final apr 2004a

Applied Science 278 April Exam - 2004

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THE UNIVERSITY OF BRITISH COLUMBIA Department of Metals and Materials Engineering

APPLIED SCIENCE 278 Engineering Materials

FINAL EXAMINATION, April 19th, 2004

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, material data and a phase diagram are given on the last 2 pages of the exam. The use of sketches or schematic diagrams are encouraged wherever these will aid in solving or discussing a problem. Show all work !! Good Luck !

Marks Question 1

(3) (a) The coefficient of thermal expansion of aluminum is larger than that of copper. The elastic modulus of copper is higher than that of aluminum. Which metal has the higher melting temperature? Why?

COPPER The interatomic bond in Cu is stronger than in Al. As a result, Tm is higher (thermal energy to break the bond), E is higher (mechanical energy to break the bond), CTE is lower (less of an asymmetry in energy well). The stress-strain curve of a 2024-T3 aluminum alloy is shown in Figures 1-3, with Figure 1 showing both the nominal and true stress strain curves, Figure 2 showing a magnification of the early part of the nominal stress-strain curve in Figure 1, and Figure 2 being the log-log plot of the true stress curve shown in Figure 1.


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0

100

200

300

400

500

600

700

0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4strain (m/m)

stre

ss (M

Pa)

nominal

true

Figure 1 – Stress-strain curves for 2024-T3 Aluminum.

0

100

200

300

400

500

0.000 0.002 0.004 0.006 0.008 0.010strain (m/m)

stre

ss (M

Pa)

Figure 2 – Early part of nominal stress-strain curve for 2024-T3 Aluminum.

331 MPa


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10

100

1000

1.E-04 1.E-03 1.E-02 1.E-01 1.E+00strain (m/m)

stre

ss (M

Pa)

(.0044,308)

(.29,632)

Figure 3 – True stress-strain curve for 2024-T3 Aluminum, note the logarithmic axes.

Determine the following for 2024-T3 aluminum: (2) (b) The elastic modulus Using Figure 2, take straight line slop of elastic part (e<0.004) resulting in E=70,000 MPa=70 GPa (3) (c ) The .2% offset yield strength Using Figure 2, offset straight line from part (b) by .002 (=.2 %), take intercept of this line with curve, for sy=331 MPa (2) (d) the UTS From Figure 1, max value of nominal stress-strain curve, UTS=487 MPa (4) (e) The strain hardening behaviour in the plastic regime, by calculating K and n in the strain

hardening equation. Two choices. (i) from Figure 3, take slopes etc.,. (ii) use values shown on curve (easier) 308=K(0.0044)^n and 632=K(.29)^n. Divide second by first to get 2.05=65.9^n Thus log 2.05 = n log 65.9. Therefore n= 0.17 Plug back into second equation, thus K=632/0.29^0.17=780 MPa Thus answer is K=780 MPa and n=0.17 (2) (f) The nominal strain to failure. Why is the nominal strain to failure larger than the true strain

to failure? From graph, last point on nominal stress-strain curve is .336, 473 MPa. Thus, if we ignore elastic springback, ef=0.336 (1 mark) If we allow for elastic springback ef=0.336-473/70,000=.329 (or do it graphically) (extra mark) Finally, ef nominal is larger than ef true since the definition of the strains means that e nominal

is larger than e true (nominal does not account for the fact that the instantaneous length is longer than the original length) (1 mark).


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(5) (g) This aluminum alloy has been precipitation hardened to this state. What does that mean, and in this context what is overaging?

Precipitation hardening is where a heat treatment has been used to create small uniformly dispersed second phase particles within the original matrix, thus enhancing strength and hardness. (Additional: this is sometimes also called age hardening, as strength develops over time. This comment is not needed for full marks).

Overaging:

(4) (h) The grain size of this material is 42 µm. Knowing the yield strength calculated in (c) and assuming that the intrinsic yield strength of this alloy series is 325 MPa, what do you estimate the yield strength of the alloy with a 5 µm grain size, all else being equal?

