A material is required for the blade of a rotary lawnmower. Cost is a consideration. For safety
Question:
A material is required for the blade of a rotary lawnmower. Cost is a consideration. For safety reasons, the designer specified a minimum fracture toughness for the blade: it is \(K_{1 c}>30 \mathrm{MPa} \mathrm{m}^{1 / 2}\). The other mechanical requirement is for high hardness, \(H\), to minimize blade wear. Hardness, in applications like this one, is related to strength:
\[H \approx 3 \sigma_{y}\]
where \(\sigma_{f}\) is the strength (Chapter 2, Engineering Materials and Their Properties gives a fuller definition). Use the \(K_{1 c}-\sigma_{f}\) chart of Fig. 3.8 to identify three materials that have \(K_{1 c}>30 \mathrm{MPa} \mathrm{m}^{1 / 2}\) and the highest possible strength. To do this, position a ' \(K_{1 c}\) ' selection line at \(30 \mathrm{MPa} \mathrm{m}{ }^{1 / 2}\) and then adjust a 'strength' selection line such that it just admits three candidates. Use the Cost chart of Fig. 3.19 to rank your selection by material cost, hence making a final selection.
Fig. 3.8
Fig. 3.19
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