(a) We elect first to estimate the thickness required to resist yielding. Since σ = P/wt, we have t = P/wσ. For the weaker alloy, we have, from Table 5–1, S_{y} = 910 MPa. Thus,
σ_{all} =\frac {S_{y}}{n} =\frac {910}{1.3} = 700 MPa
Thus
t =\frac {P}{w σ_{all}} =\frac {4.0(10)^{3}}{1.4(700)} = 4.08 mm or greater
For the stronger alloy, we have, from Table 5–1,
σ_{all}=\frac {1035}{1.3} = 796 MPa
and so the thickness is
t =\frac {P}{w σ_{all}} =\frac {4.0(10)^{3}}{1.4(796)} = 3.59 mm or greater
(b) Now let us find the thickness required to prevent crack growth. Using Fig. 5–26, we have
\frac {h}{b} =\frac {2.8/2}{1.4} = 1 \frac {a}{b} =\frac {2.7}{1.4(10^{3})} = 0.001 93
Corresponding to these ratios we find fromFig. 5–26 that β \dot {=}1.1, and K_{I} = 1.1σ\sqrt {πa}.
n =\frac {K_{I c}}{K_{I}} =\frac {115\sqrt {10^{3}}}{1.1σ\sqrt {πa}}, σ =\frac {K_{I c}}{1.1n\sqrt {πa}}
From Table 5–1, K_{I c} = 115 MPa \sqrt {m} for the weaker of the two alloys. Solving for σ with n = 1 gives the fracture stress
σ =\frac {115}{1.1 \sqrt {π(2.7 × 10^{−3})}} = 1135 MPa
which is greater than the yield strength of 910 MPa, and so yield strength is the basis for the geometry decision. For the stronger alloy S_{y} = 1035 MPa, with n = 1 the fracture stress is
σ =\frac {K_{I c}}{ nK_{I}}=\frac {55}{1 (1.1) \sqrt {π(2.7 × 10^{−3})}} = 542.9 MPa
which is less than the yield strength of 1035 MPa. The thickness t is
t =\frac {P}{w σ_{all}} =\frac{4.0(10^{3})}{1.4(542.9/1.3)} = 6.84 mm or greater
This example shows that the fracture toughness K_{I c} limits the geometry when the stronger alloy is used, and so a thickness of 6.84 mm or larger is required. When the weaker alloy is used the geometry is limited by the yield strength, giving a thickness of only 4.08 mm or greater. Thus the weaker alloy leads to a thinner and lighter weight choice since the failure modes differ.
