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
Answer:
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Answer:
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Answer:
Sisal
Explanation:
Sisal is another natural fiber that comes from plants. It is a very weak fiber that has less strength than cotton.
The design speed was used for the freeway exit ramp is 11 mph.
<h3>Design speed used in the exit ramp</h3>
The design speed used in the exit ramp is calculated as follows;
f = v²/15R - 0.01e
where;
v = ωr
v = (θ/t) r
θ = 90⁰ = 1.57 rad
v = (1.57 x 19.4)/2.5 s
v = 12.18 ft/s = 8.3 mph
<h3>Design speed</h3>
f = v²/15R - 0.01e
let the maximum superelevation, e = 1%
f = (8.3)²/(15 x 19.4) - 0.01
f = 0.22
0.22 is less than value of f which is 0.4
<h3>next iteration, try 10 mph</h3>
f = (10)²/(15 x 19.4) - 0.01
f = 0.33
0.33 is less than 0.4
<h3>next iteration, try 11 mph</h3>
f = (11)²/(15 x 19.4) - 0.01
f = 0.4
Thus, the design speed was used for the freeway exit ramp is 11 mph.
Learn more about design speed here: brainly.com/question/22279858
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