14. An engine is brought into the shop with a
1 answer:
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Answer:
0.16 micron per day
Explanation:
Given:
The initial crack length, a₁ = 0.1 micron = 0.1 × 10⁻⁶ m
Initial tensile stress, σ₁ = 120 MPa
Final stress = 30 MPa
now from Griffith's equation, we have
where,
Gc and E are the material constants
now,
for the initial stage
........{1}
and for the final case
............{2}
on dividing 1 by 2, we get
or
a₂ = 4² × 0.1 × 10⁻⁶ m
or
a₂ = 1.6 micron
Now,
the change from 0.1 micron to 1.6 micron took place in 10 days
therefore, the rate at which the crack is growing =
or
average rate of change of crack = 0.16 micron per day
Solution :
<u>Sieve Size</u> (in) <u>Weight retain</u><u>(g)</u>
3 1.62
2 2.17
3.62
2.27
1.38
PAN 0.21
Given :
Sieve weight % wt. retain % cumulative % finer
size retained wt. retain
No. 4 59.5 10.225% 10.225% 89.775%
No. 8 86.5 14.865% 25.090% 74.91%
No. 16 138 23.7154% 48.8054% 51.2%
No. 30 127.8 21.91% 70.7154% 29.2850%
No. 50 97 16.6695% 87.3849% 12.62%
No. 100 66.8 11.4796% 98.92% 1.08%
Pan <u> 6.3 </u> 1.08% 100% 0%
581.9 gram
Effective size = percentage finer 10% ()
0.149 mm, N 100, % finer 1.08
0.297, N 50 , % finer 12.62%
x , 10%
x = 0.2634 mm
Effective size,
Now, N 16 (1.19 mm) , 51.2%
N 8 (2.38 mm) , 74.91%
x, 60%
x = 1.6317 mm
Uniformity co-efficient =
Cu = 6.17
Now, fineness modulus =
= 4.41
which lies between No. 4 and No. 5 sieve [4.76 to 4.00]
So, fineness modulus = 4.38 mm