7.D8 Dividing the first of the above equations by the second, obtain: t2 _ 2450 516.34 t2 + 53,701 2 t} - 242 ( 1496 ti+ 53,701} Plot t2 vs t' t} 2 0.0988 t2 (1496 t2' + 53,701) 2 t} (516.34 t2 + 53,701) 2 0.0988(1496 ti+ 53,701} ti t2 = 0.0988(1496t2+53,701} 2 9 2.666xl0 5 t 2 2 +(5.546xl0 7 t' }~ 2 +2.884xl0 2 Given t 2, can solve quadratic equation for t 2 : ti (in} 0.5 1.0 1.5 2.0 t 2 (in} 5.45 11. 90 19.83 30.29 These results are plotted in Figure 7.D4. 10.90 11. 90 13.22 15.15 Figure 7.D4 shows that the minimum thickness of aluminum needed to meet the Leak Before Break criterion is at least 10 times she thickness of steel required. Since the mass density ratio p st is only 2.9, the use of aluminum in the hull is clearly disadvaft!ageous from the standpoint of efficient mass utilization.
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