Estimate the theoretical fracture strength (in MPa) of a brittle material if it is known that fracture occurs by the propagation
of an elliptically shaped surface crack of length 0.25 mm that has a tip radius of curvature of 0.004 mm when a stress of 1060 MPa is applied.
1 answer:
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Answer:
(a) The mean time to fail is 9491.22 hours
The standard deviation time to fail is 9491.22 hours
(b) 0.5905
(c) 3.915 × 10⁻¹²
(d) 2.63 × 10⁻⁵
Explanation:
(a) We put time to fail = t
∴ For an exponential distribution, we have f(t) =
Where we have a failure rate = 10% for 1000 hours, we have(based on online resource);
e^(1000·λ) - 0.1·e^(1000·λ) = 1
0.9·e^(1000·λ) = 1
1000·λ = ㏑(1/0.9)
λ = 1.054 × 10⁻⁴
Hence the mean time to fail, E = 1/λ = 1/(1.054 × 10⁻⁴) = 9491.22 hours
The standard deviation = √(1/λ)² = √(1/(1.054 × 10⁻⁴)²)) = 9491.22 hours
b) Here we have to integrate from 5000 to ∞ as follows;
(c) The Poisson distribution is presented as follows;
p(x = 3) = 3.915 × 10⁻¹²
d) Where at least 2 components fail in one half hour, then 1 component is expected to fail in 15 minutes or 1/4 hours
The Cumulative Distribution Function is given as follows;
p( t ≤ 1/4) .
Answer: So a 1 pound = 0.453592kg
the density of the water is D= 997 kg/ so we need to use replace the kg with pounds with 2.20462 pounds = 1kg.
then D = 2.20462*997 pounds/
and we want to make this equal to one changing the metters for other thing.
we need to replace with 2.020462*997 meters qube.
then our new metters variable x = 12.6meters.
now g = 9.8 meters for second square.
in our new lenght variable x.
g = 12.6*9.8 x for second square.
so if y is our new time variable
= 12.6*9.8 seconds square. so y = 11.1seconds
so the units of length are x = 12.6 meters and for time y = 11.1 seconds.