Essentially the problem reduces to ﬁnding
the value of the tensile stress when the critical resolved shear is reached. t¢.=ocosecos¢
9=[112] [101] ¢=[112] {111} 91 [1121o[101]=J3JEcos9 [112]» [1111: J3 J3 cos¢ [1121 [101] [1121[111]
cos0= mm, cos¢= mm cosﬂ:it2i.£=i.~ 005 :Hi—
J12 H2J3’ 3J3
J—a
m=5i=2 0——
? 2J3 37., 3‘56 B. Would have My the same stress for a BCC metal (¢ & 0 would be interchanged). 20. 21. 22. GIVEN: 0 at yield = 3.5 MPa; (111) [110] slip system [111] tensile axis
REQUIRED: Compute rm.
SOLUTION: I“. = ocosecoscb [1 11]0]]11[ 2 J3 [1 11] . [111]_
cos9 — — = , cos = www— ——~—
J3 I J‘J’z‘ 7: l’ JEJE 343
5J3x 2 2J3
”=3 —~—— —=3.5x———=1.10MPa
'5J3 x3J§ 9
9=[111] [1.1.01 ¢=[1_1_1] [111]
It_e_m Edge Screw
Linear defect? Yes Yes
Elastic Distortion? Yes Yes
Glide? Yes Yes
Climb? Yes No
Crossslip? _ No Yes
Burgers Vector (BV) J. to line I/ to line
Unique slip plane? Yes No
Offset If to BV II to BV
Motion 1/ to BV _L to BV GIVEN: BCC metal with tm— —7MPa [001] tensile axis REQUIRED: (a) Slip system that will be activated and (b) normal stress for plastic
deformation 92 SOLUTION: Recall that for BCC metals the usual slip system is <1 1 1> {110}.
Deformation occurs on the plane and direction for which casecomb is a maximum since
this will have the maximum resolved shear stress. The situation is shown below. (Note that the slip directions are shown shortened in this view)
[001] [0101 ”0"] (b) Possible slip systems are listed below: (001) [ill]
(071) [111]} sketch (also [111] on (011)) (also ﬂ11]on (011)) (also [111] 011001)) (101) [111] (701) [1}1] similar to planes shown in sketch. Also [111] on (101) We see by inspection that the resolved shear due to a tensile force in [001] will all be the
same. The resolved shear on all other {110}<111> systems is zero. B. To compute the normal stress at the onset of plastic deformation we will consider
(011) [1.1.1] 7 a:_,.....—....__..___..._...
cos9 cos¢ 'th '—' ccosﬂcosdt =' 7 e =l0011L1_111; cosd) = [001] [011] 93 cosB—  “"1’;—§§—”’ __J___, 0039, meijx J11“! Joversqrt2 = 3 xJEx7=1115MPa [00114111] Joan[101}:
1:: J. —j§. cos¢ Ix J3 Ioversqrtz Note if we considered ( 101) [1_l_]] we would have
and we would obtain exactly the same answer. cos0=———————~—— 23. GIVEN: Yielding occurs at normal stress of o = 170 MPa in [100] direction. Dislocation
moves on (101) in [1 1_] direction.
REQUIRED: 1m. and crystal structures
SOLUTION: Assume an edge dislocation tm = ocos coscb /\_ e = [100] [11.1] ¢= [160] [101] cos0=L...__—__"’0]'["“1_ =_ cos¢—m[1001'[1011 =10vers r12
IxJ— J3 1xs/3 q Netsm The  sign means that the slip direction is opposite to the motion of the dislocation.
Essentially, we have a negative edge dislocation on ( 101) as shown below: The edge dislocation moves in [11_l_] direction but the offset is in [1_l_]] direction.
The slip plane and slip direction are representative of BCC structures. 24. GIVEN: Q10)[lll] slip system. [123] tensile axis ‘ 94 1:“. == 800 psi for BCC crystal
1“,. = 80 psi for FCC crystal with
0M = 457 psi [123] tensile axis and (11 1)[110] system.