Sy(42 um) = 331 MPa. Sy=Sy0 + k d^-1/2 (Hall Petch equation) Thus 331 = 325 + K (42)^-1/2 . Thus K = 6 x 42^1/2 = 38.88 Thus at 5 um, Sy=325 + 38.88 x 5^-1/2 = 342.4 MPa. (note this is not a huge increase from the original grain size!)

Question 2 A 7075-T651 Aluminum alloy has a room temperature yield strength of 495 MPa and a fracture toughness KIC of 24 MPa m1/2. The material is tested non-destructively and guaranteed not to have any cracks larger than 2a=4 mm in length. The material is to be used in a tension application, with the plate width fixed at 1 m, but the thickness being the design variable. For the purpose of this question, you may assume that K aσ π= is the appropriate solution for a small crack in the plate. The initial design requires the structure to carry an applied load of 1000 kN, as well as have a safety factor of 3, which your team considers reasonable for a brittle fracture failure mode.


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(6) (a) Confirm that it will fail by brittle fracture, and determine the thickness B. Calculate the weight of the panel if the height of the panel is also 1 m, and the density of aluminum is 2700 kg/m3.

If Fapplied=1000 kN and SF=3, then Fdesign=3000 kN. For yield condition 3000 x 1000 = 495 x 10^6 x 1 (width) x By Thus By=3 x 10^ 6/495 x 10^6 = 6.1 mm For fracture condition, 2a=4 mm, a = 2mm. Kc=sc (pi a)^0.5, Thus sc=Kc/(pi a)^0.5 Sc=24/(pi x 2 x 10^-3)=302.8 MPa. This is < 495 MPa, so it will fail by brittle fracture. What about Bc? Bc=3 x 10^6/302.8 x 10^6 = 9.9 mm So to recap, for any thickness, sc<sy, so failure mode is brittle To take a load of 3000 kN, need B=9.9 mm (the larger of the two values calculated above). M = B x W x H x rho = 9.9E-3 x 1 x 1 x 2700 kg/m^3 = 26.75 kg (2) (b) Your team leader asks you, the materials expert, if it is true that thinner plates will have a

higher toughness than KIC. Explain why this is the case, and calculate when the transition will occur.

KIC is the plane strain fracture toughness. When a plate is thick enough, you have a state of plane strain at the crack tip. This stress state is highly constraining, and leads to a smaller plastic zone size than when the plate is thin, and there is a state of plane stress at the crack tip. In plane stress, the plastic zone size at the crack tip is larger, and as result more energy is absorbed and so Kc is higher than KIC.

Need to evaluate Btransition= 2.5 x (KIC/sy)^2 = 2.5 x (24/495)^2 = 5.87 mm (6) (c) Your team leader then says that a lower safety factor can be used if the failure mode

switches to ductile failure. If the safety factor for ductile failure is 1.5, and by using thinner plates you can get a plate toughness of double KIC, what is B now?

Let us assume Kc=2 x KIC=48 MPa m^1/2 Then sc=48/(pi x 2 x 1e-3)^0.5 = 605.6 MPa > sy, so will fail by yield If we calculate Bc2 (unnecessary step given above), then 1000 x 1000 x 1.5 (SF) = 605.6 x 1 x

Bc2


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Thus Bc2=2.47 mm But it is going to fail by yield, so we do need to calculate By2. By2 x 1 x 495E6=1000 x 1000 x 1.5 (new SF) By2=1.5E6/495E6=3.03 mm Both Bc2 and By2 are smaller than Btransition, so assumption that KC is higher, and thus

yielding is the failure mechanism is valid, thus 3.03 mm is the new plate thickness! (2) (d) How much lighter is the plate now? Is the above scenario reasonable? What else might you

consider? M=3.03E-3 x 1 x 1 x 2700 = 8.2 kg The above scenario is reasonable, in that yield failure is more benign than fracture, and so a

lower SF makes sense. However, it is an ‘advanced’ design, in that you are non-conservative. It assumes that your assumption of the max crack size being 4 mm is true, shaves off your margin of safety, etc.,. So, OK but definitely riskier, but huge benefit (27 kg -> 8 kg!)

