# Assignment on Mathematics

Mathematics
*************Which subject:MathematicsIf there is any program to use :*I need to solve 5 equations and solve 1 question via this link and you should take a screenshot of the courses that I participated inhttps://www.edx.org/course/data-models-and-decisions-in-business-analyticsCalculations or thury:both**********************Turniti report needed Similarity less than5% .All the requirements in attach file .

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ME 260                            Intro. to Engineering Materials                         SPRING 04

TEST 2 SOLUTIONS

(Time allocated = 50 min)

Student name: _________________________

This is a closed book, closed notes examination.  You may use a calculator and a scale or straight edge. Show your work for each problem clearly and concisely.  Label any figures you use.  The instructor reserves the right to deduct points for poor presentation of your answers.  Indicate answers and units by underlining them or putting them in boxes.  The point value for each problem is indicated in square brackets, e.g., [5 pts].  You have 50 minutes to work on the exam.

1. Multiple Choice [10pts]

1. Plain Carbon steels are
2. A mixture of iron and carbon
3. A mixture of iron and carbon and small amounts of alloying elements
4. A mixture of iron and carbon and a large amount of alloying elements
5. none of the above.

1. Materials with FCC unit cells
2. are very brittle
3. do not have a ductile-to-brittle transition temperature
4. can have a ductile-to-brittle transition temperature at low temperatures
5. have an atomic packing factor of 0.68

1. Hypo-eutectoid plain carbon steel alloys
2. Have compositions below 0.76wt.%C
3. Have compositions greater 0.76wt.%C
4. Are phases that form at temperatures above the eutectoid temperature
5. all of the above

1. Creep is important to consider
2. a) above a homologues temperature of 0.3.
3. b) above a homologues temperature of 0.4
4. c) above a homologues temperature of 0.5
5. d) all of the above.

5)   Pearlite

1. a) is a phase that consists of lamellar arrangement of ferrite and cementite
2. b) is harder than bainite
3. c) is softer than ferrite
4. d) all of the above.

1. Fracture toughness values of ceramics are higher than those of metals

TRUE          FALSE

lower

1. Grey cast iron is a mixture of ferrite and course cementite . TRUE FALSE

ferrite and graphite flakes

1. Martensite is a phase that is produced through a diffusion-controlled transformation of

Austenite.       TRUE          FALSE

diffusionless

1. The fracture strengths of ceramics are predictable. TRUE FALSE

random flaws

1. According to the TTT diagram if you heat martensite of eutectoid composition to a temperature of 650oC, it should transform to pearlite (Eutectoid temperature=723oC) TRUE FALSE

tempered martensite

1. One of the pre-requisites for precipitation hardening is a solid solubility limit that does not change with temperature. TRUE  FALSE

solid solubility limit increasing with temperature

1. By increasing the carbon content of plain carbon steel from 0.76 wt.%C (eutectoid composition) to 0.9 wt.%C we will increase its ductility. TRUE FALSE

decrease in ductility or increase in strength

1. Grey cast iron is less ductile in compression than in tension. TRUE FALSE

less ductile in tension

1. Binary isomorphous alloys are alloys in which both metals making up the alloy are

partially soluble in one another.      TRUE          FALSE

complete solubility

1. Coring occurs due to very slow cooling during solidification of an alloy from the melt

TRUE          FALSE

Fast cooling

1. Considering the phase diagram in Fig. 1. [36]

1. Is Fig. 1 a eutectic or eutectoid phase diagram, explain? [4]

eutectic

1. What phase(s) are present at the following conditions:

1. Temperature = 800 oC, composition 20 wt% Ag. [2]

Alpha+liquid

• Temperature = 700 oC, composition 71.9 wt% Ag.[2]

Alpha+beta

1. Consider an alloy of  composition 40 wt.% Ag at 778 oC:

• What are the compositions of the primary á and eutectic phases present respectively. [4]

Alpha 8wt%, eutectic  71.9wt%

• What are the mass fractions of both phases, explain the steps you used to arrive at your answer [6]

~ 0.5 & ~0.5

• What are the mass fraction of total á and total â, explain the steps you used to arrive at your answer [6]

Alpha =0.615

Beta = 0.385

• Is it possible to precipitation harden an alloy belonging to this system? If so what composition and heat treatment would you use? [6]

Yes, choose comp. below 8wt% Ag. Heat into alpha region, quench to room temperature, heat again to temp. below the solvus for specified time then cool to room temperature.

1. Sketch the microstructure obtained at room temperature for an alloy of composition 71.9 wt% previously cooled from 800 oC to room temperature. Label all phases present, and explain IN DETAIL why or how this particular microstructure was obtained. [6]

 See text
 beta
 alpha

1. a) i) Define fatigue & creep . [4pts]

see text

1. ii) Where do fatigue cracks usually originate? [2pts]

surface

iii) State two precautions or remedies you can do to reduce the risk of

fatigue.[4pts]

see text

Figure 2 shows S-N curves for three alloys.

1. b) Which of the alloy(s) in Figure 1 do not have a fatigue limit? [4 pts]

brass and Al alloy

• What is the fatigue life of the 2014-T6 aluminum alloy in Figure 1 at a stress amplitude of 200 MPa? If the mean stress was increased would the fatigue life of this sample increase or decrease? [4 pts]

106   decrease

• The fracture surface of a specimen that had failed by fatigue contains 20 beachmarks and 3000 striations. What do these numbers tell you about the loading pattern in service and the number of cycles to failure of this component? [6 pts]

No. of machine start and stop = 20

No. of cycles to failure=3000

Figure 2. Stress amplitude versus cycles to failure for a steel, an aluminum alloy, and a brass.

5-32
5.25 In order to solve this problem, we must first compute the value of D0 from the data given at 1200°C
(1473 K); this requires the combining of both Equations 5.3 and 5.8 as
J = − D C
x
= − D
0
C
x
exp −
Qd
RT

⎝ ⎜

⎠ ⎟
Solving for D
0 from the above expression gives
D
0 = −
J
C
x
exp
Qd
RT

⎝ ⎜

⎟⎠
= −
7.8 × 10−8 kg /m2 – s
−500 kg /m4

⎟exp
145,000 J /mol
(8.31 J /mol – K)(1200 + 273 K)

⎣ ⎢
⎤⎥⎦
= 2.18 x 10-5 m2/s
The value of the diffusion flux at 1273 K may be computed using these same two equations as follows:
J = − D
0
C
x

⎝ ⎜

⎠ ⎟exp⎛ ⎝ ⎜− Q RT d ⎞ ⎠ ⎟
= −(2.18 × 10-5 m2/s)(−500 kg/m4)exp − 145,000 J /mol
(8.31 J /mol – K)(1273 K)

⎣ ⎢

⎦ ⎥
= 1.21 x 10-8 kg/m2-s
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