REQUIRED: Normal stress at yield for BCC metal
SOLUTION: The simplest way to solve this problem is to note costMost) is the same for
the BCC and FCC crystal with the meaning oft: and 6 interchanged. Let M= BCC BCC cosBcosd).
rm. = a
(1)
FCC PUG
rm. =cr M
(2)
7:3me _ BCC
riff — am:
(3) Here crystallographically equivalent positions join ions at cube comers (by 3 ac), face
diagonals (by = «[2 an) , cube diagonals (5,, = J5 a0) 3 Jig} The most densely packed plane is the (1 10} in which we have The shortest vector that will reproduce all elements of the structure is a... Thus b = a<1 00> COMMENT: We note that this is not sufﬁcient for general defamation (e. g. a tensile
axis of the type <100> produces zero shear on the l<100> Burgers vectors. We expect
then a<110> Burgers vectors as well. atms / m2 95 26. 27. GIVEN: 0' = 1.7 MPa [100] tensile axis (111)[101] slip systems REQUIRED: rm and crystal structure. Also ﬁnd ﬂaw in problem statement.
SOLUTION: Since the slip system is of the type {lll}<110> the structure is FCC. The
problem is misstated since the Burgers vector must lie on the slip plane and [101] does not lie on (111). The slip direction would more apprOpriately be [101]. Thus the slip system
is(111)[10§ as shown below. a = [1001(107] ¢ = [100] [111] rm = a cost? cos¢ = [1001 [10?] 2 [100] [111]
0030 mlxﬁ cos¢ mlxﬁ 1 1
cosG=—~*: cos =—
5 l J? 1 I
w=1.7x—x—— 0.69M'Pa
’ J‘ 43‘” 2 0020000 00000 0y 3,
0 000
C230000 00000 x = start of Burgers circuit _L = edge dislocation _ .
y = end of Burgers Circmt b = Burgers vector 96 28. 29. _L = edge dislocation x = start of Burgers circuit
b = Burgers vector y = end of Burgers circuit FIND: Show energy/area = force/length, that is, surface energy is surface tension in
liquids. DATA: The units of energy are J = “US or N—m. The units of force are N.
SOLUTION: Energy/area = J/m2 =Nm/m2 = N/m = force/length GIVEN: Two grain sizes, 10m and 40pm
REQUIRED: A) ASTM GS# for both processes, B) Grain boundary area.
SOLUTION: Assume that the grains are in the form of cubes for ease of calculation. The ASTM GS# is deﬁned through the equation: 11 = 2"" where n = # grains/in2 at 100x.
N=ASTM GS# To solve the problem we ﬁrst convert the grain size to in. where D = length of cube edge
in um. I in
D x '4 E m..."—
[pm] 10 [pm]x2.54 cm _D x3.937x10"=D.n
pm, 910 X 3.937 x 10": 3.937x 10" in D40}; ‘9 40 x 3.937 x104: 15. 75x10‘fin Dig}: 3.937x10“ x102=3_937 x10'2in At lOOX linear magniﬁcation, the sides of the smaller grains will be:
The area of each grain at max will be 97 30. Aiga" = (3.937 x 102 2 = 1.550 x 104 m2
11 Aiﬁﬁ" = (15. 75 x 102)” 3 24.81 x 10.31.”: Similarly the area of the 40am grains at IOOX is
For the 10pm dia grain, the # of grains per in2 (at box) is _ 1 _ . . 2
163?." — W445. 16 grams/m at max 1
Similar] W=m=4a31 ainsl' 2 noox
anp 24.81x10" gr ma For the 10m grain size: 7 645.16 = 2N4
log” 645.15 = av  1) log“, 2 W + 1= N ASWMOJ
log 2 _
For the 401133 ANS W log(40. 31) + 1=N= ASW#6.3
log2 B. In computing the total g.s. area we will assume 1 in3 of materials. Since there are 6 faces cube and the area of each face is shared by 2 cubes, each cube has an area of 3x Area offace. GB. Area =[é] x 3d? = 3/d
GB Area (10“ gs) = 313.937 x 10" = 7620'm2/in3
GB Area (40“ gs) = 3115.75 1: 10“ = 1905in2/in3 GIVEN: 6,, = ZOOMPa at GS#4
