To design our manufacturing and testing capability, we must determine the maximum size of any flaws that might lead to failure within the ten-year period. The critical crack size using the fracture toughness and the maximum stress is

K_{Ic}= f \sigma\sqrt{\pi a_{c}}
80 MPa\sqrt{m} = (1.0)(500 MPa)\sqrt{\pi a_{c}}
a_{c} = 0.0081 m = 8.1 mm

The maximum stress is 500 MPa; however, the minimum stress is zero, not 60 MPa in compression, because cracks do not propagate in compression. Thus, \Delta \sigma is

\Delta \sigma =\sigma _{max} -\sigma _{min} = 500-0=500 MPa

We need to determine the minimum number of cycles that the plate must withstand:
N = [1 cycle/(5 min)](60 min/h)(24 h/day)(365 days/yr)(10 yr)
N = 1,051,200 cycles
If we assume that f = 1.0 for all crack lengths and note that C = 1.62 × 10^{-12} and n = 3.2 from Figure 7-20 in Equation 7-20, then

N= \frac{2[(a_{c})^{(2 \ – \ n)/2} \ – \ (a_{i})^{(2 \ – \ n)/2}]}{(2 \ – \ n)C f^n \Delta \sigma ^n \pi ^{n/2}}      (7-20)

1,051,200 = \frac{2[(0.0081)^{(2 \ – \ 3.2)/2} \ – \ (a_{i})^{(2 \ – \ 3.2)/2}]}{(2 \ – \ 3.2)(1.62 \ × \ 10^{-12}) (1)^{3.2} (500)^{3.2} \pi ^{3.2/2}}

1,051,200 = \frac{2[18 \ – \ a_{i})^{-0.6}]}{(-1.2)(1.62 \ × \ 10^{-12})(1) (4.332 \ × \ 10^{8})(6.244)}

a_{i} = 1.82 × 10^{-6} m = 0.00182 mm for surface flaws
2a_{i} = 0.00364 mm for internal flaws

The manufacturing process must produce surface flaws smaller than 0.00182 mm in length. In addition, nondestructive tests must be available to ensure that cracks approaching this length are not present.