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 [tex][[/tex] α₁/([tex]p_t[/tex])₁ [tex]]^{1/2[/tex] ----- let this be equation 1
where; (σ₀)₁ is the initial fracture strength
([tex]p_t[/tex])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 [tex][[/tex] α₂/([tex]p_t[/tex])₂ [tex]]^{1/2[/tex] ------- let this be equation 2
where ([tex]p_t[/tex])₂ is the etched crack tip radius
value of fracture stress of glass is constant
Now, we substitute 2(σ₀)₁ [tex][[/tex] α₁/([tex]p_t[/tex])₁ [tex]]^{1/2[/tex] from equation for (σf)₂ in equation 2.
0.84α₁ for α₂.
6( σ₀ )₁ for (σ₀)₂.
∴
2(σ₀)₁ [tex][[/tex] α₁/([tex]p_t[/tex])₁ [tex]]^{1/2[/tex] = 2(6( σ₀ )₁) [tex][[/tex] 0.84α₁/([tex]p_t[/tex])₂ [tex]]^{1/2[/tex]
divide both sides by 2(σ₀)₁
[tex][[/tex] α₁/([tex]p_t[/tex])₁ [tex]]^{1/2[/tex] = 6 [tex][[/tex] 0.84α₁/([tex]p_t[/tex])₂ [tex]]^{1/2[/tex]
[tex][[/tex] 1/([tex]p_t[/tex])₁ [tex]]^{1/2[/tex] = 6 [tex][[/tex] 0.84/([tex]p_t[/tex])₂ [tex]]^{1/2[/tex]
[tex][[/tex] 1/([tex]p_t[/tex])₁ [tex]][/tex] = 36 [tex][[/tex] 0.84/([tex]p_t[/tex])₂ [tex]][/tex]
1 / ([tex]p_t[/tex])₁ = 30.24 / ([tex]p_t[/tex])₂
([tex]p_t[/tex])₂ = 30.24([tex]p_t[/tex])₁
([tex]p_t[/tex])₂/([tex]p_t[/tex])₁ = 30.24
Therefore, the ratio of the etched to the original crack tip radius is 30.24