(6) (e) The plate goes into production using the thickness calculated in (a) above. A design change

requires that a small circular hole be cut in the middle of the plate. You are tasked with calculating how much the applied load on the panel should be reduced, assuming brittle behaviour. Assume that the brittle design safety factor does not change, and use the stress concentration factor approach to calculate the new maximum load.

Put in a hole, Kt=3 (some students may say 2, as they may use the approximation to Kt which is strictly only valid for sharp notches, -1 mark) then when sm=495 MPa (fracture condition at edge of hole) the nominal s is =sm/Kt=495/3=165 MPa.

Thus F = 165E6 x 1 x 9.9E-3=1633 kN total. Given SF=3, then Fdesign=1633/3=544.5 kN.

(3) Once the plate in (a) and (e) is in service, there is concern about fatigue behaviour. The fatigue limit of this alloy is 150 MPa. Explain what that means, and calculate the maximum load that can be applied, using the brittle design safety factor, with and without the hole from (e). The question is poorly phrased, does not specify if the 150 MPa is smax, s amplitude… Assume smax=150 MPa, but any other reasonable assumption accepted. Any overt mention of the issue +1 mark. No hole: Fmax= (1/SF=3) x (150E6 x 9.9E-3)=495 kN Hole: Fmax=(1/SF=4) x (1/Kt=3) x (150E6 x 9.9E-3)=165 kN (Observe how all these allowances can drive the initial Fmax of 1000 kN down to 165 kN! No marks for this observation, which was not asked for!)

Question 3 Consider a Sn-Pb binary eutectic alloy, whose phase diagram is shown in Figure 4, of 40 wt% Sn (density 7.24 g/cm3)-60 wt% Pb (density 11.23 g/cm3) at a temperature of 150°C. Determine: (2) (a) The phases that are present at 150°C. Two phases present: alpha and beta


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(3) (b) The composition of the phases. Composition of alpha = 10 wt% Sn, 90 wt% Pb Composition of beta = 98 wt% Sn, 2 wt% Pb

The relative amount of each phase present in terms of:

(3) (c) mass fraction

(4) (d) volume fraction

(4) (e) Explain the evolution of microstructures A and B shown in Figure 4. Make sure you

explain why do they look so different.

For microstructure A:


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For micostructure B

So the difference in the microstructures is a direct result of A being of a composition less than the maximum solid solubility at the eutectic temperature. Therefore we have a single phase system of composition alpha, which becomes supersaturate in beta, which precipitates out. This is very different to the B history. (2) (f) Using the phase diagram in Figure 4 to guide you, what composition would you use for

solder and why? The best material for solder would be the eutectic, 61.9 wt% Sn, 38.1 wt5 Pb, as this has the

lowest Tm, which is desirable for solder. However, that is normally simplified to 60 wt% Sn, 40 wt% Pb.

(3) (g) What is a eutectoid? What is the eutectoid structure in in the Fe-C system called?


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The comment regarding important in the heat treatment of steels is worth +1 mark.

(3) (h) What is martensite and why do we temper it?

Martensite is a non-equilibrium structure of steel (Fe-C) which occurs when we cool down fast enough that carbon does not have time to diffuse into the low C alpha-Fe and high C Fe3C structures. Thus when a material is ‘quenched’ a supersaturated and unstable structure (Martensite) is formed. Lattices is stuck between FCC and BCC, namely BCT (body centred tetragonal) (in terms of lattice parameters a not equal to c). Martensite is very hard and brittle and so we temper it to regain some ductility (at the expense of decreased strength and hardness).

A

B

A

B

Figure 4. Lead-Tin Equilibrium Phase Diagram


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Question 4 (2) (a) Why is glass fibre so strong relative to bulk glass? Glass is a brittle material, so its strength is very much determined by the biggest flaw present

in the structure. If you make very fine fibres, by definition the flaw size is going to be very small, and thus the strength will go up significantly.

(2) (b) What is tempered glass and what are its advantages and disadvantages?

The advantage of tempered glass is a higher strength as the compressive strength must be

overcome, but the disadvantage is that when it fails, it shatters far more, as there is more elastic energy stored in the system.

(4) (c) Sketch how the specific volume of silica varies with temperature. Consider both crystallization and no crystallization. Mark and describe all transitions.

