If the reading of mercury manometer was 728 mmHg, what is the reading for another liquid such as water in mH20 units?
1 answer:
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Answer:
the critical flaw length is 10.06 mm
Explanation:
Given the data in the question;
plane strain fracture toughness = 92 Mpa√m
yield strength σ = 900 Mpa
design stress is one-half of the yield strength ( 900 Mpa / 2 ) 450 Mpa
Y = 1.15
we know that;
Critical crack length = 1/π(
/ Yσ )²
we substitute
= 1/π( 92 Mpa√m / (1.15 × 450 Mpa )²
= 1/π( 92 Mpa√m / (517.5 Mpa )²
= 1/π( 0.177777 )²
= 1/π( 0.03160466 )
= 0.01006 m = 10.06 mm
Therefore, the critical flaw length is 10.06 mm
{ = ( 10.06 mm ) > 3 mm
The critical flow is subject to detection
Question
Determine the minimum hole diameter for the spring to operate in
Answer:
11 mm
Explanation:
Mean spring coil diameter is given from the following relationship
D=cd where C is spring index and d is the diameter of oil-tempered wire
By substitution
D=10*1 mm=10 mm
To find the outer diameter, we add the d to D hence getting the minimum hole diameter as