final dec 2000

Applied Science 278 December Exam - 2000

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THE UNIVERSITY OF BRITISH COLUMBIA Department of Metals and Materials Engineering

APPLIED SCIENCE 278 Engineering Materials

FINAL EXAMINATION, December, 2000

This is a Closed Book Examination. The use of calculators having stored information of relevance to this course is forbidden. Time: 2.5 hours Answer 4 of 5 questions. Each question is worth 25 marks. The complete exam is 7 pages in length. Some useful formulae and phase diagrams are given on the last two pages of the exam. The use of sketches or schematic diagrams are encouraged wherever these will aid in solving or discussing a problem.

Marks 1. – Testing of Materials (6) a) An unknown metal is loaded in compression. The initial diameter was 10 mm and the

diameter under a compressive load of 50 kN was 10.03 mm. Determine the modulus of elasticity for this material, given Poisson’s ratio = 0.3.

(4) b) we have seen that metals have a maximum elastic strain of 1 % whereas elastomers can sustain elastic strains of greater than 500 %. Comment on the origin of this difference, clearly identifying the relevant mechanisms for each class of material.

(5) c) a tensile sample with a gauge length of 40 mm was found to have a total elongation to failure of 50 %. The uniform elongation was observed to be 35 %. If the sample of the same material with a gauge length of 25 mm was tested, what value would you expect for the total elongation to failure.

(6) d) sketch typical S-N curves for steel and aluminum. Define the fatigue limit and fatigue strength. Discuss the important factors which affect the fatigue life of a material.

(4) e) the fracture stress of a brittle material was found to be 1.7 times higher in a 3 point bend test (modulus of rupture test) compared to a simple tension test. Rationalize the difference.


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2. Consider the following questions relating to the structure and properties of metals.

(3) a) sketch the BCC unit cell and then calculate the atomic packing factor assuming the atoms to be hard spheres.

(6) b) the metal lead has a FCC crystal structure. If the angle of diffraction for the (111) (first order reflection) is observed to be 15.63 degrees (as shown below) when monochromatic x-rays of wavelength equal to 0.1541 nm are used, calculate i) the interplanar spacings and ii) the atomic radius of the lead atom.

(4) c) with a series of atomic sketches show how an edge dislocation moves through the crystal

and then finally leaves the crystal. Clearly indicate the resulting magnitude and direction of the slip step (i.e. the Burgers vector).

(6) d) Two brass samples were tensile tested. The first sample had a grain size of 10 µm and its yield stress was found to be 170 MPa. The second sample had a grain size of 100 µm and the yield stress was 60 MPa.

i) Using this information determine the constants in the Hall-Petch equation.

ii) If a brass sample with an initial grain size of 25 µm was held at 800 oC for 2 hours and then tested at 20 oC, what value would you expect for the yield stress?


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(4) e) list 4 methods for making dislocation motion more difficult and thereby increasing the yield stress of a metal.

(2) f) why is dislocation motion so difficult in ionic or covalently bonded materials ?

3. Phase diagrams and Heat Treatment

(13) 3a. A 50 wt % Pb – 50 wt % Mg alloy is slow cooled from 700 oC.

i) Sketch the microstructure at the following temperatures: 650, 500 and 400 oC.

ii) For each of the temperatures in part i) determine the phases, their composition (in wt %) and the fraction of each phase.

iii) If the temperature were further decreased to 200 oC, what would be the equilibrium concentration and fraction of the two phases ?

(6) b) based on the Fe-C phase diagram, discuss the relationship between yield stress and carbon concentration for carbon levels between 0 and 1 wt % C. Assume slow cooling from austenization temperature. Sketches of the expected room temperature microstructures for carbon levels of 0.2 wt %, 0.76 wt % and 1.0 wt % carbon will be useful.

(6) c) compare the heat treatment of a precipitation hardening alloy and a quenched and tempered steel. Describe the changes in i) the yield stress and ductility during ageing or tempering and ii) the corresponding microstructure of the materials during ageing or tempering.

4. Polymers and Glasses

(5) a) sketch a typical stress-strain curve for polyethylene. For each characteristic region of the stress-strain curve, sketch the appropriate changes in the polymer structure for each part of the tensile sample.

(4) b) calculate the number average molecular weight for the polymer with the following ranges of molecular weights.

Molecular Weight Range

(gmol-1)

xi xw

10 000 – 20 000 0.05 0.01

20 000 – 30 000 0.2 0.16

30 000- 40 000 0.49 0.51

40 000-50 000 0.22 0.26

50 000- 60 000 0.04 0.06


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(3) c) given the mer unit is polyvinylchloride shown below; calculate the number degree of polymerization (given atomic molecular weights shown in the table below).