= 300MPa at GS#6
REQUIRED: 0,. at GS#9
SOLUTION: Recall 0,5 = 6., + kd'm
for low carbon steel. If d = grain size (assume cubes) load = grain diameter at IOOX 98 (1) 31. ’0“ (100d )2 d’ n= 2“ (ASTM GS# is N) For ASTM GS# 4; d” “£3: 16.82
For ASTM GS#6: «L = 23. 78 di”
Substituting (4) and (5) into (1) we have
200  on + 1:06.82)
300 a 0., + k(23.78) Subtracting (6) from (7):
100 = k(23.78  16.82)
.'.k = 14.37 Substituting this value of K into (6) yields 200 = 0'0 + 14.37 x 16.82 on = 41 .70 (this is not physically realistic since 0., relates to the lattice friction stress
which should not be negative) —1—=10x22=40 1/)
d9 For ASTM GS#9
Thus 0,. = 6.. + 14.37 x 40 = 41.70 + 574 = S33MPa GIVEN: bl = 0.25 pm for BCC metal tilt boundaly has angular difference of2.5°
REQUIRED: Dislocation density in tilt boundary wall (2) (3) (5) (6)
(7) 32. 33. SOLUTION: The physical situation is shown below: If in = Burgers vector, D = spacing between edge dislocation t9 (radians) = % 0= 2.5m = 0.044 radians __ loverD = leg—3 z 1. 76 x 10‘ dislocations / cm 2.5x 10 # of dislocations in boundaiy for a lcm high boundary is %(where D is in cm) FIND: ShowD =b/9. GIVEN: b is the magnitude of the Burger‘s vector; D is the spacing between dislocations,
and 6 is the tilt angle.
SKETCH: See Fig. 5.3—4. SOLUTION: We can see the geometry more clearly using the following sketch: Mb” I D l 100 9 57/2 From the Figure we can immediately write that tan; = —D—. Since the tan of a small angle is the angle itself: 34. 35. 36. £23323sothatD=b/6,asiswritteninthemargin. FIND: How can you detect a cluster of voids or a cluster of precipitates in a material?
SOLUTION: This can be a diﬂicult challenge indeed. Ifthe total void volume is large,
then the density of the sample will be lower than that of dense material. The same is true
for clusters of precipitate; however, usually the density difference between host and
precipitate is not as great as between host and air, so the technique does not work as well.
Another possible technique is microsc0py. Samples can be prepared for microscopy,
perhaps by polishing and etching and the defects observed using optical or electron
microscopy. X—ray diffraction can also be used. With a random spacing of void or
precipitate there is then an average spacing. Sometimes Bragg‘s law can be used to
calculate the spacing if an intensity maximum is observed. Note that the angle of the
maximum will be very small.
COMMENTS: There are many other potential techniques that can potentially be used.
They all rely on some property difference  magnetic, electrical, optical, or whatever. FIND: How can you ascertain whether a material contains both crystalline and
noncrystalline regions? GIVEN: Recall that the density (and other properties) of crystalline material is greater
than that of noncrystalline material of the same composition SOLUTION: There are three methods in common usage to establish crystallinity
polymers. These methods apply to all materials. 1. Density. Measure the density of your sample and compare it to the density of
noncrystalline and crystalline samples of the same composition. 2. Diﬂ’erential Scanning Calorimetry. Heat your sample in a calorimeter. Samples that are
crystalline will absorb heat at the melting temperature and show a “melting endotherm".
