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/datamodelsanddecisionsinbusinessanalyticsCalculations or thury:both**********************Turniti report needed Similarity less than5% .All the requirements in attach file .
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.
5) Pearlite
TRUE FALSE
lower
ferrite and graphite flakes
Austenite. TRUE FALSE
diffusionless
random flaws
tempered martensite
solid solubility limit increasing with temperature
decrease in ductility or increase in strength
less ductile in tension
partially soluble in one another. TRUE FALSE
complete solubility
TRUE FALSE
Fast cooling

eutectic

Alpha+liquid

Alpha+beta

Alpha 8wt%, eutectic 71.9wt%

~ 0.5 & ~0.5

Alpha =0.615
Beta = 0.385

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.





see text

surface
iii) State two precautions or remedies you can do to reduce the risk of
fatigue.[4pts]

see text
Figure 2 shows SN curves for three alloys.

brass and Al alloy

10^{6} decrease

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.
532
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 105 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 × 105 m2/s)(−500 kg/m4)exp − 145,000 J /mol
(8.31 J /mol – K)(1273 K)
⎡
⎣ ⎢
⎤
⎦ ⎥
= 1.21 x 108 kg/m2s
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20201115083601callister7e_sm_ch05_17 (2) 20201115083559callister7e_sm_ch05_12 (2) 20201115083559callister7e_sm_ch05_7 (2)
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