Use the scoring system described to calculate the alignment score for These are two lines of 14 letters aligned with each other.
1 answer:
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Answer:
i) S–N plot is attached
ii) fatigue strength = 100 MPa
iii) fatigue life = 5.62 x 10^(5) cycles
Explanation:
i) I have attached the S–N plot (stress amplitude versus logarithm of cycles to failure)
ii) The question says we should find the fatigue strength at 4 × 10^(6) cycles.
So let's find the log of this and trace it on the graph attached.
Log(4 × 10^(6)) = 6.6
From the graph attached, at log of cycle value of 6.6, the fatigue strength is approximately 100 MPa
iii) The question says we should find the fatigue life for 120 MPa.
Thus, from the graph, at stress amplitude of 120 MPa, the log of cycles is approximately 5.75.
Thus,the fatigue life will be the inverse log of 5.75.
Thus, fatigue life = 10^(5.75)
Fatigue life = 5.62 x 10^(5)
Answer:
a)σ₁ = 265.2 MPa
b)σ₂ = -172.8 MPa
c)
d)Range = 438 MPa
Explanation:
Given that
Mean stress ,σm= 46.2 MPa
Stress amplitude ,σa= 219 MPa
Lets take
Maximum stress level = σ₁
Minimum stress level =σ₂
The mean stress given as
2 x 46.2 = σ₁ + σ₂
σ₁ + σ₂ = 92.4 MPa --------1
The amplitude stress given as
2 x 219 = σ₁ - σ₂
σ₁ - σ₂ = 438 MPa --------2
By adding the above equation
2 σ₁ = 530.4
σ₁ = 265.2 MPa
-σ₂ = 438 -265.2 MPa
σ₂ = -172.8 MPa
Stress ratio
Range = 265.2 MPa - ( -172.8 MPa)
Range = 438 MPa
To resolve this problem we have,
is unknown.
With these dates we can calculate the Flexural strenght of the specimen,
After that, we can calculate the flexural strenght for the square cross section using the previously value.