Answer:
Fatigue lifetimes will be ranked as B>A>C
Explanation:
Start by calculating the mean stress for all samples
σa= (σamax + σamin)/2 = (450 - 350)/2 = 50MPa
σb= (σbmax + σbmin)/2 = (400 - 300)/2 = 50MPa
σa= (σcmax + σcmin)/2 = (340 - 340)/2 = 0MPa
Now calculate the stress amplitudes of all three samples
σa= (σamax - σamin)/2 = (450 + 350)/2 = 400MPa
σb= (σbmax - σbmin)/2 = (400 + 300)/2 = 350MPa
σc= (σcmax - σcmin)/2 = (340 + 340)/2 = 340MPa
The mean stress of samples A and B is the same which is higher than sample C. Hence samples A and B will have a higher fatigue life than sample C. However, the higher stress amplitude means lower fatigue life, hence, sample B will have a higher fatigue life than sample A. So the order will be B> A> C.
Attached picture shows the justification using and S- N plot.
Answer:
OSHA Hazard Communication Standard is responsible.
Explanation:
Answer:
Hold the tape in place and go down to the end.
Explanation:
This question is incomplete, the complete question is;
(Laminar flow) A fluid flows through two horizontal pipes of equal length which are connected together to form a pipe of length 2. The flow is laminar and fully developed. The pressure drop for the first pipe is 1.657 times greater than it is for the second pipe. If the diameter of the first pipe is D, determine the diameter of the second pipe.
D₃ = _____D.
{ the tolerance is +/-3% }
Answer:
the diameter of the second pipe D₃ is 1.13D
Explanation:
Given the data in the question;
Length = 2
pressure drop in the first pipe is 1.657 times greater than it is for the second pipe.
Now, we know that for Laminar Flow;
V' = πD⁴ΔP / 128μL
where V'₁ = V'₂ and ΔP₁₋₂ = 1.657 ΔP₂₋₃
Hence,
V'₁ = πD⁴ΔP₁₋₂ / 128μL = V'₃ = πD₃⁴ΔP₂₋₃ / 128μL
so
D₃ = D ΔP₁₋₂ / ΔP₂₋₃
we substitute
D₃ = D 1.657
D₃ = D( 1.134568 )
D₃ = 1.13D
Therefore, the diameter of the second pipe D₃ is 1.13D
Answer: 11,
Explanation: 11 •11 equals 121 which makes it a factor