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Surname and Initials: MABENGO N.D.Student Number : 48591238
Module Code : MME2601
Assignment NO. : 01
Mabengo N.D. Student#48591238July 2013 Page | 1
QUESTION 1
n
mTSA
VCT
322
39269914
20500
4mm
tDV
22
2
424115205004
5002
4
2
mm
DtD
A
2,86min
sTTS 5.171
424115
39269912
2
QUESTION 2
From table 4.1, = 24 x 10-6 mm/mm/oC
L2L1 = L1 (T2T1)
L2 = 1 + 24 x 10-6 x 1 x (650 20) = 1.01512 cm
(L2)3 = (1.01512)3 = 1.0460493cm3
at 650oC =32.581g/cm
0460493.1
70.2(assuming that the mass does not
change)
QUESTION 3
Length = 25 cm = 250mm
Increase = 12 x 10-6 mm/mm/oC x 250 mm x (260 21) = 0.717 mm
QUESTION 4
Heat = (0.21 cal/g-oC) x (103cm3) x (2.70g/cm3) x (300oC 21oC) = 158.193 cal
But 1 cal = 4.184 J 158.193 x 4.184J = 661.879 J
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Surname and Initials: MABENGO N.D.Student Number : 48591238
Module Code : MME2601
Assignment NO. : 01
Mabengo N.D. Student#48591238July 2013 Page | 2
QUESTION 5
Pultrusion is a process in which continuous fibber rovings are dipped into a resin
bath and pulled through a shaping die where the impregnated resin cures.
QUESTION 6
From table 18.1, Ac = 0.075 c = 0.075 x 2 = 0.15 mm
Die opening diameter = 75 mm
Punch diameter = 75(2 x 0.15) = 74.7 mm
QUESTION 7
From table 20.1, Ac = 0.075 c = 0.075 x 4 = 0.3 mm
Blanking punch for 85 mm Length = 85(2 x 0.3) = 84.4mm
Blanking punch for 50 mm Width = 50(2 x 0.3) = 49.4mm
Blanking punch for 25mm Top and Bottom = 25(2 x 0.3) = 24.4mm
QUESTION 8
Problems or reasons for excessive burrs
Very large clearance between punch and die for the material and stockthickness.
Worn cutting edges for punch and die with the same corollary as excessiveclearance.
Correcting the problems
Check the punch and die cutting edges to see if they are worn and regrindthe face if they really are.
If the die is not worn, measure the punch and die clearance to see if it equalsthe recommended value. If not the die maker must rebuild the punch and
die.
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Surname and Initials: MABENGO N.D.Student Number : 48591238
Module Code : MME2601
Assignment NO. : 01
Mabengo N.D. Student#48591238July 2013 Page | 3
QUESTION 9
Area A = (0.3m)2 = (30cm)2 = 900 cm2
Volume V = (30cm)3 = 27000 cm3
Surface Area of a spherical particle of D = 0.01cm is A = x D2
A = x (0.01)2 = 314.159 x 10-6 cm2 / particle
Volume of a spherical particle of D = 0.01cm is6
3DV
V= 93
105.523601.0 cm3/particle
Number of particles in 1cm3 =9
105.523
27000
= 51.576 x 109
Total surface area = (51.576 x 109) x (314.159 x 10-6) = 16 203 064.58 cm2
Percent increase = 100900
90058.16203064
= 1 800 240.51%
QUESTION 10
NpAF
mmdDA
cP
oop
062.2362074.1140
4.1140224425.025.022222
QUESTION 11
a) The tolerance band = 0.36. Wear allowance = 0.03 x 0.36 = 0.0108 mmGO gage will inspect the minimum hole diameter = 30.00
0.18 = 29.82mm
The dimension will decrease as the gage wears, so the wear allowance is
added to it.
Nominal GO size = 29.82 + 0.0108 = 29.8308mm
b) NO-GO gage will inspect the maximum hole diameter = 30.00 + 0.18 =30.18mm. In this case, no wear allowance is added because this gage
should not fit in the hole and wear away.
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7/27/2019 MME2601ASSIGN0148591238.pdf
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Surname and Initials: MABENGO N.D.Student Number : 48591238
Module Code : MME2601
Assignment NO. : 01
Mabengo N.D. Student#48591238July 2013 Page | 4
QUESTION 12
a)2
2 /8.4644125.3
14515
cmkgcm
kg
Y
b) S = E x e, subtracting the 0.2 offset, 018.0002.05
0.51.5
e
2/4.258044
018.0
8.4644cmkg
e
SE
c) TS = 2/8.8708125.3
27215cmkg
d) EL = %3232.05
56.6
e) AR = %505008.0125.3
56.1125.3
QUESTION 13
a)
306.93MPa
5.25.372
100067802
22
tR
T
b) 0.0130875
5
360
2175.3
L
R
c) a23452.14MP0130875.0
93.306G,
G
d)
491.187MPa
5.25.372
10108502
3
S
QUESTION 14
From Fig 6.2, the compositions are as follows:
Liquid phase composition = 65%Ni 35%Cu.
Solid phase composition = 83% - 17%Cu.
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Surname and Initials: MABENGO N.D.Student Number : 48591238
Module Code : MME2601
Assignment NO. : 01
Mabengo N.D. Student#48591238July 2013 Page | 5
QUESTION 15
From Fig 6.2 measured values of CL and CS are: CL = 5mm, CS = 12mm
Liquid phase proportion = 12/(12+5) = 0.71
Solid phase proportion = 5/7 = 0.29
QUESTION 16
From Fig 6.3, the compositions are as follows:
Liquid phase composition = 56% Sn44%Pb.
phase composition = 18%Sn
82%Pb.
QUESTION 17
From Fig 6.3 measured values of CL and CS are: CL = 10.5mm, CS = 15mm.
Liquid phase proportion = 15/(15+10.5) = 0.59
phase proportion = 10.5/25.5 = 0.41
QUESTION 18
From Fig 6.3 the compositions are observed as follows:
Liquid phase composition = 78%Sn22%Pb.
phase composition = 98%Sn2%Pb.
QUESTION 19
From Fig 6.3 measured values of CL and CS are: CL = 7.8mm, CS = 4.2mm
Liquid phase proportion = 4.2/(7.8+4.2) = 0.35
phase proportion = 7.8/12 = 0.65