a metal rod 24mm diameter and 2m long is subjected to an axial pull of 40 kN. If the rod is 0.5mm, then find the stress-induced
1 answer:
You might be interested in
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