A semi-infinite plate has an edge crack of length 0.4 mm. If the plate is subjected to a cyclic repeated stress loading of 180 N/mm^2, its fracture toughness is 1800 N/mm^3/2 and the rate of crack growth is 30×10^-15 (ΔK)^4 mm/cycle determine the crack length at failure and the number of cycles

Question

A semi-infinite plate has an edge crack of length 0.4 mm. If the plate is subjected to a cyclic repeated stress loading of 180 N/mm² , its fracture toughness is [latex] 1800 N / mm ^{3 / 2} [/latex] and the rate of crack growth is [latex] 30 imes 10^{-15}(\Delta K)^{4} [/latex] mm/cycle determine the crack length at failure and the number of cycles to failure.

Accepted Answer

From Eq. (8.7)

[latex] K=S(\pi a)^{1 / 2} \alpha [/latex] (8.7)

[latex] K=S(\pi a)^{1 / 2} imes 1.12 [/latex]

so that

[latex] a=\frac{1800^{2}}{\pi imes 180^{2} imes 1.12^{2}} [/latex]

which gives

[latex] a=25.4 mm [/latex]

From the above data and Eq. (8.14) it can be seen that [latex] C=30 imes 10^{-15} [/latex] and n=4. Substituting these values in Eq. (8.17)

[latex] ( d a / d N)=C(\Delta K)^{n} [/latex] (8.14)

[latex]N_{ f }=2\left[\left(1 / a_{ i }^{(n-2) / 2} ight)-\left(1 / a_{ f }^{(n-2) / 2} ight) ight] /\left\{C(n-2)\left[\left(S_{\max }-S_{\min } ight) \pi^{1 / 2} ight]^{n} ight\} [/latex] (8.17)

[latex] N_{ f }=\frac{1}{30 imes 10^{-15}\left(180 imes \pi^{1 / 2} ight)^{4}}[(1 / 0.4)-(1 / 25.4)] [/latex]

which gives

[latex] N_{ f }=7916 ext { cycles } . [/latex]

Question 8.9: A semi-infinite plate has an edge crack of length 0.4 mm. If...

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