Element Molecular Weight

H 1.008

Cl 35.45

C 12.01

F 19.00

(5) d) Identify 3 characteristics of a polymer chain that promote crystallinity in the bulk material. Identify two properties that are improved with increasing crystallinity.

(4) e) The fracture stress of glass slide sample was measured to be 125 MPa. After the test, it was determined an ellipitically shaped crack initiated failure. The critical crack length was measured in an electron microscope to be 2 x 10-3 mm (i.e. a = 1.x 10-3). If you assume that the theoretical strength of glass equals E/10, calculate the radius of curvature at the crack tip. (note: E is the modulus of elasticity. For glass, E = 70 GPa).

(4) f) i) describe the procedure used for tempering glass, ii) how does this procedure lead to the development of residual stresses in the glass and iii) compare the strength and nature of the fracture process for tempered glass and normal glass.


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5. Wood, concrete and composites

a) the following reactions are important for strength development in cement.

C3A + 6H → C3AH6 + heat

2C2S + 4H → C3S2H3 + CH + heat

2C3S + 6H → C3S2H3 + 3CH + heat

Type I cement has the following composition: C3S = 50 % C2S = 25 % C3A = 12 % C4AF= 8 %

(3) i) in this nomenclature, what is meant by C, S, and H ?

(4) i) discuss the role of each of the constituents in the development of strength in cement.

(4) b) We have seen that the cell walls are similar for many types of wood. Given that the properties of the cell wall below, calculate the modulus of elasticity parallel and perpendicular to the wood grain for balsa and oak wood. Cell properties: density =1.5 gcm-3, modulus = 35 GPa. Density of balsa wood = 0.2 gcm-3 , density of oak = 0.75 gcm-3.

(4) c) discuss the fracture of wood under tensile loading parallel and perpendicular to the grain of the wood. Use of diagrams to show the fracture path would be helpful.

(6) d) a continuous fibre composite has a volume fraction of fibres equal to 40 %. The fibres are medium quality carbon fibre and have a modulus of elasticity equal to 500 GPa. The matrix is an epoxy with a modulus of elasticity equal to 4 GPa.

1. calculate the elastic modulus of the composite for loading parallel to the fibre axis.

2. calculate the elastic modulus of the composite for loading perpendicular to the fibre axis.

3. sketch a diagram of how you expect the modulus of elasticity to vary with the angle between the loading axis and the fibre axis (i.e. 0o is when the fibres and the loading axis are parallel).

(4) e) discuss the advantages and disadvantages of using non-oriented discontinuous fibre composites.


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Useful Formulae

z

y

z

x

εε

εεν −=−=

( )E G= +2 1 ν

22

1 2

EU y

yr

σεσ =×=

2

toughness fuy ε

σσ×

+≈

( )εσσ += 1T

( )εε += 1lnT

σ εT TnK=

c

A

VNnA / = ρ

n dλ θ= 2 sin

( )d a

h k lhkl =

+ +2 2 21

2

rate Ae Q RT= −

RTQ

recrx eAt ′=

21−+= dkyoσσ

n

s K σε 1=& R = 8.314 Jmol-1K-1

& expε σsnK Q

RT=

−

2

N N QRTv

v=−

exp

σ theoreticalE

≅10

λφστ coscos=R

+=

21

21t

omaρ

σσ

σ σρm o

t

a=

2

12

σ γπc

sEa

=

21

2

( ) 21

2

+=

aE ps

c πγγ

σ

( )θπ

σ yy fr

K2

=

aYK πσ=

B KIc

y≥

2 5

2

.σ

max

min

σσ

=R

iin MxM ∑=

iiw MwM ∑=

mMnDP n

nn ==

mMnDP w

ww ==

σ στt ot

=−

exp

( )E tt

ro

( ) =σε

E Ew ss

ll =

ρρ

E Ew ss

⊥ =

ρρ

2

mmffc VEVEE +=

mffm

fmc VEVE

EEE

+=

FF

= E VE V

f

m

f f

m m

FF

= E / EE / E V / V

f

c

f m

f m m f+

c

fc

dl

τσ2

*

=

Ec=KEfVf + EmVm


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Phase Diagrams

Figure A.1 – Lead – Magnesium phase diagram

Figure A.2 – Iron – Carbon Phase diagram

final dec 2000

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