Tm is the melt temperature, structure shifts from disordered liquid to ordered crystals. Tg is

the glass transition temperature, below which the structure has significantly less mobility (glassy) than above that temperature.

(3) (d) What are network modifiers and why are they added to silica glass?


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(3) (d) What microstructural features encourage crystallization in polymers? - simple mer structures - small side groups - lack of branching -symmetry (isotactic or syndiotactic) -lack of randomness in copolymers Any three gets full marks (5) (e) A polyethylene (C2H4) polymer has a number average molecular weight of 300,000 g/mol.

The atomic weight of C is 12.01 g/mol and of H is 1.008 g/mol. The C-C bond angle is 109 degrees. Calculate the average extended chain length of this polymer as well as it’s average end-to-end distance.

M= 2 x AC + 4 x AH = 2 * 12.01 + 4 * 1.008 = 28.05 g/mol Thus n = Mn/m = 300,000/28.05 = 10,695 Thus number of chain bonds = 2 * 10,695 = 21,390 Each bond length d = .154 nm (information not in question, but provided in the exam), thus

L=N x d x sin (theta/2) = 21390 x 0.154 nm x sin (109/2) = 2682 nm Average chain end-to-end distance is r = d root(N) = .154 nm x root(21,390)=22.5 nm! (2) (f) What is the role of water in concrete? - provides for the hydration reaction - provides a fluid mixture which can be poured and is workable (4) (g) What are roles of alite, belite, aluminate, and ferrite in cement?


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1 main point per material gets 1 mark Question 5 (3) (a) Sketch the cellular structure of wood. Define the alignment of this structure relative to the

tree axis.

The majority of the cells are aligned with the main tree axis (longitudinal). (3) (b) What are the cell walls of wood made of? Cellulose fibres in a matrix of lignin, hemicellulose, and water and extractives (oils and salts).


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(2) (c) From a microstructural perspective, what is the primary difference between different tree species, eg balsa and oak?

Different densities of the cell structure (4) (d) Using Figure 5, estimate the axial and transverse strength of the solid material in the cell

walls. Many students appear to have difficulty with this question. Looking at figure 5, the easiest

way to estimate the values is to extend the two lines to a relative density of 1! This is a very inaccurate process, but a value of 150 MPa for axial strength and 50 MPa for transverse strength is correct. Anything close, with axial>transverse should get high marks, if the procedure is correct (realization that solid wall means extrapolation to relative density of 1, or evaluation of the line by anchoring on a point or two).

(4) (e) You are designing the dimensions of the legs of a table for use on wooden floors. Assuming that the four-legged table weighs 50 kg, and that the legs are square in cross-section, approximately what is the smallest size of footprint one can have on a floor made of (a) a harder wood such as oak and (b) a softer wood such as mahogany

The correct strength to use is the transverse strength (as the floor beams are obviously cut parallel to the axis for strength). Very approximately, s(oak)transverse = 4 MPa, and s(mahogany)transverse=0.5 MPa.

Thus, 50 kg ~ 500 N. Per leg = 500/4=125 N. thus Oak A=125/4E6=3.125e-5 thus d=5.6 mm Mahogany A=125/.5E6=2.5E-4 thus d=15.8 mm (5) (f) For an aligned long fibre composite, loaded perpendicular to the fibre direction such that the

fibre and matrix are in an isostress condition, show that mffm

fmc VEVE

EEE

+= .

(5) A glass fibre (E=80 GPa) is combined at a volume fraction of 50% with a polyester (E=3

GPa) matrix. The fibre has a critical length of 1 mm, and is made available in short fibre form with length 10 mm. What is the composite axial modulus, assuming that all the fibres


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can be aligned in the loading direction, such that the isostrain equation can be used with the K (efficiency) factor solely a function of fibre length.

Efficiency of fibre=9.5/10=.95 Thus axial modulus= .95 * 80 *.5 + 3 * 0.5= 39.5 GPa

Figure 5. Strength as a function of density for wood


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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 ==

sin( )2

L Nd θ=

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


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Phase Diagram

Figure A.1 – Iron – Carbon Phase diagram

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

final apr 2004a

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