If a shear stress acts in one plane of an element, there must be an equal and opposite shear stress acting on a plane that is
2 answers:
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Answer:
the ratio of the etched to the original crack tip radius is 30.24
Explanation:
Given the data in the question;
we determine the initial fracture stress using the following expression;
(σf)₁ = 2(σ₀)₁ α₁/(
)₁
----- let this be equation 1
where; (σ₀)₁ is the initial fracture strength
()₁ is the original crack tip radius
α₁ is the original crack length.
first, we determine the final crack length;
α₂ = α₁ - 16% of α₁
α₂ = α₁ - ( 0.16 × α₁)
α₂ = α₁ - 0.16α₁
α₂ = 0.84α₁
next, we calculate the final fracture stress;
the fracture strength is increased by a factor of 6;
(σ₀)₂ = 6( σ₀ )₁
Now, expression for the final fracture stress
(σf)₂ = 2(σ₀)₂ α₂/(
)₂
------- let this be equation 2
where ()₂ is the etched crack tip radius
value of fracture stress of glass is constant
Now, we substitute 2(σ₀)₁ α₁/(
)₁
from equation for (σf)₂ in equation 2.
0.84α₁ for α₂.
6( σ₀ )₁ for (σ₀)₂.
∴
2(σ₀)₁ α₁/(
)₁
= 2(6( σ₀ )₁)
0.84α₁/(
)₂
divide both sides by 2(σ₀)₁
α₁/(
)₁
= 6
0.84α₁/(
)₂
1/(
)₁
= 6
0.84/(
)₂
1/(
)₁
= 36
0.84/(
)₂
1 / ()₁ = 30.24 / (
)₂
()₂ = 30.24(
)₁
()₂/(
)₁ = 30.24
Therefore, the ratio of the etched to the original crack tip radius is 30.24