Some noncrystalline samples (such as amorphous metals) will crystallize in the calorimeter
and show a huge release of heat prior to melting. This is a "crystallization exotherm". 3. X—ray diffraction. Crystalline materials show welldeﬁned peaks. COMMENTS: Knowing whether a material is crystalline or noncrystalline is a common
challenge to polymers scientists. We often need to quantify the ﬁ'action or percent
crystallinity. Can you suggest a method for each of the 3 techniques outlined? FIND: State examples of materials' applications that require the material to behave in a
purely elastic manner. SOLUTION: There are many such possible examples. Since plastic deformation is
nonrecoverable deformation, any application that requires repeated stressing and
dimensional stability is a good example. Here are some examples: 1. Springs in automobiles  leaf and coil springs 2. A diving board 3. Trusses in a bridge 10} 37. 38. 39. 40. 41. 4. The walls in a building
5. A bicycle frame 6. Piano wire 7. Airplane wings As the dislocation density T, there are more dislocation/dislocation interactions and the
strength goes up. At the same time, the degree of “damage” also increases and the _ ductility decreases. If the point defect concentration T, the strength will go up as well. This is because the
defects may migrate to edge dislocations where they cause jogs on the dislocations. A
jogged dislocation is much harder to move and may itself require the generation of point
defects to move. In addition the point defects may collapse to form dislocation loops
which also impede the motion of other dislocations making the materials stronger. Ifthe
defects are interstitials, they may migrate to areas around the dislocations in which the
system energy is reduced. For the dislocation to move away from the interstitial an
increase in the system energy is required which means the stress to move the dislocation
must increase. Ifthe point defect is a substitutional atom, similar considerations apply.
However, the magnitude of the energy reduction is less because of the less severe
distortion. Thus the strength increase is not as high as for intersitital. As dloyﬁ since this means the path over which a dislocation moves Jv. This means that
the stress will have to increase to either nucleate or unlock dislocations in adjacent grains.
The relationship quantifying this behavior is the Hall—Fetch equation: 6,. = e. + kd'm The strength may increase as a result of: 1. decreasing grain size  should not be too (see previous questions) temperature
dependent. 2. Adding impurities (e. g. C in Fe). The impurities “lock” the dislocation by associating
with the dislocation to lower the system energy. This will be very temperature dependent
for dilute concentrations of impurities as the impurities will diffuse away at high
temperatures. 3. Adding precipitates  blocks the motion of dislocations through either having a
diﬁ‘erent crystal structure or a large strain ﬁeld. Since the precipitates are usually large
compared to the atomistic dimension, strong temperature dependence is not expected. 4. Cold work  increase quantity of dislocations. GIVEN: p = Ion/cm2 for low C steel
REQUIRED: concentration of C atoms (at %) to lock all dislocations
SOLUTION: Recalling the At. weight of Fe is 55.85 and the density is about 7.8 g1n/cm3 N _ 7.8
we may write F” 5.5.85 6.02 x 10” 102 42. 7.8
55.85 NF, = x 6. 02 x [0” arms / cm’ = 8.41:: 1022 (assume lC atom for every Fe atom along dislocations) 10” cm dp, _2.5x 10'scm
ch Ne: 10,2 10 I9 I9
o=m=ux =4x
N 2.5x10“” 2.5 10 10 4x10” x100 4x102! ..______._. ——=4.76x '2at"/
8.41x10”+ 4x10” " 8.41:: 10” I0 0 at %C= FIND: Why can you not bend the bar of tin? GIVEN: The bar has been well annealed, so the initial dislocation density is low. You
are required to re—bend the bar after cold working. SOLUTION: The deformation has increased the dislocation density and the bar now
requires much more stress, or force, to deform it. You are not necessarily a weakling, but
you have been taken. Reanneal the bar and bend it back or use brute force.
COMMENTS: It is often difﬁcult to bend a metal back to its original shape and this is
just one of many possible reasons that depend on the metal and its memo—mechanical 103 ...
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 Fall '08
 Tannebaum

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