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Tutorial 5 - Solution

Fall 2023 Structure & Properties of Materials - MANE2220
Tutorial #5
1. Determine the carburizing time necessary to necessary to achieve a carbon concentration of 0.30 wt% at a
position 4 mm into an iron-carbon alloy that initially contains 0.10 wt% C. The surface concentration is to be
maintained at 0.90 wt% C, and the treatment is to be conducted at 1100℃.
Hint: Use the diffusion data for γ-Fe in the table of diffusion data in the textbook.
Solution:
We are asked to compute the carburizing (i.e. diffusion) time required for a specific nonsteady-state
diffusion situation. It is necessary to use Fick’s second law with the condition of constant surface
concentration.
𝐢π‘₯ − πΆπ‘œ
𝐢𝑠 − πΆπ‘œ
= 1 - erf (
π‘₯
2√𝐷𝑑
)
Cx = 0.30, x = 4 mm = 4 x 10-3 m;
Co = 0.10, Cs = 0.90
𝐢π‘₯ − πΆπ‘œ
=
0.30 − 0.10
𝐢𝑠 − πΆπ‘œ 0.90 − 0.10
π‘₯
erf (
2√𝐷𝑑
Let z =
= 0.25 = 1 - erf (
π‘₯
2√𝐷𝑑
)
) = 1 - 0.25 = 0.75
π‘₯
2√𝐷𝑑
By linear interpolation using data from the table of error function values:
Which leads to the following:
𝒁 − 0.80
0.85 − 0.80
=
0.75 − 0.7421
0.7707 − 0.7421
z = 0.814 =
π‘₯
2√𝐷𝑑
From the table of diffusion data in the textbook, diffusion of carbon into γ-iron requires activation energy
of 148,000 J/mol, and the temperature-independent pre-exponential is 2.3 x 10-5 m2/s.
The diffusion coefficient is given by D = Do exp (
−𝑄𝑑
𝑅𝑇
)
Therefore, at 1100℃ (1373K):
D = (2.3 x 10-5 m2/s){ exp (
−148000 𝐽/π‘šπ‘œπ‘™
(8.31 𝐽/π‘šπ‘œπ‘™⋅𝐾) (1373 𝐾)
)
D = 5.357 x 10-11 m2/s
Thus, 0.814 =
0.004 π‘š
2√(5.357 π‘₯ 10−11 π‘š2/𝑠) (𝑑)
t = 1.13 x 105 s
t = 31.3 hours
2. A specimen of aluminum has a rectangular cross section of 20 mm (width) x 40 mm (thickness), an elastic
modulus of 70 GPa, and a Poisson’s ratio of 0.35. If this specimen is pulled in tension with a force of 50,000
N, what is the change in width if deformation is totally elastic?
Solution:
3. A cylindrical metal specimen having an original diameter of 12.8 mm and gauge length of 50.80 mm is pulled
in tension until fracture occurs. The diameter at the point of fracture is 6.60 mm, and the fractured gauge
length is 72.14 mm. Calculate the ductility in terms of percent reduction in area and percent elongation.
Solution: