mme2601assign0148591238.pdf

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 | 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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Assignment NO. : 01


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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Assignment NO. : 01


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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Assignment NO. : 01


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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Assignment NO. : 01


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

mme2601assign0148591238.pdf

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