final dec 2005

Applied Science 278 December Exam - 2005

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THE UNIVERSITY OF BRITISH COLUMBIA Department of Materials Engineering

APPLIED SCIENCE 278 Engineering Materials

FINAL EXAMINATION, December, 2005

This is a Closed Book Examination. The use of calculators containing stored information of relevance to this course is forbidden. Time: 2.5 hours Answer 4 of 5 questions. Only 4 questions will be marked. Each question is worth 25 marks. The complete exam is 10 pages in length. Some useful formulae, material data and phase diagrams are given on the last 5 pages of the exam. The use of sketches or schematic diagrams is 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 atomic mechanisms for each class of material.

(5) c) A 6.4 mm diameter cylindrical rod fabricated form 1045 steel is subjected to fully reversed tensile-compression loading. If the maximum tensile and compressive loads are 10 300 N, determine the fatigue life.

(4) d) 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.

(6) e) In the diagram of a tensile sample at the top of the next page:

i) compute the magnitude of the stress at point P, when the externally applied stress is 140 MPa

ii) how much would the radius of curvature at the point P need to be increased to reduce the stress by 25 %


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

(4) 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) sketch the following crystal planes: ( )011 , ( )111 and ( )221

(Note: please do a separate sketch for each crystal plane)

(3) e) list 3 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 ?


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3. Phase diagrams and Heat Treatment

3a. A steel with a carbon concentration of 0.4 wt% is slow cooled from 1000 oC to 700 oC. Given that at 1000 oC, the microstructure consists of 100 % austentite with an equiaxed grain size of 50 µm.

(6) i) sketch the expected microstructures at 900 oC, 728 oC, and 726 oC.

(6) ii) in the final microstructure (i.e. at 726 oC), calculate:

1) the weight percentage of α and Fe3C

2) the weight percentage of proeutectoid ferrite or iron carbide (which ever is appropriate) and pearlite

3) the weight percentage of α and Fe3C in pearlite

(7) b) describe the heat treatment procedure for an Al-Cu precipitation hardened alloy and the resulting change in the yield stress of the alloy. Sketch the microstructure at each point in the heat treatment and relate the evolution of microstructure to the change in the yield stress. (note: you should also include overageing).

(3) c) what is martensite and how can it be formed ?

(3) d) what is the purpose of tempering and describe the structural changes that occur in the steel during tempering ?

4. Polymers and Glasses

(6) a) Sketch a typical stress-strain curve for an amorphous polymer tested above the glass transition temperature. For each characteristic region of the stress-strain curve, sketch the appropriate changes in the macroscopic of the sample and the change in the polymer structure at the atomic level.

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

Molecular Weight Range

(gmol-1)

xi wi

10 000 – 20 000 0.04 0.01

20 000 – 30 000 0.25 0.16

30 000- 40 000 0.48 0.51

40 000-50 000 0.19 0.26

50 000- 60 000 0.04 0.06


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(3) c) given the mer unit for the polymer in part b) is polyvinylchloride as shown below; calculate the number degree of polymerization.

(3) d) Identify 3 characteristics of a polymer chain that promote crystallinity in the bulk material.

Identify two properties that are improved with increasing crystallinity.

(3) 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.

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.


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(6) d) a continuous fibre composite has a volume fraction of fibres equal to 45 %. The fibres are medium quality carbon fibre and have a modulus of elasticity equal to 610 GPa. The matrix is an epoxy with a modulus of elasticity equal to 4.5 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=&

223

bdLFf

fs =σ

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 ==

ndr =

or nr λ= i.e. d ≡ λ

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

σc* l = σm' (1 - Vf) +σf* Vf


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Figure A.1 – Fe-C Phase diagram

Figure A.2 - Aluminum-copper phase diagram


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Figure A.3 - S-N curve for several different engineering materials (mean stress = 0).

Figure A.4 - Rupture life plot for a low carbon-nickel steel.


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Figure A5 – stress concentration factor for geometry shown.

Table A1 – Characteristics of selected elements

Element Symbol Atomic Number

Atomic Weight (gmol-1)

Density (gcm-3)

Aluminum Al 13 26.98 2.71 Bromine Br 35 79.90 - Carbon C 6 12.011 2.25 Chlorine Cl 17 35.45 - Copper Cu 29 63.55 8.94 Fluorine F 9 19.00 - Gold Au 79 196.97 19.32 Hydrogen H 1 1.008 - Nickel Ni 28 58.69 8.90 Nitrogen N 7 14.007 - Oxygen O 8 16.00 - Titanium Ti 22 47.88 4.51 Tungsten W 74 183.85 19.3 Zinc Zn 30 65.39 7.13


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final dec 2005

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